Can You Prove √2 Is Not a Fraction?

Why compare top² with 2 × bottom²? If $p/q = \sqrt{2}$, square both sides: $p^2/q^2 = 2$, so $p^2 = 2q^2$. The question about a never-ending decimal turns into a question about whole numbers: are there two whole numbers, one of whose squares is exactly twice the other’s? Whole numbers are exact, so there is no rounding to hide behind.

Start small. With bottom 1, you need a square equal to 2; but $1^2 = 1$ and $2^2 = 4$ jump right over it. With bottom 2, you need a square equal to 8, which lies between 4 and 9. Here are the best tops for the first few bottoms, and for 12:

Bottom q	Best top p	$p^2$	$2q^2$	Off by
1	1	1	2	1 too few
2	3	9	8	1 too many
3	4	16	18	2 too few
4	6	36	32	4 too many
5	7	49	50	1 too few
6	8	64	72	8 too few
12	17	289	288	1 too many

Never zero. Most bottoms miss by a few, but a handful miss by exactly 1, as close as two different whole numbers can be. Sweep the widget all the way to 1,000 and you find nine of them:

$$\frac{1}{1},\ \frac{3}{2},\ \frac{7}{5},\ \frac{17}{12},\ \frac{41}{29},\ \frac{99}{70},\ \frac{239}{169},\ \frac{577}{408},\ \frac{1393}{985}$$

Look at them as a list, not one at a time. Can you see how each comes from the one before? Take 7/5. Its bottom plus its top, 5 + 7 = 12, is the next bottom. The next top is the old top plus twice the old bottom, 7 + 2 × 5 = 17. Check it on 17/12: the next is (17 + 24)/(17 + 12) = 41/29. And the misses take turns, 1 too few, then 1 too many, then 1 too few.

The near misses get closer and closer. 99/70 is 1.4142857…, off by less than one ten-thousandth; 577/408 is 1.4142157…, off by about two millionths. Each one is a better and better fraction for √2, but each misses by that stubborn 1. Will the hunt ever land a 0?

Here is the way to be sure about infinitely many fractions without checking any of them. Suppose there were a fraction that squares to exactly 2. Follow what it would force to be true, step by step, using only facts you are certain of. If the steps lead to something impossible, like a number that is both odd and even, then the supposition was wrong: there is no such fraction.

This is called proof by contradiction. You may have used it already on the island of truth-tellers and liars (Who Is Telling the Truth?): “if Asha told the truth, she would be a liar” rules Asha out. A detective uses it too: “if the butler did it, he was in two places at nine o’clock; nobody can be, so the butler didn’t do it.”

One more tool first. A fraction can always be canceled down until top and bottom have no common factor: 18/12 becomes 3/2 when you divide both by 6. A fraction that can’t be canceled any further is in its lowest terms. If √2 were a fraction, it would be some fraction in lowest terms, so we may start from one.

Proof 1: the last digits

This proof needs nothing but multiplication. Suppose $p^2 = 2q^2$, with p/q in lowest terms. The last digit of a square depends only on the last digit of the number squared:

A number ending in	0	1	2	3	4	5	6	7	8	9
its square ends in	0	1	4	9	6	5	6	9	4	1
twice its square ends in	0	2	8	8	2	0	2	8	8	2

$p^2$ is a square, so its last digit is in the middle row: 0, 1, 4, 5, 6 or 9. $2q^2$ is twice a square, so its last digit is in the bottom row: 0, 2 or 8. They are the same number, so they end in the same digit, and the only digit in both rows is 0.

So $p^2$ ends in 0, which the table says happens only when p ends in 0. And $2q^2$ ends in 0 only when q ends in 0 or 5. Either way, both p and q are divisible by 5. But then the fraction could be canceled by 5, and we said it was in lowest terms. Contradiction. So no fraction squares to exactly 2.

Proof 2: odd and even

This is the proof the ancient Greeks knew. Aristotle, in his book on reasoning, the Prior Analytics, used it as his example of proof by contradiction: if the diagonal and side of a square could be measured by one unit, he said, odd numbers would have to equal even ones.

Suppose $p^2 = 2q^2$, in lowest terms.

• $p^2 = 2q^2$ is 2 times a whole number, so $p^2$ is even. An odd number times itself is odd, so p can’t be odd: p is even. Write it as p = 2k.
• Then $(2k)^2 = 2q^2$, that is $4k^2 = 2q^2$, so $q^2 = 2k^2$. Now $q^2$ is even, and by the same reasoning q is even.
• Both p and q are even, so the fraction could be canceled by 2. But it was in lowest terms. Contradiction.

The same wall, odd never equal to even, is why twelve fifths never land exactly on an octave in Why Does a Scale Have Twelve Notes?.

Proof 3: count the 2s

Every whole number bigger than 1 breaks into primes in only one way (the fundamental theorem of arithmetic, in Is There a Biggest Prime Number?): 360 = 2 × 2 × 2 × 3 × 3 × 5, whatever order you find them in.

Now count the 2s. Squaring a number doubles every prime in it: 12 = 2 × 2 × 3 has two 2s, and $12^2 = 144$ has four. So a square always has an even number of 2s (perhaps none). In $p^2 = 2q^2$, the left side has an even number of 2s. The right side has the even number in $q^2$, plus one more: an odd number. One number can’t have both an even and an odd number of 2s in its only breakdown into primes. Contradiction, and this time without even needing lowest terms.

This is the proof that grows. Nothing in it is special to 2:

• $\sqrt{3}$: in $p^2 = 3q^2$, count the 3s. Even on the left, odd on the right. Not a fraction.
• ∛2, the cube root of 2: in $p^3 = 2q^3$, the 2s on the left come in threes, and on the right there are a multiple of three plus one. Not a fraction.
• $\sqrt{n}$ for any whole number n that isn’t a perfect square: some prime in n appears an odd number of times, and counting that prime gives the contradiction.

The fourth proof uses no odd or even and no primes, only a picture. It is said to have been found by Stanley Tennenbaum while he was a student, in the early 1950s.

Suppose $p^2 = 2q^2$ for some whole numbers, and take the smallest such pair. In pictures: a square of side p has exactly the area of two squares of side q. Put the two small squares inside the big one, in opposite corners. They overlap in a middle square, which is covered twice, and they leave two corner squares bare. If the two small squares really have the same total area as the big one, then the area covered twice must exactly equal the area left bare.

But look at the sides. The middle square has side 2q − p, and each bare corner has side p − q. Both are whole numbers bigger than 0, since $p^2 = 2q^2$ lies between $q^2$ and $(2q)^2$, so p lies between q and 2q; and 2q − p is less than p. “Middle = two corners” says $(2q - p)^2 = 2(p - q)^2$: a smaller pair with the same property. Yet we took the smallest pair. Contradiction. (Or without “smallest”: each pair gives a smaller one, forever, and whole numbers can’t shrink forever. This is called infinite descent.)

From 99/70 the steps go 41/29, 17/12, 7/5, 3/2, 1/1: the near-miss list, backwards, each still missing by exactly 1, until it bottoms out at 1/1. An exact pair would give another exact pair at every step and never reach a bottom. In algebra, for any whole numbers,

$$p^2 - 2q^2 = -\left[\,(2q - p)^2 - 2(p - q)^2\,\right].$$

The step down keeps the size of the miss, so a miss of 1 stays 1 and a miss of 0 would stay 0 forever.

√2 is a real length

None of this says √2 doesn’t exist. It is a perfectly real length: the diagonal of your chalk square. Here is a square whose area is exactly 2, and whose side is therefore exactly √2.

The tilted square’s side is the diagonal of a unit square, and its area is 2, so the diagonal is √2. (This is the drawing in Plato’s dialogue Meno, where Socrates, only by asking questions, leads a slave boy who has never been taught geometry to find the square twice the size of a given one.)

What the proof says is that the side and the diagonal have no common measure. However tiny a unit you choose, it can’t fit a whole number of times along both the side and the diagonal, or their ratio would be a fraction. The Greeks called such lengths incommensurable; today we call √2 an irrational number: a number that is not a ratio of whole numbers.

“Its decimals never end” is not the reason

And the calculator? 1.41421356 is a fraction (141,421,356 over 100,000,000), and its square is 1.9999999932…, close but not 2. Every number a calculator stores is a fraction, fine for building and measuring. The proof is about exactly.

Mixing rational and irrational

Is $1 + \sqrt{2}$ a fraction? Suppose it were, $1 + \sqrt{2} = r$ with r a fraction. Then $\sqrt{2} = r - 1$, and a fraction minus 1 is a fraction. But √2 isn’t. Contradiction: $1 + \sqrt{2}$ is irrational. The same argument shows:

• a fraction plus an irrational number is always irrational;
• a fraction (other than 0) times an irrational number is always irrational: if $3\sqrt{2} = r$, then $\sqrt{2} = \frac{r}{3}$, a fraction.

But two irrational numbers can combine to give a fraction. $\sqrt{2} + (-\sqrt{2}) = 0$, and $\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4$. One such example, a counterexample, is enough to show that “irrational + irrational is always irrational” is false. A proof by contradiction shows something is true in every case; a counterexample shows a claim fails in at least one.

The twist: when the proof should fail

A good test of any proof is to try it where the answer is “yes”, and see where it breaks. Try Proof 2 on $\sqrt{4}$, which is the fraction $\frac{2}{1}$. Suppose $p^2 = 4q^2$ in lowest terms. Then $p^2$ is even, so p is even, p = 2k. Then $4k^2 = 4q^2$, so $k^2 = q^2$: nothing forces q to be even, and there is no contradiction. Good: the proof stops exactly where it should. Counting primes agrees: 4 = 2 × 2 has an even number of 2s.

• What’s Wrong with This Proof That 1 = 2? A chain of steps that each look right and lead somewhere impossible, and the one step that breaks the rules.
• Can You Square a Circle? Another “it can’t be done”, proved: π turns out to be even stranger than √2.
• Are Some Infinities Bigger Than Others? The fractions and the irrational numbers are both endless, but not equally endless.
• How Did Bhaskara Solve Fermat’s Challenge 500 Years Early? Equations like $p^2 - 2q^2 = 1$, whose ladders of answers Indian mathematicians knew how to climb.
• How Do You Find Every Right Triangle with Whole Sides? Why no right triangle with whole sides has two equal short sides: this page’s proof at work.
• Is There a Biggest Prime Number? (Grades 6–8) The one-way breakdown into primes that Proof 3 uses, and another famous proof that covers infinitely many cases.
• Who Is Telling the Truth? (Grades 6–8) Proof by contradiction on an island of truth-tellers and liars.
• Grade 10 Optional Math: Surd Working with √2 and its relatives in calculations, and the short proof that it is irrational.
• Computing, Going Further: How Does a Calculator Find a Square Root? How a calculator gets its fraction for √2, a few correct digits at a time. (Coming soon.)