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Algebra

Unit 1 · CDC textbook pages 1–88

Hide the stories, facts, photos, activities and interactives, for revising before an exam
In this unit 27 sections
  1. 1.1 Relation and function
  2. Exercise 1.1 (A)
  3. Exercise 1.1 (B)
  4. Exercise 1.1 (C)
  5. Exercise 1.1 (D)
  6. Exercise 1.1 (E)
  7. Exercise 1.1 (F)
  8. Exercise 1.1 (G)
  9. Exercise 1.1 (H): types of function
  10. Exercise 1.1 (H): functional values
  11. Exercise 1.1 (I)
  12. Back to the boiling pot
  13. 1.2 Polynomials
  14. Exercise 1.2 (A)
  15. Exercise 1.2 (B)
  16. Exercise 1.2 (C)
  17. 1.3 Equation and inequality
  18. Exercise 1.3 (A)
  19. Exercise 1.3 (B)
  20. 1.4 Number system
  21. Exercise 1.4 (A)
  22. Exercise 1.4 (B)
  23. 1.5 Matrix
  24. Exercise 1.5 (A)
  25. Exercise 1.5 (B)
  26. Exercise 1.5 (C)
  27. How the exam asks it

In 1849 the botanist Joseph Hooker stood on the Tunkra pass, on the border of Sikkim and Tibet, and wanted to know how high he was. He carried a barometer: a glass tube of mercury nearly three feet long, cased in metal, heavy and easy to break. He also carried something far simpler, a thermometer and a little water to boil. He boiled the water, read the thermometer, and worked out the height: 16,137 feet. The barometer said 16,083 feet. A pot of boiling water had measured a mountain pass to within 54 feet, about 16 m.

How can boiling water tell you how high you are? And why could Hooker trust it? The answer is a function, a rule that gives exactly one output for each input, and a function that can be read backwards. This unit builds that idea from pairs of numbers, then uses it for polynomials, inequalities, surds and matrices. In Back to the boiling pot you will answer Hooker’s question.

1.1 Relation and function

The word function was first used in 1673 by the German mathematician Gottfried Wilhelm Leibniz, for a quantity that depends on another. In 1734 Leonhard Euler began writing f(x) for “a function of x”, the way we still do.

Functions are everywhere once you look. The interest on a fixed sum at a fixed rate depends on the time it is invested. The distance a bus covers at a steady speed depends on the time it has been driving. The fare of a taxi depends on the distance, the price of rice on how many kilograms you buy. In each case one quantity decides another, and a function is the mathematician’s name for that. Functions are used in science, computer science, economics, engineering and medicine; every formula you meet in Science is one.

A painted portrait of a man in a long curled wig and a dark coat
Figure 1.1 Gottfried Wilhelm Leibniz (1646–1716), who first used the word “function”. He also invented calculus, at the same time as Newton, and built one of the first calculating machines that could multiply. Photo: Christoph Bernhard Francke, Public domain

To get there we need three things first: pairs of numbers in a fixed order, all the pairs two sets can make, and the pairs picked out by a rule.

1.1.1 Ordered pair

A pair whose meaning changes when you swap its elements is an ordered pair. The first element is called the antecedent and the second the consequent: in (x, y), x comes first and y second.

Ordered pair: two elements written in a fixed order, inside round brackets and separated by a comma, such as (x, y). The order is part of its meaning.

You already use ordered pairs every time you plot a point. The point (4, 5) is 4 across and 5 up; the point (5, 4) is 5 across and 4 up. They are different points.

1234512345XYOP(4, 5)Q(5, 4)
Figure 1.2 The same two numbers in the other order: (4, 5) and (5, 4) are different points.

Equal ordered pairs. Two ordered pairs are equal only when their first elements are equal and their second elements are equal:

(a,b)=(c,d)meansa=c and b=d. (a, b) = (c, d) \quad \text{means} \quad a = c \ \text{and} \ b = d.

For example, (4,62)=(123,3) \left(4, \frac{6}{2}\right) = \left(\frac{12}{3}, 3\right) , because 4=123 4 = \frac{12}{3} and 62=3 \frac{6}{2} = 3 . And (a, b) = (b, a) can only happen when a = b.

Which ordered pair is equal to (3, −2)?

Pick an answer to see why.

Worked examples

Exercise 1.1 (A)

Question 1

a.

Define an ordered pair, with an example.

Answer
b.

When are two ordered pairs equal? Give an example.

Answer
c.

If (a, b) = (m, n), how are a and m related?

Answer

Question 2

Find x and y.

a.

(x, 4) = (5, y)

Answer
b.

(x − 1, y + 2) = (6, 7)

Answer
c.

(x − 3, y + 7) = (2, −5)

Answer
d.

(2x − 5, 4) = (9, y + 4)

Answer
e.

(x3+1,y−23)=(53,23) \left(\frac{x}{3} + 1, y - \frac{2}{3}\right) = \left(\frac{5}{3}, \frac{2}{3}\right)

Answer
f.

(2x + y, 4) = (6, 3x − y)

Answer

Question 3

a.

For what values of p and q are the ordered pairs (2p + 4, p − q) and (8, q) equal?

Answer
b.

For what values of m and n are (3m − 5, 2n + m) and (7, m + 2) equal?

Answer

Question 4

The ordered pairs satisfy y = 2x + 4. Find the missing elements.

a.

(1, …)

Answer
b.

(3, …)

Answer
c.

(…, 0)

Answer
d.

(…, −2)

Answer

Question 5

a.

The ordered pairs (3, p) and (q, 4) satisfy 4x + 3y = 24. Find the ordered pair (p, q).

Answer
b.

The ordered pairs (3, m) and (n, 7) satisfy 7x − 3y = 21. Find m and n.

Answer

1.1.2 Cartesian product

Cartesian product: for two non-empty sets A and B, the set of every ordered pair whose first element is from A and second element from B:

A×B={(x,y):x∈A, y∈B} A \times B = \{(x, y) : x \in A, \ y \in B\}

It is read “A cross B”, and named after René Descartes, who joined algebra to geometry with exactly these pairs as points on a grid. Since each of the n(A) first elements pairs with each of the n(B) second ones,

n(A×B)=n(A)×n(B). n(A \times B) = n(A) \times n(B).

That is why it is called a product.

Ways of showing a Cartesian product

Take A = {1, 2} and B = {3, 4, 5}. Then A × B = {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)}, and n(A × B) = 2 × 3 = 6. The same six pairs can be shown in five ways.

1. As a set of ordered pairs, as just written.

2. In a table: A down the side, B along the top, a pair in every cell.

× 3 4 5
1 (1, 3) (1, 4) (1, 5)
2 (2, 3) (2, 4) (2, 5)

3. In a mapping (arrow) diagram: an arrow from each element of A to each element of B.

A12B345
Figure 1.4 A × B as arrows: every element of A to every element of B.

4. In a tree diagram: each element of A branches to each element of B.

A B A × B
1 → 3, 4, 5 (1, 3), (1, 4), (1, 5)
2 → 3, 4, 5 (2, 3), (2, 4), (2, 5)

5. On a graph: each pair is a point.

12312345XYO(1, 3)(1, 4)(1, 5)(2, 3)(2, 4)(2, 5)
Figure 1.5 A × B on a graph: a column of three points above each element of A.

Worked examples

n(A) = 4 and n(B) = 3. How many ordered pairs are in B × A?

Pick an answer to see why.

Exercise 1.1 (B)

Question 1

a.

What is a Cartesian product? Give an example.

Answer
b.

The cardinal numbers of sets A and B are 3 and 2. What is the cardinal number of A × B?

Answer
c.

If n(A × B) = x, n(A) = y and n(B) = z, how are x, y and z related?

Answer

Question 2

a.

If A = {2, 3} and B = {7}, find A × B and show it on a graph and in a mapping diagram.

Answer
b.

If A = {2, 3} and B = {4, 5, 6}, find B × A and show it in a table and a mapping diagram.

Answer

Question 3

a.

If A × B = {(a, 1), (a, 5), (a, 2), (b, 2), (b, 1), (b, 5)}, find A, B, n(A), n(B), B × A and n(B × A).

Answer
b.

If A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}, find A, B, n(A), n(B), B × A and n(B × A).

Answer

Question 4

a.

If A = {x:x < 4, x ∈ N} and B={x:x2−5x+6=0, x∈N} B = \{x : x^2 - 5x + 6 = 0, \ x \in N\} , find A × B and B × A, and show them in a table, a mapping diagram and a tree diagram.

Answer
b.

If M = {x:x ≤ 3, x ∈ N} and N = {y:y = x − 1, x ∈ M}, find M × N and show it on a graph.

Answer

1.1.3 Relation

A × B is every pair you could make. A rule, “is the daughter of”, “is less than”, “is double of”, “is the square of”, keeps only the pairs it is true for.

Relation: a subset of the Cartesian product of two sets. A relation R from A to B has R ⊆ A × B. If (x, y) ∈ R, y is called the image of x under R.

Ways of showing a relation

Take A = {1, 2, 3}, B = {2, 7}, and the relation “an element of A is less than an element of B”. From all the pairs
A×B={(1,2),(1,7),(2,2),(2,7),(3,2),(3,7)} A \times B = \{(1, 2), (1, 7), (2, 2), (2, 7), (3, 2), (3, 7)\}
the relation keeps those whose first element is smaller:
R={(1,2),(1,7),(2,7),(3,7)} R = \{(1, 2), (1, 7), (2, 7), (3, 7)\}
(2, 2) is out because 2 is not less than 2, and (3, 2) because 3 is more than 2.

As a set of ordered pairs: R = {(1, 2), (1, 7), (2, 7), (3, 7)}.

In set-builder form: R = {(x, y):x < y}.

In a table:

x 1 1 2 3
y 2 7 7 7

In a mapping diagram, and on a graph:

A123RB27
Figure 1.11 “Is less than” from A to B: 1 is less than both, 2 and 3 only than 7.
2246XYO(1, 2)(1, 7)(2, 7)(3, 7)
Figure 1.12 The same relation as points.

Worked examples

A = {1, 2, 3} and B = {1, 4, 9}. Which is the relation “y is the square of x” from A to B?

Pick an answer to see why.

Types of relation

When both elements of every pair come from the same set A, R ⊆ A × A, and R is called a relation in A (or on A). For example, in A = {1, 2, 3},
A×A={(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)} A \times A = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}
and the relation “y=x2 y = x^2 ” in A is R = {(1, 1)}: 22=4 2^2 = 4 and 32=9 3^2 = 9 are not in A.

A relation in a set can have three properties, and each has a name.

A. Reflexive. Every element is related to itself.

  • “Is parallel to” among lines: every line l is parallel to itself (we count a line as parallel to itself).
  • “Is congruent to” among figures: △ABC≅△ABC \triangle ABC \cong \triangle ABC ; the same for angles, line segments and polygons.
  • “Is equal to” among numbers: a = a for every number a.

Reflexive relation: R in A is reflexive if (x, x) ∈ R for every x ∈ A.

B. Symmetric. If x is related to y, then y is related to x.

  • “Is equal to”: if x = y, then y = x.
  • “Is parallel to”: if l∥m l \parallel m , then m∥l m \parallel l .
  • “Is congruent to”: if △ABC≅△DEF \triangle ABC \cong \triangle DEF , then △DEF≅△ABC \triangle DEF \cong \triangle ABC .

“Is less than” is not symmetric: 2 < 5 but 5≮2 5 \not< 2 . Nor is “is the daughter of”.

Symmetric relation: R in A is symmetric if, whenever (x, y) ∈ R, also (y, x) ∈ R. (We write xRy for (x, y) ∈ R: if xRy then yRx.)

C. Transitive. If x is related to y and y to z, then x is related to z.

  • “Is equal to”: if x = y and y = z, then x = z.
  • “Is parallel to”: if l∥m l \parallel m and m∥n m \parallel n , then l∥n l \parallel n .
  • “Is congruent to”: if △ABC≅△DEF \triangle ABC \cong \triangle DEF and △DEF≅△GHI \triangle DEF \cong \triangle GHI , then △ABC≅△GHI \triangle ABC \cong \triangle GHI . So is “is similar to” among triangles.
  • “Is taller than”: if Dolma is taller than Kiran and Kiran than Nabin, Dolma is taller than Nabin.

“Is the daughter of” is not transitive: a daughter’s father’s father is her grandfather, not her father.

Transitive relation: R in A is transitive if, whenever (x, y) ∈ R and (y, z) ∈ R, also (x, z) ∈ R.

Equivalence relation: a relation that is reflexive, symmetric and transitive. Equality, parallelism and congruence are all equivalence relations. An equivalence relation sorts a set into groups of things that count as “the same”: all lines with the same direction, all fractions equal to 12 \frac{1}{2} .

In the set of all the students of your school, which relation is not symmetric?

Pick an answer to see why.

Exercise 1.1 (C)

Question 1

a.

Define a relation, with an example.

Answer
b.

What are the ways of showing a relation?

Answer
c.

Define, with examples: i. a symmetric relation, ii. a reflexive relation, iii. a transitive relation, iv. an equivalence relation.

Answer
d.

In the set {a, b}, R = {(a, a), (b, b)}. What type of relation is R?

Answer
e.

If R = {(a, b)}, write a symmetric relation that contains R.

Answer

Question 2

A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}. Find these relations and show them in arrow diagrams.

a.

R₁ = {(x, y):x + y = 6}

Answer
b.

R₂ = {(x, y):x < y}

Answer
c.

R3={(x,y):y=x2} R_3 = \{(x, y) : y = x^2\}

Answer

Question 3

A = {1, 2, 3} and B = {2, 3, 4}. Find A × B, and the relation from A to B given by each rule (“x is greater than y” and so on).

a.

is greater than

Answer
b.

is equal to

Answer
c.

is double of

Answer
d.

is half of

Answer
e.

is the square of

Answer

Question 4

A = {6, 7, 8, 10} and B = {2, 4, 6}. Show each relation from A to B as:

  • a.a set of ordered pairs
  • b.a mapping diagram
  • c.a graph
  • d.a table
i.

R₁ = {(x, y):x + y < 12, x ∈ A, y ∈ B}

Answer
ii.

R₂ = {(x, y):2x + y > 10, x ∈ A, y ∈ B}

Answer

Question 5

5.

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is a relation in A = {1, 2, 3}. Is it an equivalence relation?

Answer

Question 6

6.

If A = {4, 5, 6}, find a. A × A, b. a reflexive relation R₁ in A, c. a symmetric relation R₂ in A, d. a transitive relation R₃ in A.

Answer

1.1.4 Domain, range and co-domain of a relation

Let R be a relation from A to B.

  • The domain of R is the set of first elements of its pairs: {x ∈ A:(x, y) ∈ R}.
  • The range of R is the set of second elements of its pairs: {y ∈ B:(x, y) ∈ R}.
  • The co-domain of R is the whole set B, the set the images are allowed to come from. The range is always a subset of the co-domain.

For example, from A = {1, 2, 3} to B = {2, 3, 4}, the relation y = 2x is R = {(1, 2), (2, 4)}. Its domain is {1, 2} (3 is left out: 2 × 3 = 6 isn’t in B), its range is {2, 4}, and its co-domain is B = {2, 3, 4}.

A123RB234
Figure 1.21 Domain {1, 2}, range {2, 4}, co-domain {2, 3, 4}.

Think of a bus route. The co-domain is every stop the bus could stop at; the range is the stops where somebody actually got off.

Worked examples

For R = {(2, 5), (3, 5), (4, 7)} from A = {2, 3, 4} to B = {5, 6, 7}, what is the range?

Pick an answer to see why.

Exercise 1.1 (D)

Question 1

a.

Define the domain and the range of a relation, with examples.

Answer
b.

The relation R is given in a table. Write its domain and range.

x 1 3 5
y 5 7 9
Answer
c.

If R = {(a, b), (c, d)}, find its domain and range.

Answer

Question 2

Find the domain and range of each relation.

a.

R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5)}

Answer
b.

R = {(2, 4), (2, 6), (3, 6), (3, 9), (4, 8), (4, 12)}

Answer
c.

R = {(3, −1), (4, −2), (5, −3), (6, −4)}

Answer
d.

R = {(4, −2), (4, 2), (1, −1), (1, 1)}

Answer
e.

R = {(5, 8), (6, 9), (7, 10), (8, 11)}

Answer

Question 3

Write the range of each relation.

a.

y = 2x, x ∈ {0, 1, 5}

Answer
b.

y = 2x + 3, x ∈ {−1, 1, 2}

Answer
c.

y=3x2−4 y = 3x^2 - 4 , −2 < x ≤ 3, x ∈ W

Answer

Question 4

4.

The figure shows a relation R from A to B.

A1234RB567
Figure 1.23

a. Write R as a set of ordered pairs. b. Write its domain and range.

Answer

Question 5

Find the domain and range of each relation, and show it as a set of ordered pairs and in a mapping diagram.

a.

R = {(x, y):y = 2x + 1, 0 ≤ x ≤ 3, x ∈ W}

Answer
b.

R={(x,y):y=x2−2, 0≤x≤3, x∈W} R = \{(x, y) : y = x^2 - 2, \ 0 \le x \le 3, \ x \in W\}

Answer
c.

R={(x,y):y=3x2−2x−1, 1≤x≤4, x∈W} R = \{(x, y) : y = 3x^2 - 2x - 1, \ 1 \le x \le 4, \ x \in W\}

Answer
d.

R={(x,y):y=x2−x, 0≤x≤3, x∈W} R = \{(x, y) : y = x^2 - x, \ 0 \le x \le 3, \ x \in W\}

Answer

1.1.5 Inverse relation

Inverse relation: the relation made by swapping the first and second elements of every ordered pair of R. It is written R−1 R^{-1} (“R inverse”):
R−1={(y,x):(x,y)∈R} R^{-1} = \{(y, x) : (x, y) \in R\}
The domain of R−1 R^{-1} is the range of R, and the range of R−1 R^{-1} is the domain of R. If R goes from A to B, R−1 R^{-1} goes from B back to A.

The −1 is not a power here; it only means “backwards”. A relation and its inverse say the same thing from opposite ends: “Shila is the daughter of Resham” and “Resham is the father of Shila”.

Worked examples

R = {(2, 7), (3, 9), (4, 11)}. What is the domain of R−1 R^{-1} ?

Pick an answer to see why.

Exercise 1.1 (E)

Question 1

a.

What is an inverse relation? Give an example.

Answer
b.

If R = {(a, b), (c, d)}, write R−1 R^{-1} .

Answer

Question 2

Write the inverse of each relation.

a.

R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5)}

Answer
b.

R = {(1, 3), (1, 6), (2, 3), (2, 6), (3, 3), (3, 6)}

Answer
c.

R = {(3, −1), (4, −2), (5, −3), (6, −4)}

Answer
d.

R = {(8, 6), (7, 5), (6, 4), (5, 3), (4, 2), (3, 1)}

Answer

Question 3

Write each relation as ordered pairs, and find its inverse.

a.

R = {(x, y):y = 3x, x ∈ {1, 2, 3}}

Answer
b.

R = {(x, y):y = 2x + 3, x ∈ {1, 2, 3}}

Answer
c.

R = {(x, y):y = x − 2, x ∈ {6, 7, 8}}

Answer
d.

R={(x,y):y=x2, x∈{0,1,−1}} R = \{(x, y) : y = x^2, \ x \in \{0, 1, -1\}\}

Answer

Question 4

4.

The figure shows the relation R−1 R^{-1} from B to A.

B123R−1A369
Figure 1.30
  • a.Write R−1 R^{-1} as ordered pairs, with its domain and range.
  • b.Write the domain and range of R.
  • c.Write R−1 R^{-1} and R in set-builder form.
Answer

Question 5

5.

In the set A = {2, 3, 4, 5, 6, 7, 8, 9}, R = {(x, y):y is a multiple of x}. Write R as ordered pairs, and find a. its domain and range, b. R−1 R^{-1} , c. the domain and range of R−1 R^{-1} .

Answer

Question 6

6.

In A = {2, 3, 4, 5}, R = {(x, y):x is a factor of y}. Write R as ordered pairs and find a. its domain, b. its range, c. R−1 R^{-1} , shown in a mapping diagram.

Answer

Question 7

Find the inverse of each relation with its domain and range, and draw its mapping diagram.

a.

{(x, y):y = x − 1, 4 ≤ x ≤ 7, x ∈ N}

Answer
b.

{(x,y):y=x2−1, 0≤x≤3, x∈W} \{(x, y) : y = x^2 - 1, \ 0 \le x \le 3, \ x \in W\}

Answer
c.

{(x,y):y=3x2−2x−1, 1≤x≤4, x∈W} \{(x, y) : y = 3x^2 - 2x - 1, \ 1 \le x \le 4, \ x \in W\}

Answer

1.1.6 Function

A function works like a machine. You feed in an input x; the machine follows its rule f; out comes exactly one output y. Feed in the same input again and you get the same output again. A vending machine that sometimes gave tea and sometimes coffee for the same button would be broken, and a rule that gave two outputs for one input would not be a function.

Function: a relation from a non-empty set A to a non-empty set B in which every element of A has exactly one image in B. It is written f:A → B, and y = f(x), read “y is a function of x”, where x ∈ A and y ∈ B.

So a function has two rules, and a relation can fail either one:

  1. No element of A is left out: each has an image.
  2. No element of A has two images.

Worked examples

Ways of showing a function

Take A = {1, 2, 3}, B = {3, 4, 5}, and f(1) = 3, f(2) = 4, f(3) = 5. The same function can be shown in four ways.

a. Visually, in an arrow (mapping) diagram, or as points on a graph:

A123fB345
Figure 1.39
224XYO(1, 3)(2, 4)(3, 5)
Figure 1.40

b. With numbers, as a set of ordered pairs, f = {(1, 3), (2, 4), (3, 5)}, or in a table:

x 1 2 3
y = f(x) 3 4 5

c. In words: for every x, f(x) is 2 more than x.

d. Algebraically, by a formula: f(x) = x + 2. This is the most useful way, because it works for numbers not in the table too: f(100) = 102.

The vertical line test

On a graph, the first element of a pair is its x, how far across. A function gives each x only one y, so a function’s graph can never have two points straight above one another. That gives a test you can do with a ruler:

Vertical line test: imagine vertical lines (parallel to the Y-axis) drawn all across the graph.

  • If every vertical line meets the graph at most once, the graph is the graph of a function.
  • If any vertical line meets it twice or more, it is not.

(Here, and in the graphs below, x can be any real number in an interval, not just whole numbers. Intervals come in 1.4.2.)

Try it

The vertical line test

Pick a graph and drag the yellow line across it, or press Sweep. Which graphs are functions? What do the others have in common?

−22−22XYO
Line at
x = −1.5
Meets the graph
1 time
Most so far
0

The line misses the graph here. Drag it across, or press Sweep.

A relation has the pairs (1, 5), (2, 5) and (3, 7). Is it a function?

Pick an answer to see why.

Exercise 1.1 (F)

Question 1

1.

Define a function, with an example from daily life.

Answer

Question 2

a.

What does y = f(t) mean?

Answer
b.

D is the depth below the surface of water and P the pressure there. What does P = f(D) mean?

Answer
c.

What does y = f(x) mean?

Answer

Question 3

a.

Write the perimeter P of a square as a function of its area A.

Answer
b.

Write the area A of a circle as a function of its circumference C.

Answer
c.

Write the volume V of a cube as a function of its surface area A.

Answer

Question 4

Which of these relations are functions? Give reasons.

a.
A123R1​B345
Figure 1.43
Answer
b.
A1−12−2R2​B149
Figure 1.44
Answer
c.
A123R3​Babc
Figure 1.45
Answer
d.
A579R4​B6−1810
Figure 1.46
Answer
e.
1234561234XYO
Figure 1.47 R₅
Answer
f.
12341234XYO
Figure 1.48 R₆
Answer
g.

R₇ = {(2, 4), (3, 6), (4, 8)}

Answer
h.

R₈ = {(3, 6), (2, 4), (3, 9), (4, 16)}

Answer
i.

R₉:

x 3 4 5
y 2 3 4
Answer
j.

R10 R_{10} :

x 2 1 2
y 7 6 3
Answer

Question 5

5.

Write each function of Question 4 as a formula.

Answer

Question 6

6.

Which of these graphs are graphs of functions? Give reasons.

(a)(b)(c)(d)(e)(f)(g)(h)(i)(j)(k)(l)
Figure 1.49
Answer

Question 7

Show each function in a table, as a formula and on a graph.

a.

f₁ = {(1, 1), (2, 4), (3, 9)}

Answer
b.

f₂ = {(1, 5), (2, 7), (3, 9)}

Answer
c.

f₃ = {(3, 1), (5, 3), (7, 5)}

Answer
d.

f₄ = {(0, 3), (1, 4), (2, 5)}

Answer

1.1.7 Domain, co-domain, range, image and pre-image of a function

For a function f:A → B:

  • A is the domain: everything that goes in.
  • B is the co-domain: everything allowed to come out.
  • The range is the set of outputs that actually come out; it is always a subset of the co-domain.
  • If f(x) = y, then y is the image of x, and x is a pre-image of y.
A123fB2345
Figure 1.55 Domain {1, 2, 3}, co-domain {2, 3, 4, 5}, range {2, 3, 4}.

For example, with A = {1, 2, 3}, B = {2, 3, 4, 5} and f = {(1, 2), (2, 3), (3, 4)}: the domain is {1, 2, 3}, the co-domain {2, 3, 4, 5}, and the range {2, 3, 4}. The image of 1 is 2, written f(1) = 2, and the pre-image of 2 is 1; likewise f(2) = 3 and f(3) = 4. 5 is in the co-domain but has no pre-image.

Worked examples

f:{1, 2, 3} → {1, 2, 3, 4, 5, 6} is f(x) = 2x. What is its range?

Pick an answer to see why.

Exercise 1.1 (G)

Question 1

a.

Explain the domain, co-domain and range of a function, with an example.

Answer
b.

Explain image and pre-image, with an example.

Answer

Question 2

2.

The table shows a function f.

x 1 2 3 4
y 1 4 9 16

a. Write f as a set of ordered pairs. b. Write its domain and range. c. What is the pre-image of 9? d. Write f as a formula.

Answer

Question 3

3.

Answer from the mapping diagram.

A357gB681013
Figure 1.57

a. Is g a function? Why? b. What is the pre-image of 8? c. Write the domain, co-domain and range of g. d. Find g(5) + g(7). e. Write g as a formula.

Answer

Question 4

Find the range of each function.

a.

f = {(2, 4), (3, 6), (4, 8)}

Answer
b.

g = {(x, y):y = x − 3, 2 < x < 5, x ∈ N}

Answer
c.

h(x)=x2−2x h(x) = x^2 - 2x , domain = {0, 1, 2}

Answer

Question 5

Find the domain (the set of pre-images) of each function.

a.

f(x) = 3x + 5, range = {2, 8}

Answer
b.

g(x) = 4x − 5, range = {−1, 7}

Answer
c.

h(x)=x2 h(x) = x^2 , range = {9, 16}

Answer
d.

The images of g(x) = 2x + 5 are {7, 11, 15}. Find the set of pre-images.

Answer
e.

The images of f(x) = 4x − 3 are {1, 5, 9}. Find the set of pre-images.

Answer

Question 6

a.

For which element of the domain of f(x)=2x−35 f(x) = \frac{2x - 3}{5} is the image 7?

Answer
b.

For which element of the domain of f(x) = 3x + 5 is the image 8?

Answer
c.

Which element of the domain has the image 12 \frac{1}{2} under f(x)=2x−4 f(x) = \frac{2}{x - 4} ?

Answer
d.

Which element of the domain has the image 9 under f(x) = 2x − 5?

Answer

Question 7

7.

A function f:A → Z, where A = {x:0 ≤ x ≤ 4, x ∈ W}, is f(x) = 2x − 3.

  • a.Write the domain.
  • b.Find the range.
  • c.Draw its arrow diagram and say what type of function it is.

The textbook gives the co-domain as W, the whole numbers, but f(0) = −3 and f(1) = −1 are not whole numbers, so f would not be a function into W. So the book’s numbers don’t work; here the co-domain is Z, the integers.

Answer

Types of function

Functions are sorted by two questions about the co-domain. Is every element of the co-domain reached? And is any element reached twice?

A357fB6810
Figure 1.59 Onto: every element of B is reached.

Onto function: every element of the co-domain has a pre-image, so the range equals the co-domain.

Into function: some element of the co-domain has no pre-image, so the range is a proper subset of the co-domain.

A357fB681013
Figure 1.60 Into: 13 is never reached.

One-to-one function: different elements of the domain have different images; every element of the range has exactly one pre-image. On a graph, no horizontal line meets a one-to-one function more than once (the horizontal line test).

Many-to-one function: at least two elements of the domain share an image.

A1−12fB14
Figure 1.61 Many-to-one: 1 and −1 share the image 1.

The two questions are independent, so a function has one answer to each: one-to-one onto, one-to-one into, many-to-one onto, or many-to-one into. A one-to-one onto function pairs the two sets off perfectly, each element with exactly one partner, like dancers in a line dance.

Some special functions

Constant function: every element of the domain has the same image. Its form is f(x) = c, where c is a constant; its graph is a horizontal line y = c.

A123fB4
Figure 1.62 A constant function: everything goes to 4.
−2−112−2−11XYOy = 2
Figure 1.63 The constant function f(x) = 2.

Identity function: a function from a set to itself in which every element is its own image: f(x) = x. Its graph is the line y = x through the origin.

A123fA123
Figure 1.64 The identity function on {1, 2, 3}.
−2−112−2−112XYO(1, 1)(2, 2)y = x
Figure 1.65 The identity function f(x) = x.

Linear function: f(x) = mx + c. Its graph is always a straight line: m is its slope (how steeply it climbs) and c its y-intercept (where it crosses the Y-axis). With c = 0: m = 1 gives y = x, going up; m = −1 gives y = −x, going down; m = 0 gives the flat line y = 0, a constant function.

−2−112−2−112XYOm = 1m = −1m = 0
Figure 1.66 Linear functions through the origin, c = 0, with slopes 1, −1 and 0.

f:{1, 2, 3} → {4, 5} is f = {(1, 4), (2, 5), (3, 5)}. What type is it?

Pick an answer to see why.

Exercise 1.1 (H): types of function

Question 1

1.

Define, with examples: a. an onto function, b. an into function, c. a linear function, d. a constant function.

Answer

Question 2

State the type of each function, with reasons.

a.
AabcdfB1234
Figure 1.68
Answer
b.
AabcgB123
Figure 1.69
Answer
c.
A123hBabc
Figure 1.70
Answer
d.
A12fBabc
Figure 1.71
Answer
e.
A123jBabc
Figure 1.72
Answer
f.
AabckBd
Figure 1.73
Answer

Question 3

3.

State the type of function each graph shows.

−212−2−12XYOy = f(x)
Figure 1.74 (a)
−2−112−2−12XYOy = g(x)
Figure 1.75 (b)
−2−112−2−11XYOy = h(x)
Figure 1.76 (c)
Answer

Question 4

4.

A = {1, 3} and B = {4, 6, 8}. Make functions from A to B that are a. one-to-one, b. one-to-one into, c. many-to-one into.

Answer

Question 5

5.

P = {3, 4, 5} and Q = {4, 5, 6}. Make these functions:

  • a.a one-to-one onto function from P to Q
  • b.a constant function from P to Q
  • c.a many-to-one into function from P to Q
  • d.the identity function on P

The textbook gives d. as the identity function from P to Q, but the identity function sends each element to itself, and 3 is not in Q. So the book’s numbers don’t work; here it is the identity function on P.

Answer

Question 6

Draw each function’s mapping diagram and state its type.

a.

f = {(1, 2), (2, 4), (3, 6)}

Answer
b.

g = {(1, 1), (−1, 1), (2, 4), (3, 9)}

Answer
c.

h = {(1, 0), (2, 0), (3, 0)}

Answer

1.1.8 Problems on functions

If f(x) is a function, the output y for a given x is the functional value of f at x. To find it, put the number in place of x everywhere in the formula. For f(x) = x + 5 and x = 2: f(2) = 2 + 5 = 7. Here f(2) = 7 is the image of 2, and 2 is the pre-image of 7.

Brackets matter: f(−2) for f(x)=x2 f(x) = x^2 is (−2)2=4 (-2)^2 = 4 , not −22=−4 -2^2 = -4 . Put the number in with its brackets.

Worked examples

If f(x)=x2−3x f(x) = x^2 - 3x , what is f(−2)?

Pick an answer to see why.

Exercise 1.1 (H): functional values

Question 1

a.

If f(x) = 4x + 5, find f(2), f(3), f(5).

Answer
b.

If f(x)=2x2−1 f(x) = 2x^2 - 1 , find f(−1), f(0), f(2).

Answer
c.

If h(x)=3x2+2x−1 h(x) = 3x^2 + 2x - 1 , find h(0), h(2), h(4).

Answer
d.

If g(x)=x3−2 g(x) = x^3 - 2 , find g(1), g(−1), g(2), g(−2).

Answer
e.

If f(x)=x2+3x−2x−2 f(x) = \frac{x^2 + 3x - 2}{x - 2} , x ≠ 2, find f(4) − f(1) + f(0).

Answer

Question 2

a.

If f(2x + 3) = 2x − 1, find f(x) and f(−2).

Answer
b.

If f(3x − 4) = 6x − 5, find f(x) and f(2).

Answer
c.

If f(3x − 5) = 6x − 13, find f(5x) and f(−3).

Answer
d.

If f(3x + 2) = 12x − 5, find f(x) and f(6).

Answer

Question 3

a.

If f(y)=ay f(y) = a^y , prove that f(a − b) = f(a) ÷ f(b).

Answer
b.

If f(x)=5x f(x) = 5^x , prove that f(a + b) = f(a) × f(b).

Answer
c.

If f(m)=7m f(m) = 7^m , prove that f(xy)=(f(y))x f(xy) = (f(y))^x .

Answer

Question 4

a.

f:R → R satisfies f(5x) = f(5) + f(x). Find f(1), f(5) and f(25).

Answer
b.

If f(x + 3) = f(3) + f(x), prove that f(0) = 0 and f(−3) = −f(3).

Answer
c.

If f(x + 7) = f(10) + f(x), prove that f(3) = 0 and f(−4) = −f(10).

Answer
d.

If f(x + a) = f(x) + f(a) for a ∈ R, prove that f(0) = 0 and f(−a) = −f(a).

Answer

Question 5

a.

For which values of x are f(x)=2x2−1 f(x) = 2x^2 - 1 and g(x) = 1 − 3x equal?

Answer
b.

f(x)=3x2+4x−5 f(x) = 3x^2 + 4x - 5 and g(x)=2x2−3x−17 g(x) = 2x^2 - 3x - 17 . Find the values of x for which f(x) = g(x).

Answer
c.

h(t)=5t2−3t+4 h(t) = 5t^2 - 3t + 4 and g(t)=t2+4t+1 g(t) = t^2 + 4t + 1 . If h(t) = g(t), find t.

Answer

Question 6

a.

f(x) = ax + b, with f(2) = 2 and f(−3) = −13. i. Find a and b. ii. Write f(x). iii. Find f(13) f\left(\frac{1}{3}\right) .

Answer
b.

f(x) = mx + c, with f(3) = 9 and f(5) = 13. Find m and c.

Answer

1.1.9 Relation and function: the difference

Relation Function
Any subset of the Cartesian product of two sets. A special relation in which every element of the domain has exactly one image in the co-domain.
Written R:A → B; its domain may be only part of A. Written f:A → B; its domain is the whole of A.
Example: R = {(1, 1), (1, −1), (4, 2), (4, −2)} (1 and 4 have two images each). Example: f = {(1, 1), (2, 4), (3, 9)}.

Which picture shows how relations and functions fit together?

Pick an answer to see why.

1.1.10 The graphs of y = x, y = x² and y = x³

The graph of a function y = f(x) is the set of all the points (x, f(x)). It shows the whole function at once: where it rises and falls, where it crosses the axes, how fast it grows. Here x can be any real number, so the points join into a smooth line or curve.

All three pass the vertical line test. Only y = x and y=x3 y = x^3 pass the horizontal line test too: y=x2 y = x^2 gives 4 for both 2 and −2, so it is many-to-one.

Exercise 1.1 (I)

Question 1

Which relations are functions? Give reasons.

a.

R = {(1, 2), (3, 5), (5, 7), (7, 9)}

Answer
b.

R = {(1, 2), (2, 3), (3, 5), (3, 6)}

Answer
c.

R = {(1, 3), (1, 5), (4, 3)}

Answer

Question 2

Which relations are functions? Give reasons.

a.
A1−12RB149
Figure 1.85
Answer
b.
1234123XYO
Figure 1.86
Answer
c.
1234123XYO
Figure 1.87
Answer
d.
x 1 2 3
y 1 4 9
Answer
e.
x 1 1 2
y 1 −1 4
Answer
f.
A1−12RB14
Figure 1.88
Answer

Question 3

3.

Draw the graphs of a. f(x) = x, b. f(x)=x2 f(x) = x^2 , c. f(x)=x3 f(x) = x^3 .

Answer

Question 4

From the graphs, find the missing values.

a.
−3−2−1123−3−2−1123XYOy = x
Figure 1.89
x 3 −1
y … …
Answer
b.
−2−1121234XYOy = x2​
Figure 1.90
x … …
y 4 1
Answer

Question 5

5.

The graph of a function f(x) is given. Find f(0), f(−1), f(2).

−22−8−6−4−22468XYOy = f(x)
Figure 1.91
Answer

Back to the boiling pot

Water boils when its vapor can push back as hard as the air pressing down on it. At sea level that happens at 100 °C. Higher up there is less air above you, the pressure is lower, and water boils at a lower temperature. So the boiling point is a function of height: at each height there is exactly one boiling point.

012345678965707580859095100height above sea level (km)water boils at (°C)sea level, 100.0 °CKathmandu, 95.4 °CTunkra pass, 83.7 °CEverest, 70.3 °C
Figure 1.92 The boiling point of water at each height, worked out from the standard atmosphere’s air pressure. It falls by about 1 °C for every 300 m you climb.
Place Height Water boils at
Sea level 0 m 100.0 °C
Kathmandu 1,400 m 95.4 °C
Tunkra pass 4,902 m (16,083 ft) 83.7 °C
Top of Everest 8,849 m 70.3 °C

Hooker needed the function the other way round: he knew the boiling point and wanted the height. Reading a function backwards is its inverse relation, and the inverse is only useful if it is a function too, with one height for each boiling point. That is true here because the boiling-point function is one-to-one: as you climb, the boiling point always falls, so no two heights share a boiling point. Draw any horizontal line across the graph and it meets the curve once.

012345678965707580859095100height above sea level (km)water boils at (°C)boils at 83.7 °Cso about 4.9 km up
Figure 1.93 Read backwards: water boiling at 83.7 °C means a height of about 4.9 km.

Compare a rule that is many-to-one, like y=x2 y = x^2 . Its inverse is not a function: told that y = 4, you can’t say whether x was 2 or −2. If the boiling point first fell and then rose again as you climbed, a boiling point couldn’t tell you your height either.

In his book Hooker wrote that the boiling-point thermometer was “much underrated” for measuring heights, that he always had three of them in use and for months five, and that it beat the barometer for practical use, because carrying “a glass tube full of mercury, nearly three feet long” over the mountains was “a great drawback”. Below about 4,000 feet he found it not to be depended on, and his heights by boiling water were off by about 100 feet on average. Instruments that measure height this way are called hypsometers, from the Greek for “height measure”. The same function works in your kitchen: in Kathmandu, water boils at about 95 °C, which is why rice and lentils take longer to cook there than in the Tarai, and why a pressure cooker, which raises the pressure and so the boiling point, saves so much time on the hills.

Why could Hooker work out his height from the temperature of boiling water?

Pick an answer to see why.

1.2 Polynomials

The Greek mathematician Diophantus of Alexandria, probably in the 3rd century, was the first to study equations systematically and to use symbols for an unknown. Around 820 CE, in Baghdad, al-Khwarizmi wrote the book whose title gave us the word algebra (al-jabr, “restoring”), with methods for quadratic equations such as completing the square. Our word algorithm comes from his name.

A drawing of a bearded man in a turban and robe
Figure 1.94 Al-Khwarizmi (about 780–850), on a Soviet postage stamp of 1983. No picture of him from his own time survives, so this face is an artist’s idea. Photo: Unknown author, Public domain

Polynomials are the simplest functions there are: only adding, subtracting and multiplying by x. Yet they describe a surprising amount. A car’s stopping distance is a polynomial in its speed: the distance covered while the driver reacts grows in step with the speed u, and the braking distance grows with u2 u^2 . With a reaction time of 1 s and brakes that slow the car by 5 m/s every second, the stopping distance in meters is
d=u+u210 d = u + \frac{u^2}{10}
for a speed of u m/s. At 10 m/s (36 km/h) that is 10 + 10 = 20 m; at 20 m/s (72 km/h) it is 20 + 40 = 60 m. Twice the speed, three times the distance. The u2 u^2 term is why speed limits near schools are so low.

1.2.1 Polynomials and their roots

An expression built from terms of the form c xn c\,x^n , where c is a constant and the power n is a whole number (0, 1, 2, …), is a polynomial in x. Polynomials are named p(x), q(x), r(x), …

For example, p(x)=5x3−3x2+2x−6 p(x) = 5x^3 - 3x^2 + 2x - 6 and q(x) = 3x − 2 are polynomials. But x−3+2x2−5x x^{-3} + 2x^2 - 5x is not (the power −3 isn’t a whole number), nor is 1x+5 \frac{1}{x} + 5 , since 1x=x−1 \frac{1}{x} = x^{-1} . A coefficient can be anything, 5 \sqrt{5} or 12 \frac{1}{2} ; only the powers must be whole numbers.

Standard form. A polynomial is in standard form when its terms are in order of their powers, usually from the highest down: p(x)=2x4−5x3+3x2+5x−4 p(x) = 2x^4 - 5x^3 + 3x^2 + 5x - 4 . Missing powers can be shown with zero coefficients: q(x)=2x4+3x2−4=2x4+0x3+3x2+0x−4 q(x) = 2x^4 + 3x^2 - 4 = 2x^4 + 0x^3 + 3x^2 + 0x - 4 . (Synthetic division, below, needs them.)

Degree and leading coefficient. The degree is the highest power of x; the coefficient of that term is the leading coefficient.

Which of these is not a polynomial?

Pick an answer to see why.

Exercise 1.2 (A)

Question 1

1.

Define a polynomial, with an example.

Answer

Question 2

2.

Write the coefficients of x3 x^3 and x2 x^2 in 4x5+2x3−7x+6 4x^5 + 2x^3 - 7x + 6 .

Answer

Question 3

3.

How many terms has 4x6−7x+6 4x^6 - 7x + 6 ? What is its degree?

Answer

Question 4

Which are polynomials? Give reasons.

a.

x2+5 x^2 + \sqrt{5}

Answer
b.

y+3 \sqrt{y} + 3

Answer
c.

z+5z z + \frac{5}{z}

Answer
d.

y+y3 \sqrt{y} + y\sqrt{3}

Answer
e.

x3−7x2+2x−4 x^3 - 7x^2 + 2x - 4

Answer
f.

3x2−11x+6 3x^2 - 11x + 6

Answer
g.

4x3+x3 4x^3 + \sqrt[3]{x}

Answer

Question 5

Write the degree of each polynomial.

a.

2x4−5x2−1 2x^4 - 5x^2 - 1

Answer
b.

x10+2x−4+3x3 x^{10} + 2x - 4 + 3x^3

Answer
c.

x5+2x3−7x7+6 x^5 + 2x^3 - 7x^7 + 6

Answer
d.

−2x3+5x2+7x+x5 -2x^3 + 5x^2 + 7x + x^5

Answer

Question 6

Write each polynomial in standard form.

a.

2x3+5x2+7x+9x4 2x^3 + 5x^2 + 7x + 9x^4

Answer
b.

x4+2x+1+3x3 x^4 + 2x + 1 + 3x^3

Answer
c.

x5+2x3−7x+6 x^5 + 2x^3 - 7x + 6

Answer
d.

2x4+7x2−3 2x^4 + 7x^2 - 3

Answer

Roots of a polynomial

Root of a polynomial: a value a of the variable for which the polynomial is zero, p(a) = 0. The roots of x2−7x+12 x^2 - 7x + 12 are 3 and 4.

What a root means on a graph: at a root, y = p(x) = 0, so the graph meets the X-axis there. The roots are where the graph crosses (or touches) the X-axis.

1123456XYO(3, 0)(4, 0)y = x2 − 7x + 12
Figure 1.96 y=x2−7x+12 y = x^2 - 7x + 12 meets the X-axis at its roots, 3 and 4.

The graph of a polynomial crosses the X-axis at (−2, 0) and (5, 0). What are its roots?

Pick an answer to see why.

Exercise 1.2 (B)

Question 1

1.

What is a root of a polynomial? Explain with an example, and give its meaning on a graph.

Answer

Question 2

a.

One root of p(x) is 2. Where does its graph meet the X-axis?

Answer
b.

One root of p(x) is −1. Where does its graph meet the X-axis?

Answer
c.

The roots of p(x) are 3 and −1. Where does its graph meet the X-axis?

Answer

Question 3

Find the roots.

a.

p(x) = x + 3

Answer
b.

p(x)=x2−9 p(x) = x^2 - 9

Answer
c.

p(x)=x2−1 p(x) = x^2 - 1

Answer
d.

p(x)=2x2−7x+6 p(x) = 2x^2 - 7x + 6

Answer
e.

p(x)=x2−5x+6 p(x) = x^2 - 5x + 6

Answer

Question 4

a.

One root of p(x) = kx + 3 is 3. Find k.

Answer
b.

One root of p(x)=x2−kx+6 p(x) = x^2 - kx + 6 is 3. Find k.

Answer
c.

One root of p(x)=kx2−25 p(x) = kx^2 - 25 is 5. Find k.

Answer

Question 5

Find the roots of p(x) from each graph.

a.
−1123−212XYOy = p(x)
Figure 1.98
Answer
b.
24−1123XYOy = p(x)
Figure 1.99
Answer
c.
−212−3−2−1234XYOy = p(x)
Figure 1.100
Answer

1.2.2 Division of polynomials

Dividing 4x3−3x2+x+9 4x^3 - 3x^2 + x + 9 by x − 2 the long way works just like long division of numbers: divide the leading terms, multiply back, subtract, bring down, repeat.

4x2+5x+11x−24x3−3x2+x+94x3−8x2‾  5x2+x  5x2−10x‾  11x+9  11x−22‾  31 \begin{array}{r|l} & 4x^2 + 5x + 11 \\ \hline x - 2 & 4x^3 - 3x^2 + x + 9 \\ & \underline{4x^3 - 8x^2} \\ & \qquad\ \ 5x^2 + x \\ & \qquad\ \ \underline{5x^2 - 10x} \\ & \qquad\qquad\ \ 11x + 9 \\ & \qquad\qquad\ \ \underline{11x - 22} \\ & \qquad\qquad\qquad\ \ 31 \end{array}

The quotient is 4x2+5x+11 4x^2 + 5x + 11 and the remainder 31. Look at what actually happens to the numbers: the x’s just keep the columns in line. Synthetic division keeps only the numbers.

Synthetic division by x − a:

  1. Write the polynomial in standard form, with a 0 for every missing power, and list its coefficients.
  2. Write a at the left (for x − 2, a = 2; for x + 3, a = −3).
  3. Bring the first coefficient straight down.
  4. Multiply it by a, write the result under the next coefficient, and add. Repeat to the end.
  5. The last number is the remainder; the others are the quotient’s coefficients, one degree lower than the dividend.

For (4x3−3x2+x+9)÷(x−2) (4x^3 - 3x^2 + x + 9) \div (x - 2) : bring down 4; 4 × 2 = 8, −3 + 8 = 5; 5 × 2 = 10, 1 + 10 = 11; 11 × 2 = 22, 9 + 22 = 31.

24−31981022451131 \begin{array}{r|rrrr} 2 & 4 & -3 & 1 & 9 \\ & & 8 & 10 & 22 \\ \hline & 4 & 5 & 11 & \boxed{31} \end{array}

Quotient 4x2+5x+11 4x^2 + 5x + 11 , remainder 31: the same as the long way, in three lines. The method is also called Horner’s method, after William Horner, who published it in 1819; Chinese mathematicians such as Qin Jiushao used the same steps over 500 years earlier. Computers still use it to work out a polynomial’s value, because no general method needs fewer multiplications.

Try it

Synthetic division, one step at a time

Step through the division. Change a: when is the remainder 0?

Divide
Remainder
31
Putting x = 2 in
f(−3) = 4·(−3)3 − 3·(−3)2 + (−3) + 9 = −129f(−2) = 4·(−2)3 − 3·(−2)2 + (−2) + 9 = −37f(−1) = 4·(−1)3 − 3·(−1)2 + (−1) + 9 = 1f(0) = 4·03 − 3·02 + 0 + 9 = 9f(1) = 4·13 − 3·12 + 1 + 9 = 11f(2) = 4·23 − 3·22 + 2 + 9 = 31f(3) = 4·33 − 3·32 + 3 + 9 = 93f(−3) = (−3)3 − 3·(−3)2 − 2·(−3) + 5 = −43f(−2) = (−2)3 − 3·(−2)2 − 2·(−2) + 5 = −11f(−1) = (−1)3 − 3·(−1)2 − 2·(−1) + 5 = 3f(0) = 03 − 3·02 − 2·0 + 5 = 5f(1) = 13 − 3·12 − 2·1 + 5 = 1f(2) = 23 − 3·22 − 2·2 + 5 = −3f(3) = 33 − 3·32 − 2·3 + 5 = −1f(−3) = 2·(−3)3 − 7·(−3)2 + (−3) + 10 = −110f(−2) = 2·(−2)3 − 7·(−2)2 + (−2) + 10 = −36f(−1) = 2·(−1)3 − 7·(−1)2 + (−1) + 10 = 0f(0) = 2·03 − 7·02 + 0 + 10 = 10f(1) = 2·13 − 7·12 + 1 + 10 = 6f(2) = 2·23 − 7·22 + 2 + 10 = 0f(3) = 2·33 − 7·32 + 3 + 10 = 4

Quotient q(x) = 4x2 + 5x + 11

Bring down the first coefficient, 4.

Multiply: 4 × −3 = −12, written under −3.

Add: −3 + (−12) = −15.

Multiply: −15 × −3 = 45, written under 1.

Add: 1 + 45 = 46.

Multiply: 46 × −3 = −138, written under 9.

The last number, −129, is the remainder, and it is f(−3), the value from putting −3 in.

Bring down the first coefficient, 4.

Multiply: 4 × −2 = −8, written under −3.

Add: −3 + (−8) = −11.

Multiply: −11 × −2 = 22, written under 1.

Add: 1 + 22 = 23.

Multiply: 23 × −2 = −46, written under 9.

The last number, −37, is the remainder, and it is f(−2), the value from putting −2 in.

Bring down the first coefficient, 4.

Multiply: 4 × −1 = −4, written under −3.

Add: −3 + (−4) = −7.

Multiply: −7 × −1 = 7, written under 1.

Add: 1 + 7 = 8.

Multiply: 8 × −1 = −8, written under 9.

The last number, 1, is the remainder, and it is f(−1), the value from putting −1 in.

Bring down the first coefficient, 4.

Multiply: 4 × 0 = 0, written under −3.

Add: −3 + 0 = −3.

Multiply: −3 × 0 = 0, written under 1.

Add: 1 + 0 = 1.

Multiply: 1 × 0 = 0, written under 9.

The last number, 9, is the remainder, and it is f(0), the value from putting 0 in.

Bring down the first coefficient, 4.

Multiply: 4 × 1 = 4, written under −3.

Add: −3 + 4 = 1.

Multiply: 1 × 1 = 1, written under 1.

Add: 1 + 1 = 2.

Multiply: 2 × 1 = 2, written under 9.

The last number, 11, is the remainder, and it is f(1), the value from putting 1 in.

Bring down the first coefficient, 4.

Multiply: 4 × 2 = 8, written under −3.

Add: −3 + 8 = 5.

Multiply: 5 × 2 = 10, written under 1.

Add: 1 + 10 = 11.

Multiply: 11 × 2 = 22, written under 9.

The last number, 31, is the remainder, and it is f(2), the value from putting 2 in.

Bring down the first coefficient, 4.

Multiply: 4 × 3 = 12, written under −3.

Add: −3 + 12 = 9.

Multiply: 9 × 3 = 27, written under 1.

Add: 1 + 27 = 28.

Multiply: 28 × 3 = 84, written under 9.

The last number, 93, is the remainder, and it is f(3), the value from putting 3 in.

Bring down the first coefficient, 1.

Multiply: 1 × −3 = −3, written under −3.

Add: −3 + (−3) = −6.

Multiply: −6 × −3 = 18, written under −2.

Add: −2 + 18 = 16.

Multiply: 16 × −3 = −48, written under 5.

The last number, −43, is the remainder, and it is f(−3), the value from putting −3 in.

Bring down the first coefficient, 1.

Multiply: 1 × −2 = −2, written under −3.

Add: −3 + (−2) = −5.

Multiply: −5 × −2 = 10, written under −2.

Add: −2 + 10 = 8.

Multiply: 8 × −2 = −16, written under 5.

The last number, −11, is the remainder, and it is f(−2), the value from putting −2 in.

Bring down the first coefficient, 1.

Multiply: 1 × −1 = −1, written under −3.

Add: −3 + (−1) = −4.

Multiply: −4 × −1 = 4, written under −2.

Add: −2 + 4 = 2.

Multiply: 2 × −1 = −2, written under 5.

The last number, 3, is the remainder, and it is f(−1), the value from putting −1 in.

Bring down the first coefficient, 1.

Multiply: 1 × 0 = 0, written under −3.

Add: −3 + 0 = −3.

Multiply: −3 × 0 = 0, written under −2.

Add: −2 + 0 = −2.

Multiply: −2 × 0 = 0, written under 5.

The last number, 5, is the remainder, and it is f(0), the value from putting 0 in.

Bring down the first coefficient, 1.

Multiply: 1 × 1 = 1, written under −3.

Add: −3 + 1 = −2.

Multiply: −2 × 1 = −2, written under −2.

Add: −2 + (−2) = −4.

Multiply: −4 × 1 = −4, written under 5.

The last number, 1, is the remainder, and it is f(1), the value from putting 1 in.

Bring down the first coefficient, 1.

Multiply: 1 × 2 = 2, written under −3.

Add: −3 + 2 = −1.

Multiply: −1 × 2 = −2, written under −2.

Add: −2 + (−2) = −4.

Multiply: −4 × 2 = −8, written under 5.

The last number, −3, is the remainder, and it is f(2), the value from putting 2 in.

Bring down the first coefficient, 1.

Multiply: 1 × 3 = 3, written under −3.

Add: −3 + 3 = 0.

Multiply: 0 × 3 = 0, written under −2.

Add: −2 + 0 = −2.

Multiply: −2 × 3 = −6, written under 5.

The last number, −1, is the remainder, and it is f(3), the value from putting 3 in.

Bring down the first coefficient, 2.

Multiply: 2 × −3 = −6, written under −7.

Add: −7 + (−6) = −13.

Multiply: −13 × −3 = 39, written under 1.

Add: 1 + 39 = 40.

Multiply: 40 × −3 = −120, written under 10.

The last number, −110, is the remainder, and it is f(−3), the value from putting −3 in.

Bring down the first coefficient, 2.

Multiply: 2 × −2 = −4, written under −7.

Add: −7 + (−4) = −11.

Multiply: −11 × −2 = 22, written under 1.

Add: 1 + 22 = 23.

Multiply: 23 × −2 = −46, written under 10.

The last number, −36, is the remainder, and it is f(−2), the value from putting −2 in.

Bring down the first coefficient, 2.

Multiply: 2 × −1 = −2, written under −7.

Add: −7 + (−2) = −9.

Multiply: −9 × −1 = 9, written under 1.

Add: 1 + 9 = 10.

Multiply: 10 × −1 = −10, written under 10.

The remainder is 0, and so is f(−1): x + 1 is a factor.

Bring down the first coefficient, 2.

Multiply: 2 × 0 = 0, written under −7.

Add: −7 + 0 = −7.

Multiply: −7 × 0 = 0, written under 1.

Add: 1 + 0 = 1.

Multiply: 1 × 0 = 0, written under 10.

The last number, 10, is the remainder, and it is f(0), the value from putting 0 in.

Bring down the first coefficient, 2.

Multiply: 2 × 1 = 2, written under −7.

Add: −7 + 2 = −5.

Multiply: −5 × 1 = −5, written under 1.

Add: 1 + (−5) = −4.

Multiply: −4 × 1 = −4, written under 10.

The last number, 6, is the remainder, and it is f(1), the value from putting 1 in.

Bring down the first coefficient, 2.

Multiply: 2 × 2 = 4, written under −7.

Add: −7 + 4 = −3.

Multiply: −3 × 2 = −6, written under 1.

Add: 1 + (−6) = −5.

Multiply: −5 × 2 = −10, written under 10.

The remainder is 0, and so is f(2): x − 2 is a factor.

Bring down the first coefficient, 2.

Multiply: 2 × 3 = 6, written under −7.

Add: −7 + 6 = −1.

Multiply: −1 × 3 = −3, written under 1.

Add: 1 + (−3) = −2.

Multiply: −2 × 3 = −6, written under 10.

The last number, 4, is the remainder, and it is f(3), the value from putting 3 in.

Worked examples

In the synthetic division of 2x3+5x−4 2x^3 + 5x - 4 by x + 1, what goes in the top row and at the left?

Pick an answer to see why.

Exercise 1.2 (C)

Question 1

1.

What is the degree of the divisor in synthetic division?

Answer

Question 2

  1. In synthetic division, what is the difference between the degrees of the dividend and the quotient?

Pick an answer to see why.

Question 3

Find the quotient and remainder by synthetic division.

a.

(x3−3x2+4x−3)÷(x−1) (x^3 - 3x^2 + 4x - 3) \div (x - 1)

Answer
b.

(2x3+4x2−5x+6)÷(x−2) (2x^3 + 4x^2 - 5x + 6) \div (x - 2)

Answer
c.

(2x3−11x2−19x−10)÷(x−5) (2x^3 - 11x^2 - 19x - 10) \div (x - 5)

Answer
d.

(y3−3y2−4y+12)÷(y+2) (y^3 - 3y^2 - 4y + 12) \div (y + 2)

Answer
e.

(m3−7m2+7m+15)÷(m+1) (m^3 - 7m^2 + 7m + 15) \div (m + 1)

Answer
f.

(5x2−6x−9)÷(x−2) (5x^2 - 6x - 9) \div (x - 2)

Answer

Question 4

Find the quotient and remainder by synthetic division.

a.

(2x3+9x2+7x−6)÷(2x−1) (2x^3 + 9x^2 + 7x - 6) \div (2x - 1)

Answer
b.

(2x3+3x2−11x−6)÷(2x+1) (2x^3 + 3x^2 - 11x - 6) \div (2x + 1)

Answer
c.

(6x3−5x2−3x+2)÷(3x+2) (6x^3 - 5x^2 - 3x + 2) \div (3x + 2)

Answer
d.

(3x3−13x2+16)÷(3x−4) (3x^3 - 13x^2 + 16) \div (3x - 4)

Answer
e.

(6x3−4+13x2)÷(3x+2) (6x^3 - 4 + 13x^2) \div (3x + 2)

Answer

Question 5

a.

If 2x3−7x2+x+10=(x−1)⋅Q(x)+R 2x^3 - 7x^2 + x + 10 = (x - 1) \cdot Q(x) + R , find Q(x) and R.

Answer
b.

If x3−19x−30=(x+2)⋅Q(x)+R x^3 - 19x - 30 = (x + 2) \cdot Q(x) + R , find Q(x) and R.

Answer
c.

If x3−21x−20=(x+1)⋅Q(x)+R x^3 - 21x - 20 = (x + 1) \cdot Q(x) + R , find Q(x) and R.

Answer

1.3 Equation and inequality

Much of daily life is not “exactly” but “at most” or “at least”. The bus takes at most two hours. The bargain is at least fifty rupees off. “I can pay at most a thousand.” To vote in Nepal you must be at least 18 years old. These are inequalities.

For example: Mohan earns at most Rs 90 an hour, and Sarina earns at least Rs 80 an hour. If Mohan’s hourly earning is x and Sarina’s is y, then
x≤90andy≥80. x \le 90 \qquad \text{and} \qquad y \ge 80.

Inequality: a statement that uses <, >, ≤ or ≥. One whose expressions are of the first degree is a linear inequality: x ≤ 4, y > 10, x + 2y ≥ 5.

Sign Read Includes the number itself?
< is less than no
> is greater than no
≤ is at most (less than or equal to) yes
≥ is at least (greater than or equal to) yes

1.3.1 Linear inequality in one variable

Statement Inequality
x is less than a x < a, or a > x
x is greater than a x > a, or a < x
x is at least a (a or more) x ≥ a, or a ≤ x
x is at most a (a or less) x ≤ a, or a ≥ x

Rules for solving an inequality. They are the rules for equations, with one exception:

  • Adding or subtracting the same number on both sides keeps the sign.
  • Multiplying or dividing both sides by the same positive number keeps the sign.
  • Multiplying or dividing both sides by the same negative number turns the sign round: < becomes >, ≤ becomes ≥.

Why the exception? 2 < 5 is true. Multiply both by −1: −2 and −5. But −2 > −5: on the number line, multiplying by −1 flips everything to the other side of 0, so the order flips too.

Inequalities on a number line and a graph

When you draw an inequality’s solution:

  • with ≤ or ≥, the boundary belongs to the solution: a filled dot, a solid line;
  • with < or >, it does not: an open dot, a dashed line.

Solve 4 − 2x > 10.

Pick an answer to see why.

Exercise 1.3 (A)

Question 1

1.

What is a linear inequality? Give an example.

Answer

Question 2

2.

Show each inequality on a graph and shade its solution set.

  • a.x ≥ 0 b. x ≤ 0 c. x ≥ 3 d. x > 1 e. x ≥ −2 f. x > −3 g. x < 2
  • h.y ≥ 0 i. y ≤ 0 j. y ≥ 1 k. y > 3 l. y ≥ −4 m. y > −5 n. y < 2
Answer

Question 3

Solve, and show the solution set on a number line.

a.

2x + 5 ≤ 15

Answer
b.

2x + 5 ≥ 15

Answer
c.

x + 3 > 7

Answer
d.

5x ≤ 20

Answer
e.

7−x2≥1 \frac{7 - x}{2} \ge 1

Answer
f.

2+x3<1 \frac{2 + x}{3} < 1

Answer
g.

3x+1≥23(4x−3) 3x + 1 \ge \frac{2}{3}(4x - 3)

Answer
h.

54(x+1)≤32(x−1) \frac{5}{4}(x + 1) \le \frac{3}{2}(x - 1)

Answer
i.

13(x+2)≥15(x+6) \frac{1}{3}(x + 2) \ge \frac{1}{5}(x + 6)

Answer
j.

3x+12≤2x−13 \frac{3x + 1}{2} \le \frac{2x - 1}{3}

Answer
k.

4x+13>3x−12 \frac{4x + 1}{3} > \frac{3x - 1}{2}

Answer

Question 4

Write each as an inequality and solve it.

a.

Three times a number, added to 4, is less than 13.

Answer
b.

If 3 is subtracted from twice a number, the result is greater than or equal to 5.

Answer
c.

A motorcycle covers 120 km at a speed of at least 40 km/h and at most 60 km/h. Between what times does the journey take?

Answer
d.

Sahabuddin scored 72, 85 and 75 in his first three Optional Mathematics exams. He wants an average of at least 80 over four exams. What must he score in the fourth?

Answer
e.

Lakpa has Rs 500. He buys half a dozen pens at Rs 20 each, and then pens of another kind at Rs 40 each. At most how many of the second kind can he buy, and how much money is left?

Answer
f.

Sanam has Rs 2,000 for prize packets. One kind costs Rs 300 a packet and the other Rs 500. She needs 3 packets of the first kind. At most how many of the second kind can she buy, and how much is left?

Answer
g.

A triangle’s base is 5 cm, and its area is between 20 cm² and 30 cm². Between what values is its height?

Answer
h.

A square’s perimeter is between 40 cm and 200 cm. What can the length of its side be?

Answer
i.

A batter scored 30 and 85 runs in the first two matches of a T20 series. At least how many must she score in the third match for an average of at least 50?

Answer

Question 5

Write the inequality each graph shows.

a.
−2−123−2−112XYO
Figure 1.117
Answer
b.
−3−212−2−112XYO
Figure 1.118
Answer
c.
−2−112−1134XYO
Figure 1.119
Answer

1.3.2 Linear inequality in two variables

An inequality with two variables, such as 5x + 10y ≤ 30, is a linear inequality in two variables. Its solutions are not a stretch of a line but a whole region of the plane, on one side of the boundary line.

To graph it:

  1. Draw the boundary line (the inequality with “=”), from two points on it: solid for ≤, ≥, dashed for <, >.
  2. Test a point not on the line, usually (0, 0). If it satisfies the inequality, shade its side; if not, shade the other side. (If the line passes through the origin, test another point, such as (1, 0).)

Why does one test point decide a whole side? Moving across the line is the only way to change 5x + 10y from below 30 to above it. So every point on one side gives the same answer.

Try it

One test point decides the side

Drag the yellow point about. Each grid point it visits is marked true or false. Where do the true points lie?

−4−224−4−224XYO2x + 3y = 6
2x + 3y 6

2 × 0 + 3 × 0 = 0: is 0 ≤ 6? Yes, so (0, 0) is a solution.

Which point is in the solution region of 3x + 2y > 12?

Pick an answer to see why.

Exercise 1.3 (B)

Question 1

1.

What is a linear inequality in two variables? Give an example.

Answer

Question 2

Show each inequality on a graph, shading its solution region.

a.

y ≤ x + 5

Answer
b.

y ≥ 3x + 2

Answer
c.

y ≤ 2x − 3

Answer
d.

2y ≥ x − 4

Answer
e.

3x + 2y ≥ 6

Answer
f.

2x − y ≤ 10

Answer
g.

3x − 5y + 15 ≥ 0

Answer
h.

3x + 4y ≤ 24

Answer

Question 3

Write each as an inequality.

a.

Raman has Rs 5,000 for pants at Rs 1,000 each and T-shirts at Rs 500 each. Write the possible numbers he can buy as an inequality.

Answer
b.

Shristi has Rs 1,000 for two kinds of exercise books, at Rs 50 and Rs 80 each. Write the possible numbers she can buy as an inequality.

Answer
c.

3 kg of apples and 2 kg of pomegranates together cost less than Rs 1,200. Write the cost as an inequality.

Answer

1.4 Number system

1.4.1 Surds and their operations

Some roots come out exactly: 16=4 \sqrt{16} = 4 , 9=3 \sqrt{9} = 3 , 643=4 \sqrt[3]{64} = 4 , 4=2 \sqrt{4} = 2 . Others don’t: 2 \sqrt{2} , 3 \sqrt{3} and 43 \sqrt[3]{4} have no exact value as a fraction or a terminating decimal. Each is the root of a rational number that can’t itself be written as a rational number. Such a number is a surd.

Surd: a root an \sqrt[n]{a} of a positive rational number a that is not itself rational. The natural number n is its order: 53 \sqrt[3]{5} is a surd of order 3 (a cube root, or third-order surd). A square root a \sqrt{a} has order 2; the 2 is not written.

Every surd is irrational, but not every irrational number is a surd. π=3.14159… \pi = 3.14159\ldots and Euler’s number e = 2.71828… are irrational but are not roots of any rational number.

Changing the order. an=a1/n \sqrt[n]{a} = a^{1/n} , and multiplying the power’s top and bottom by the same number doesn’t change it: a1/n=am/mn=ammn a^{1/n} = a^{m/mn} = \sqrt[mn]{a^m} . So a surd of order 2 can be rewritten with order 4, 6, 12, …, by raising the number inside to the matching power.

Operations on surds

1. Adding and subtracting. Only like surds (the same order and the same number inside) can be added or subtracted, by adding or subtracting their coefficients, just as 2x + 5x = 7x:
23+53=73,23−53=−33 2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}, \qquad 2\sqrt{3} - 5\sqrt{3} = -3\sqrt{3}
2+3 \sqrt{2} + \sqrt{3} can’t be made into one surd. (It is not 5 \sqrt{5} : 1.414 + 1.732 = 3.146, but 5=2.236 \sqrt{5} = 2.236 .)

2. Multiplying and dividing surds of the same order. Multiply (or divide) the coefficients together and the numbers inside together:
72×25=(7×2)2×5=1410,81025=82105=42 7\sqrt{2} \times 2\sqrt{5} = (7 \times 2)\sqrt{2 \times 5} = 14\sqrt{10}, \qquad \frac{8\sqrt{10}}{2\sqrt{5}} = \frac{8}{2}\sqrt{\frac{10}{5}} = 4\sqrt{2}

3. Surds of different orders must first be given the same order. For 53÷3 \sqrt[3]{5} \div \sqrt{3} , the orders are 3 and 2, with L.C.M. 6:
53÷3=526÷336=25276 \sqrt[3]{5} \div \sqrt{3} = \sqrt[6]{5^2} \div \sqrt[6]{3^3} = \sqrt[6]{\frac{25}{27}}

Laws of radicals. For positive a, b and natural numbers m, n:

Law Example
1. an=a1n \sqrt[n]{a} = a^{\frac{1}{n}} 83=813=2 \sqrt[3]{8} = 8^{\frac{1}{3}} = 2
2. an×bn=abn \sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{ab} 2×18=36=6 \sqrt{2} \times \sqrt{18} = \sqrt{36} = 6
3. anbn=abn \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\dfrac{a}{b}} 502=25=5 \dfrac{\sqrt{50}}{\sqrt{2}} = \sqrt{25} = 5
4. anm=amn=amn \sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a} = \sqrt[n]{\sqrt[m]{a}} 643=646=2 \sqrt{\sqrt[3]{64}} = \sqrt[6]{64} = 2

All four are the laws of indices in disguise, since a root is a fractional power.

Rationalizing

Before calculators, dividing by 2=1.41421… \sqrt{2} = 1.41421\ldots by hand was miserable, but multiplying by it was easy. So mathematicians learned to move surds out of the denominator. Making the denominator rational is called rationalizing it.

35=35×55=355,332=332×22=66 \frac{3}{\sqrt{5}} = \frac{3}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{5}, \qquad \frac{\sqrt{3}}{3\sqrt{2}} = \frac{\sqrt{3}}{3\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{6}

A number that multiplies a surd to give a rational number is its rationalizing factor. For 3 \sqrt{3} it is 3 \sqrt{3} itself, since 3×3=3 \sqrt{3} \times \sqrt{3} = 3 . For 5−3 \sqrt{5} - \sqrt{3} it is 5+3 \sqrt{5} + \sqrt{3} , because
(5−3)(5+3)=(5)2−(3)2=5−3=2 (\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3}) = (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2
In general, a−b \sqrt{a} - \sqrt{b} and a+b \sqrt{a} + \sqrt{b} are rationalizing factors of each other (they are called conjugates), by (p−q)(p+q)=p2−q2 (p - q)(p + q) = p^2 - q^2 .

What is the rationalizing factor of 27−3 2\sqrt{7} - 3 ?

Pick an answer to see why.

Exercise 1.4 (A)

Question 1

1.

Define a surd, with an example.

Answer

Question 2

Write the order of each surd.

a.

6 \sqrt{6}

Answer
b.

93 \sqrt[3]{9}

Answer
c.

367 3\sqrt[7]{6}

Answer
d.

an \sqrt[n]{a}

Answer

Question 3

a.

For what value of m are x3 x^3 and xm−13 \sqrt[3]{x^{m - 1}} equal?

Answer
b.

If y4 y^4 and y2x4 \sqrt[4]{y^{2x}} are equal, find x.

Answer

Question 4

Simplify.

a.

3×5 \sqrt{3} \times \sqrt{5}

Answer
b.

2×8 \sqrt{2} \times \sqrt{8}

Answer
c.

155 \dfrac{\sqrt{15}}{\sqrt{5}}

Answer
d.

27336 \dfrac{\sqrt[3]{27}}{\sqrt{36}}

Answer
e.

403×253 \sqrt[3]{40} \times \sqrt[3]{25}

Answer
f.

64316 \dfrac{\sqrt[3]{64}}{\sqrt{16}}

Answer

Question 5

Simplify.

a.

473+273−73 4\sqrt[3]{7} + 2\sqrt[3]{7} - \sqrt[3]{7}

Answer
b.

243−183 \dfrac{24}{\sqrt{3}} - \dfrac{18}{\sqrt{3}}

Answer
c.

33+63 3\sqrt{3} + \dfrac{6}{\sqrt{3}}

Answer
d.

168074 \sqrt[4]{16807}

Answer
e.

2503 \sqrt[3]{250}

Answer
f.

32+25004+644+68 3\sqrt{2} + \sqrt[4]{2500} + \sqrt[4]{64} + 6\sqrt{8}

Answer
g.

50+18−82 \sqrt{50} + \sqrt{18} - 8\sqrt{2}

Answer
h.

42503−81283+4543 4\sqrt[3]{250} - 8\sqrt[3]{128} + 4\sqrt[3]{54}

Answer
i.

4243−111923+66483 4\sqrt[3]{24} - 11\sqrt[3]{192} + 6\sqrt[3]{648}

Answer
j.

125−45+5 \sqrt{125} - \sqrt{45} + \sqrt{5}

Answer

Question 6

Simplify.

a.

5×2×10 \sqrt{5} \times \sqrt{2} \times \sqrt{10}

Answer
b.

543×43 \sqrt[3]{54} \times \sqrt[3]{4}

Answer
c.

643×7123 6\sqrt[3]{4} \times 7\sqrt[3]{12}

Answer
d.

40×18 \sqrt{40} \times \sqrt{18}

Answer
e.

(25+32)(35−42) (2\sqrt{5} + 3\sqrt{2})(3\sqrt{5} - 4\sqrt{2})

Answer
f.

(2+43)(12−23) (\sqrt{2} + 4\sqrt{3})(\sqrt{12} - 2\sqrt{3})

Answer
g.

(3−23)2 (3 - 2\sqrt{3})^2

Answer
h.

(5+22)2 (5 + 2\sqrt{2})^2

Answer
i.

(75−27)2 (7\sqrt{5} - 2\sqrt{7})^2

Answer
j.

(a+b)(a−b) (\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})

Answer
k.

(5+3)(5−3) (\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})

Answer
l.

1283−163223 \dfrac{\sqrt[3]{128} - \sqrt[3]{16}}{2\sqrt[3]{2}}

Answer
m.

12+273 \dfrac{\sqrt{12} + \sqrt{27}}{\sqrt{3}}

Answer
n.

645−280−3202180 \dfrac{6\sqrt{45} - 2\sqrt{80} - 3\sqrt{20}}{2\sqrt{180}}

Answer

Question 7

Rationalize the denominators and simplify.

a.

127−5 \dfrac{12}{\sqrt{7} - \sqrt{5}}

Answer
b.

2+72−7 \dfrac{2 + \sqrt{7}}{2 - \sqrt{7}}

Answer
c.

73−5248−18 \dfrac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} - \sqrt{18}}

Answer
d.

3+13−1 \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}

Answer
e.

5+35−3+5−35+3 \dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}

Answer
f.

3+23−2+3−23+2 \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}

Answer
g.

x−x2−1x+x2−1 \dfrac{x - \sqrt{x^2 - 1}}{x + \sqrt{x^2 - 1}}

Answer
h.

x+ax−a+x−ax+a \dfrac{\sqrt{x} + \sqrt{a}}{\sqrt{x} - \sqrt{a}} + \dfrac{\sqrt{x} - \sqrt{a}}{\sqrt{x} + \sqrt{a}}

Answer
i.

1x+x2−1+1x−x2−1 \dfrac{1}{x + \sqrt{x^2 - 1}} + \dfrac{1}{x - \sqrt{x^2 - 1}}

Answer
j.

x+x2−1x−x2−1−x−x2−1x+x2−1 \dfrac{x + \sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}} - \dfrac{x - \sqrt{x^2 - 1}}{x + \sqrt{x^2 - 1}}

Answer

1.4.2 Real numbers

Real numbers: the union of the rational and the irrational numbers. Every real number is a point on the number line, and every point on the number line is a real number. Some real numbers on the line:

−12−10−8−6−4−202468−√125−√27−π/2√3/22 + √52π8.13133…
Figure 1.131 −2502=−125≈−11.18 -\frac{\sqrt{250}}{\sqrt{2}} = -\sqrt{125} \approx -11.18 , −27≈−5.20 -\sqrt{27} \approx -5.20 , −π2≈−1.57 -\frac{\pi}{2} \approx -1.57 , 32≈0.87 \frac{\sqrt{3}}{2} \approx 0.87 , 2+5≈4.24 2 + \sqrt{5} \approx 4.24 , 2π≈6.28 2\pi \approx 6.28 , and 8.13133…

Intervals

Interval: the set of all real numbers between two real numbers a and b (with a < b), its endpoints, without any gap. An interval holds infinitely many numbers. Its endpoints may or may not be included, so there are four kinds.

Interval Name Means Endpoints
[a, b] closed {x:a ≤ x ≤ b, x ∈ R} both in
(a, b) open {x:a < x < b, x ∈ R} both out
[a, b) left-closed, right-open {x:a ≤ x < b, x ∈ R} a in, b out
(a, b] left-open, right-closed {x:a < x ≤ b, x ∈ R} a out, b in

A square bracket hugs its endpoint: it is in. A round bracket leaves it out. On a number line, “in” is a filled dot and “out” an open dot. Half-open intervals are everywhere in real rules: “for ages 13 to under 18” is [13, 18).

Try it

Make an interval

Drag the two ends and choose each bracket. Which kind of interval has no largest number in it?

−5−4−3−2−1012345
Left end:
Right end:
Interval
[−1, 3)
Inequality
−1 ≤ x < 3
Name
half-open interval

Every real number from −1 to 3, not just whole ones: 0.33, 1, 2.999… −1 is in, 3 is not, so it has no biggest number: 2.999 is in, and so is 2.9999.

Nepal’s Citizenship Act lets a person apply for a citizenship certificate at 16. Which interval is “at least 16 but under 18 years old”?

Pick an answer to see why.

Exercise 1.4 (B)

Question 1

1.

What is a closed interval? Give an example.

Answer

Question 2

2.

What is an open interval? Give an example.

Answer

Question 3

  1. Which interval is left-closed and right-open?

Pick an answer to see why.

Question 4

Show each interval on a number line, with its meaning.

a.

[1, 3]

Answer
b.

(−2, 4)

Answer
c.

(−3, 0)

Answer
d.

(0, 5)

Answer

Question 5

Write the interval each number line shows.

a.
a.−10123
Figure 1.141
Answer
b.
b.−10123
Figure 1.142
Answer
c.
c.−10123
Figure 1.143
Answer
d.
d.−10123
Figure 1.144
Answer

1.5 Matrix

A photo on your phone is a grid of numbers. A black-and-white photo 4,000 dots wide and 3,000 high is 3,000 rows of 4,000 numbers, each saying how bright one dot is, from 0 (black) to 255 (white); a color photo keeps three such grids, for red, green and blue. When an app brightens a photo, it adds the same number to every entry; when it blends two photos, it adds the grids entry by entry. That is matrix arithmetic, the subject of this section.

A rectangle of numbers in rows and columns is a matrix (plural matrices). The name was given in 1850 by the English mathematician James Joseph Sylvester, from the Latin word for a womb, because he saw such an array as the place from which determinants (numbers worked out from its square parts) are “born”. In 1858 his friend Arthur Cayley wrote the first paper on matrices as things in their own right, with rules for adding them, multiplying them and multiplying them by a number.

1.5.1 Introduction to matrix

The prices (Rs per kg) of three kinds of rice in three shops:

Shop Basmati Jira masino Sona mansuli
A 2200 2000 2450
B 2210 2050 2500
C 2190 2010 2480

Keep only the numbers, in their places, and put them in brackets:
(220020002450221020502500219020102480) \begin{pmatrix} 2200 & 2000 & 2450 \\ 2210 & 2050 & 2500 \\ 2190 & 2010 & 2480 \end{pmatrix}
Each row is a shop, each column a kind of rice. Reading across row B tells you everything shop B charges; reading down the second column compares the price of Jira masino in all three shops.

Matrix: an arrangement of numbers in rows and columns inside square (or round) brackets. The numbers are its elements (or entries). A matrix is named by a capital letter, A, B, P, and its elements by small letters.

A matrix is a way of keeping numbers in order; it has no single numerical value of its own, just as a table in a newspaper has no single value. (Grade 10 meets the determinant, a number worked out from a square matrix; that is something else.)

Order of a matrix

X=(123126) X = \begin{pmatrix} 1 & 2 \\ 3 & 1 \\ 2 & 6 \end{pmatrix}
X has 3 rows and 2 columns. We say its order is 3 × 2, read “3 by 2”, and write X3×2 X_{3 \times 2} . It has 3 × 2 = 6 elements.

Order of a matrix: (number of rows) × (number of columns), rows always first. A matrix with m rows and n columns is of order m × n and has mn elements.

Rows go across and columns go down, like the rows of seats in a hall and the columns that hold up its roof.

Where an element sits

Every element has an address: its row and its column. In a matrix A, the element in row i and column j is called aij a_{ij} (read “a, i, j”), row first again.

row 10−153row 22450row 31−380col. 1col. 2col. 3col. 4a23= 5: row 2, column 3
Figure 1.145 In row 2 and column 3 of this 3 × 4 matrix is 5, so a23=5 a_{23} = 5 . (a13 a_{13} is 5 too: two elements can be equal and still have different addresses.)

So a 2 × 3 matrix written in general is
P=(p11p12p13p21p22p23) P = \begin{pmatrix} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \end{pmatrix}
and a matrix of order m × n is written in short as A=[aij]m×n A = [a_{ij}]_{m \times n} .

A matrix M has order 4 × 2. Which is true?

Pick an answer to see why.

Exercise 1.5 (A)

Question 1

a.

What is a matrix? Give an example, and say how a matrix is named.

Answer
b.

What is meant by the order of a matrix? Explain with an example.

Answer

c. In the matrix A=(2−110) A = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} , what is a22 a_{22} ?

Pick an answer to see why.

d. In a matrix P, where is the element p23 p_{23} ?

Pick an answer to see why.

e. How many elements does a matrix of order 3 × 2 have?

Pick an answer to see why.

Question 2

Write the order of each matrix.

a.

P=(142) P = \begin{pmatrix} 1 & 4 & 2 \end{pmatrix}

Answer
b.

X=(04−702−31−5) X = \begin{pmatrix} 0 & 4 \\ -7 & 0 \\ 2 & -3 \\ 1 & -5 \end{pmatrix}

Answer
c.

M=(1224−3−2) M = \begin{pmatrix} 1 & 2 \\ 2 & 4 \\ -3 & -2 \end{pmatrix}

Answer

Question 3

A=(103210201) A = \begin{pmatrix} 1 & 0 & 3 \\ 2 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix}

a.

Find a32 a_{32} .

Answer
b.

Find (a32+a21)2 (a_{32} + a_{21})^2 .

Answer
c.

Find a31−a22+a13 a_{31} - a_{22} + a_{13} .

Answer

Question 4

Construct each matrix from its rule.

a.

A=[aij]2×3 A = [a_{ij}]_{2 \times 3} , where aij=3i+2j a_{ij} = 3i + 2j .

Answer
b.

B=[bij]3×2 B = [b_{ij}]_{3 \times 2} , where bij=2i2−j b_{ij} = 2i^2 - j .

Answer
c.

X=[xij]3×3 X = [x_{ij}]_{3 \times 3} , where xij=∣2i−j∣ x_{ij} = |2i - j| .

Answer
d.

M=[mij]2×4 M = [m_{ij}]_{2 \times 4} , where mij=(i−j)22 m_{ij} = \dfrac{(i - j)^2}{2} .

Answer
e.

P=[pij]2×3 P = [p_{ij}]_{2 \times 3} , where pij=(i×j)2 p_{ij} = (i \times j)^2 .

Answer

Types of matrix

Matrices are named by their shape and by where their zeros are.

Row matrix: a matrix with only one row, of order 1 × n. Example: A=(123) A = \begin{pmatrix} 1 & 2 & 3 \end{pmatrix} .

Column matrix: a matrix with only one column, of order m × 1. Example: B=(−415) B = \begin{pmatrix} -4 \\ 1 \\ 5 \end{pmatrix} .

Rectangular matrix: a matrix whose number of rows and number of columns are different (m ≠ n). Example: C=(79−8536) C = \begin{pmatrix} 7 & 9 \\ -8 & 5 \\ 3 & 6 \end{pmatrix} , of order 3 × 2.

Square matrix: a matrix with as many rows as columns (m = n). Example: D=(104012312) D = \begin{pmatrix} 1 & 0 & 4 \\ 0 & 1 & 2 \\ 3 & 1 & 2 \end{pmatrix} , of order 3 × 3, or just “of order 3”.

The elements a11,a22,a33,… a_{11}, a_{22}, a_{33}, \ldots of a square matrix, from top left to bottom right, make its main diagonal (or leading diagonal). In D they are 1, 1 and 2.

Zero (null) matrix: every element is 0. It is written O: O=(0000) O = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} , or (000) \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} . A zero matrix can be of any order.

Diagonal matrix: a square matrix whose elements off the main diagonal are all 0. Examples: M=(1002) M = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} , N=(10005000−4) N = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & -4 \end{pmatrix} .

Scalar matrix: a diagonal matrix whose diagonal elements are all equal. Examples: (1001) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} , (500050005) \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} .

Identity (unit) matrix: a scalar matrix whose diagonal elements are all 1. It is written I (I₂, I₃ for its order): I2=(1001) I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} , I3=(100010001) I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} . It does for matrices what 1 does for numbers, as Grade 10 shows.

Triangular matrix: a square matrix with only zeros on one side of the main diagonal.

  • Upper triangular: every element below the diagonal is 0, so the numbers sit in the upper triangle. (2401) \begin{pmatrix} 2 & 4 \\ 0 & 1 \end{pmatrix} , (212038006) \begin{pmatrix} 2 & 1 & 2 \\ 0 & 3 & 8 \\ 0 & 0 & 6 \end{pmatrix}
  • Lower triangular: every element above the diagonal is 0. (2051) \begin{pmatrix} 2 & 0 \\ 5 & 1 \end{pmatrix} , (2005308−36) \begin{pmatrix} 2 & 0 & 0 \\ 5 & 3 & 0 \\ 8 & -3 & 6 \end{pmatrix}

Symmetric matrix: a square matrix that is its own mirror image in the main diagonal: aij=aji a_{ij} = a_{ji} for every i and j. Example: (212138286) \begin{pmatrix} 2 & 1 & 2 \\ 1 & 3 & 8 \\ 2 & 8 & 6 \end{pmatrix} : a12=a21=1 a_{12} = a_{21} = 1 , a13=a31=2 a_{13} = a_{31} = 2 , a23=a32=8 a_{23} = a_{32} = 8 .

Equal matrices

A=(2351) A = \begin{pmatrix} 2 & 3 \\ 5 & 1 \end{pmatrix} and C=(2351) C = \begin{pmatrix} 2 & 3 \\ 5 & 1 \end{pmatrix} have the same order and the same element in every place: they are equal, A = C. So are B=(79−8536) B = \begin{pmatrix} 7 & 9 \\ -8 & 5 \\ 3 & 6 \end{pmatrix} and D=(79−8536) D = \begin{pmatrix} 7 & 9 \\ -8 & 5 \\ 3 & 6 \end{pmatrix} . But (2351) \begin{pmatrix} 2 & 3 \\ 5 & 1 \end{pmatrix} and (3251) \begin{pmatrix} 3 & 2 \\ 5 & 1 \end{pmatrix} are not equal: the same numbers in other places say something different, just as (2, 3) ≠ (3, 2).

Equal matrices: two matrices are equal when they are of the same order and their corresponding elements (the elements in the same place) are equal.

So an equation between matrices is many equations at once, one for each place. If (zxba)=(2351) \begin{pmatrix} z & x \\ b & a \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 5 & 1 \end{pmatrix} , then z = 2, x = 3, b = 5 and a = 1.

Exercise 1.5 (B)

Question 1

a. If A=[aij]m×n A = [a_{ij}]_{m \times n} , when is A a square matrix?

Pick an answer to see why.

b.

Define each type of matrix with an example: (i) square matrix, (ii) diagonal matrix, (iii) scalar matrix, (iv) symmetric matrix.

Answer
c.

Compare an identity matrix and a scalar matrix.

Answer

Question 2

Name the type of each matrix.

a.

(−300−3) \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix}

Answer
b.

(100010001) \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

Answer
c.

(200330026) \begin{pmatrix} 2 & 0 & 0 \\ 3 & 3 & 0 \\ 0 & 2 & 6 \end{pmatrix}

Answer
d.

(1235) \begin{pmatrix} 1 & 2 & 3 & 5 \end{pmatrix}

Answer
e.

(5225) \begin{pmatrix} 5 & 2 \\ 2 & 5 \end{pmatrix}

Answer
f.

(200030006) \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6 \end{pmatrix}

Answer

Question 3

Find the values of x, y and z.

a.

(x−122yz) \begin{pmatrix} x - 1 & 2 \\ 2y & z \end{pmatrix} and (72105) \begin{pmatrix} 7 & 2 \\ 10 & 5 \end{pmatrix} are equal matrices.

Answer
b.

(5x−102y3z+5) \begin{pmatrix} 5x - 1 & 0 \\ 2y & 3z + 5 \end{pmatrix} is an identity matrix.

Answer
c.

(4x25xy)=(4162z−38) \begin{pmatrix} 4 & x^2 \\ 5 & xy \end{pmatrix} = \begin{pmatrix} 4 & 16 \\ 2z - 3 & 8 \end{pmatrix}

Answer
d.

(3x+100z3) \begin{pmatrix} 3x + 1 & 0 \\ 0 & \frac{z}{3} \end{pmatrix} is a scalar matrix with 7 on its main diagonal.

Answer
e.

(5y+52y4) \begin{pmatrix} 5 & y + 5 \\ 2y & 4 \end{pmatrix} is a symmetric matrix.

Answer
f.

(x+y524)=(452x−y) \begin{pmatrix} x + y & 5 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 4 & 5 \\ 2 & x - y \end{pmatrix}

Answer
g.

(2x+yx−3y22z−6)=(101225z+3) \begin{pmatrix} 2x + y & x - 3y \\ 2 & 2z - 6 \end{pmatrix} = \begin{pmatrix} 10 & 12 \\ 2 & 5z + 3 \end{pmatrix}

Answer
h.

(sin⁡xcos⁡y31)=(1213tan⁡z) \begin{pmatrix} \sin x & \cos y \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 1 \\ 3 & \tan z \end{pmatrix} , where 0∘≤x,y,z≤90∘ 0^\circ \le x, y, z \le 90^\circ

Answer

Question 4

4.

If (a−b538)=(253ab) \begin{pmatrix} a - b & 5 \\ 3 & 8 \end{pmatrix} = \begin{pmatrix} 2 & 5 \\ 3 & ab \end{pmatrix} , find the possible values of a and b.

Answer

Question 5

5.

Matrix A is of order (2x − 3) × (y + 5) and matrix B of order (x + 2) × (3y − 1). If A and B are equal, find the order of A.

Answer

1.5.2 Operations on matrices

Addition of matrices

Adding matrices is adding tables that have the same layout: each place is added to the same place.

Addition and subtraction of matrices: only matrices of the same order can be added or subtracted. Add (or subtract) the corresponding elements, the elements in the same place; the result has the same order.
(abcd)±(pqrs)=(a±pb±qc±rd±s) \begin{pmatrix} a & b \\ c & d \end{pmatrix} \pm \begin{pmatrix} p & q \\ r & s \end{pmatrix} = \begin{pmatrix} a \pm p & b \pm q \\ c \pm r & d \pm s \end{pmatrix}

A 2 × 3 matrix and a 3 × 2 matrix can’t be added: some places in one have no partner in the other. It is like adding a two-day sales table to a three-day one; the answer would mean nothing.

1.5.3 Properties of matrix addition

Adding matrices is adding numbers place by place, so the laws of adding numbers carry over.

a. Closure property. If P=(3524) P = \begin{pmatrix} 3 & 5 \\ 2 & 4 \end{pmatrix} and Q=(1325) Q = \begin{pmatrix} 1 & 3 \\ 2 & 5 \end{pmatrix} , then P+Q=(4849) P + Q = \begin{pmatrix} 4 & 8 \\ 4 & 9 \end{pmatrix} : again a 2 × 2 matrix. The sum of two matrices of the same order is a matrix of the same order.

b. Commutative property. Take A=(1123) A = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} and B=(2014) B = \begin{pmatrix} 2 & 0 \\ 1 & 4 \end{pmatrix} .
A+B=(1+21+02+13+4)=(3137),B+A=(2+10+11+24+3)=(3137) A + B = \begin{pmatrix} 1+2 & 1+0 \\ 2+1 & 3+4 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 3 & 7 \end{pmatrix}, \quad B + A = \begin{pmatrix} 2+1 & 0+1 \\ 1+2 & 4+3 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 3 & 7 \end{pmatrix}
For any matrices A and B of the same order, A + B = B + A. Each place holds a sum of two numbers, and the order of adding numbers doesn’t matter.

c. Associative property. Take A=(1032) A = \begin{pmatrix} 1 & 0 \\ 3 & 2 \end{pmatrix} , B=(5122) B = \begin{pmatrix} 5 & 1 \\ 2 & 2 \end{pmatrix} , C=(1101) C = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} .
(A+B)+C=(6154)+(1101)=(7255) (A + B) + C = \begin{pmatrix} 6 & 1 \\ 5 & 4 \end{pmatrix} + \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ 5 & 5 \end{pmatrix}
A+(B+C)=(1032)+(6223)=(7255) A + (B + C) = \begin{pmatrix} 1 & 0 \\ 3 & 2 \end{pmatrix} + \begin{pmatrix} 6 & 2 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ 5 & 5 \end{pmatrix}
For any three matrices of the same order, (A + B) + C = A + (B + C): adding A and B first and then C gives the same as adding A to the sum of B and C.

d. Existence of additive identity. With A=(1123) A = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} and O=(0000) O = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} :
A+O=(1+01+02+03+0)=(1123)=A,O+A=A A + O = \begin{pmatrix} 1+0 & 1+0 \\ 2+0 & 3+0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} = A, \quad O + A = A
For any matrix A there is a zero matrix O of the same order with A + O = O + A = A. O is the additive identity: it does for matrix addition what 0 does for adding numbers.

e. Existence of additive inverse. Take A=(0−52153) A = \begin{pmatrix} 0 & -5 & 2 \\ 1 & 5 & 3 \end{pmatrix} . Is there a matrix that, added to A, gives the zero matrix? Yes: change the sign of every element, B=(05−2−1−5−3) B = \begin{pmatrix} 0 & 5 & -2 \\ -1 & -5 & -3 \end{pmatrix} .
A+B=(0+0−5+52−21−15−53−3)=(000000)=O,B+A=O A + B = \begin{pmatrix} 0+0 & -5+5 & 2-2 \\ 1-1 & 5-5 & 3-3 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = O, \quad B + A = O
For any matrix A there is a matrix of the same order, written −A, with A + (−A) = (−A) + A = O. A and −A are additive inverses of each other; −A is A with every sign changed.

1.5.4 Scalar multiplication

If A=(1234) A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} , what is 2A? It should be A + A:
2A=A+A=(1+12+23+34+4)=(2468)=(2×12×22×32×4) 2A = A + A = \begin{pmatrix} 1+1 & 2+2 \\ 3+3 & 4+4 \end{pmatrix} = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix} = \begin{pmatrix} 2 \times 1 & 2 \times 2 \\ 2 \times 3 & 2 \times 4 \end{pmatrix}
Doubling the matrix doubled every element.

Scalar multiplication: to multiply a matrix A by a number k (a scalar), multiply every element of A by k: if A=(abcd) A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} , then kA=(kakbkckd) kA = \begin{pmatrix} ka & kb \\ kc & kd \end{pmatrix} .

A shop that raises every price by 10% multiplies its price matrix by the scalar 1.1. And −A, the additive inverse, is just (−1)A.

A and B are both of order 2 × 3. Which of these can’t be found?

Pick an answer to see why.

Exercise 1.5 (C)

Question 1

1.

What are the conditions for adding and subtracting matrices?

Answer

Question 2

a.

What are the properties of matrix addition? Write them.

Answer

b. If A and B are matrices of the same order, which property of matrix addition is A + B = B + A?

Pick an answer to see why.

c.

Write the associative property of matrix addition.

Answer

Question 3

3.

Which of these matrices can be added? Give a reason.
A=(1120) A = \begin{pmatrix} 1 & 1 \\ 2 & 0 \end{pmatrix} , P=(1120−42) P = \begin{pmatrix} 1 & 1 \\ 2 & 0 \\ -4 & 2 \end{pmatrix} , Q=(12−2) Q = \begin{pmatrix} 1 & 2 & -2 \end{pmatrix} , R=(3725−42) R = \begin{pmatrix} 3 & 7 \\ 2 & 5 \\ -4 & 2 \end{pmatrix} , X=(1570) X = \begin{pmatrix} 1 & 5 \\ 7 & 0 \end{pmatrix}

Answer

Question 4

If X=(4680) X = \begin{pmatrix} 4 & 6 \\ 8 & 0 \end{pmatrix} and Y=(1570) Y = \begin{pmatrix} 1 & 5 \\ 7 & 0 \end{pmatrix} , find:

a.

X + Y

Answer
b.

X − Y

Answer
c.

2X + 3Y

Answer
d.

12X−2Y \frac{1}{2}X - 2Y

Answer

Question 5

a.

If X=(−2286) X = \begin{pmatrix} -2 & 2 \\ 8 & 6 \end{pmatrix} and Y=(−4−570) Y = \begin{pmatrix} -4 & -5 \\ 7 & 0 \end{pmatrix} , verify the commutative property of matrix addition. Does it hold for subtraction too? Write and verify your conclusion.

Answer
b.

If P=(4286) P = \begin{pmatrix} 4 & 2 \\ 8 & 6 \end{pmatrix} , Q=(4561) Q = \begin{pmatrix} 4 & 5 \\ 6 & 1 \end{pmatrix} and R=(4830) R = \begin{pmatrix} 4 & 8 \\ 3 & 0 \end{pmatrix} , verify the associative property of matrix addition.

Answer
c.

If M=N=H=(10−1325) M = N = H = \begin{pmatrix} 1 & 0 & -1 \\ 3 & 2 & 5 \end{pmatrix} , prove: (i) M + H = H + M, (ii) (M + H) + N = M + (H + N), (iii) M + (−M) = O, where O is the zero matrix of the same order.

Answer

Question 6

Find x and y in each case.

a.

(14x03)+(4246)=(5843y) \begin{pmatrix} 1 & 4x \\ 0 & 3 \end{pmatrix} + \begin{pmatrix} 4 & 2 \\ 4 & 6 \end{pmatrix} = \begin{pmatrix} 5 & 8 \\ 4 & 3y \end{pmatrix}

Answer
b.

(13x03)+(020y)=I \begin{pmatrix} 1 & 3x \\ 0 & 3 \end{pmatrix} + \begin{pmatrix} 0 & 2 \\ 0 & y \end{pmatrix} = I , the identity matrix of the same order.

The textbook gives the second matrix as (420y) \begin{pmatrix} 4 & 2 \\ 0 & y \end{pmatrix} , but then the top left of the sum is 1 + 4 = 5, and an identity matrix has 1 there, whatever x and y are. So the book’s numbers don’t work; with 0 in that place they give the book’s answer.

Answer
c.

The additive inverse of (x+5202y−3) \begin{pmatrix} x + 5 & 2 \\ 0 & 2y - 3 \end{pmatrix} is (4−20−7) \begin{pmatrix} 4 & -2 \\ 0 & -7 \end{pmatrix} .

The textbook gives the inverse as (420−7) \begin{pmatrix} 4 & 2 \\ 0 & -7 \end{pmatrix} , but the additive inverse changes the sign of every element, so the 2 must become −2. So the book’s numbers don’t work; with −2 in that place they give the book’s answer.

Answer
d.

2(54x7y−3)+3(41−26)=(22883y) 2\begin{pmatrix} 5 & 4x \\ 7 & y - 3 \end{pmatrix} + 3\begin{pmatrix} 4 & 1 \\ -2 & 6 \end{pmatrix} = \begin{pmatrix} 22 & 8 \\ 8 & 3y \end{pmatrix}

The textbook gives the right side as (5843y) \begin{pmatrix} 5 & 8 \\ 4 & 3y \end{pmatrix} , but the left side’s first column is 2 × 5 + 3 × 4 = 22 and 2 × 7 + 3 × (−2) = 8, not 5 and 4. So the book’s numbers don’t work; with 22 and 8 there they give the book’s answer.

Answer
e.

2(3x00y−3)+3(4006) 2\begin{pmatrix} 3x & 0 \\ 0 & y - 3 \end{pmatrix} + 3\begin{pmatrix} 4 & 0 \\ 0 & 6 \end{pmatrix} is the zero matrix of the same order.

Answer

Question 7

a.

If P=(304−2) P = \begin{pmatrix} 3 & 0 \\ 4 & -2 \end{pmatrix} and 3P+X=(8696) 3P + X = \begin{pmatrix} 8 & 6 \\ 9 & 6 \end{pmatrix} , find the matrix X.

Answer
b.

If P=(103−2) P = \begin{pmatrix} 1 & 0 \\ 3 & -2 \end{pmatrix} and R=(4830) R = \begin{pmatrix} 4 & 8 \\ 3 & 0 \end{pmatrix} , find the matrix X such that 2P − 3X = 3R.

Answer

Question 8

8.

If P+Q=(503−2) P + Q = \begin{pmatrix} 5 & 0 \\ 3 & -2 \end{pmatrix} and P−Q=(8426) P - Q = \begin{pmatrix} 8 & 4 \\ 2 & 6 \end{pmatrix} , find P and Q.

Answer

Question 9

9.

If X=(−3−167) X = \begin{pmatrix} -3 & -1 \\ 6 & 7 \end{pmatrix} , find the additive inverse of X.

Answer

How the exam asks it

Grade 9’s final exam follows CDC’s Specification Grid and Model Question for Optional Mathematics, Grade 9 (2082). There, Algebra is 17 marks:

Where Questions Marks Levels
Group A, multiple choice 2 2 × 1 knowledge, application
Algebra’s own questions 3 (with parts) 10 knowledge 2, understanding 2, application 4, higher ability 2
Questions joining areas parts of 2 5 understanding 2, application 2, higher ability 1

In the grid’s model paper the multiple-choice questions ask the radical form of x2/3 x^{2/3} and the remainder of −a3+6a−7 -a^3 + 6a - 7 divided by a + 1 (put a = −1: 1 − 6 − 7 = −12). Then come a function question (degree, a wrong ordered pair, the graph of an inequality), a matrix question (order, can two be added, a difference) and a polynomial and surd question (a root read from a graph, rationalizing a denominator). The joined question links a function to statistics, or an inequality to a straight line.

Here is a question in the shape of the model paper’s function question, with how it is marked.

Where these facts come from

A companion to your textbook, not a replacement for it: study from the book your school uses, and where the two differ, follow your teacher. Found a mistake? Please tell us.