The Square Root of Two Is Irrational

The claim

$\sqrt{2}$√2 is the positive number that multiplies by itself to make 2. It is a little more than 1.4, because $1.4^2=1.96$1.4² = 1.96, and a little less than 1.5, because $1.5^2=2.25$1.5² = 2.25.

The claim is that no fraction equals $\sqrt{2}$√2. However the whole numbers a and b are chosen, ${\left(\frac{a}{b}\right)}^2$(a/b)² is never exactly 2. A number that cannot be written as a fraction of whole numbers is irrational, so the claim is that $\sqrt{2}$√2 is irrational.

Trying fractions one by one could never prove this, because there are infinitely many of them. The proof is a proof by contradiction: suppose that $\sqrt{2}$√2 is a fraction, and show that this leads to something impossible.

First, odd and even squares

The proof needs one fact: an even number squares to an even number, and an odd number squares to an odd number.

An even number is 2 times a whole number, so it can be written 2m. Its square is ${\left(2m\right)}^2=4m^2=2\times 2m^2$(2m)² = 4m² = 2 × 2m², which is 2 times a whole number, so it is even.

An odd number is one more than an even number, so it can be written 2m + 1. Its square is ${\left(2m+1\right)}^2=4m^2+4m+1=2\left(2m^2+2m\right)+1$(2m + 1)² = 4m² + 4m + 1 = 2(2m² + 2m) + 1, which is one more than an even number, so it is odd.

Turn this round. If $a^2$a² is even, then a cannot be odd, because an odd a would give an odd $a^2$a². So if $a^2$a² is even, a is even.

Along the top, n runs from 1 to 8. The odd values 1, 3, 5 and 7 have odd squares, and the even values 2, 4, 6 and 8 have even squares.

Suppose $\sqrt{2}$√2 is a fraction

Suppose $\sqrt{2}=\frac{a}{b}$√2 = a/b, where a and b are whole numbers. Any fraction can be canceled down until the top and the bottom have no common factor, so choose $\frac{a}{b}$a/b in lowest terms. In particular, a and b are not both even.

Square both sides: $2=\frac{a^2}{b^2}$2 = a²/b². Multiply both sides by $b^2$b²: $a^2=2b^2$a² = 2b².

Then a is even

$a^2=2b^2$a² = 2b² is 2 times a whole number, so $a^2$a² is even. By the fact above, a is even too.

Then b is even too

Since a is even, write a = 2k for some whole number k. Put 2k in place of a in $a^2=2b^2$a² = 2b²: ${\left(2k\right)}^2=2b^2$(2k)² = 2b², so $4k^2=2b^2$4k² = 2b². Divide both sides by 2: $b^2=2k^2$b² = 2k².

Now $b^2$b² is 2 times a whole number, so $b^2$b² is even, and so b is even. Both a and b are even. But $\frac{a}{b}$a/b was chosen in lowest terms, so a and b cannot both be even. That is a contradiction.

The supposition that $\sqrt{2}$√2 is a fraction has led to something impossible, so it is false. No fraction equals $\sqrt{2}$√2: $\sqrt{2}$√2 is irrational.

For these values of b, $2b^2$2b² misses a square by exactly 1, so the fractions $\frac{3}{2}$3/2, $\frac{7}{5}$7/5, $\frac{17}{12}$17/12 and $\frac{41}{29}$41/29 come closer and closer to $\sqrt{2}$√2. The proof shows that $2b^2$2b² never equals a square.

Why lowest terms matters

Without lowest terms, finding that a and b are both even would be no contradiction: $\frac{4}{2}$4/2 has both even. Choosing lowest terms at the start is what makes it one.

The same idea can be seen another way. If a and b were both even, the fraction would cancel by 2 to $\frac{a}{2}$a/2 over $\frac{b}{2}$b/2, with smaller numbers. The same argument would make those even too, and they would cancel again, forever. Whole numbers cannot be halved forever and stay whole, so no such a and b exist.

Where the argument breaks for $\sqrt{4}$√4

A proof should fail where its claim is false. $\sqrt{4}=2=\frac{2}{1}$√4 = 2 = 2/1 is a fraction, so try the same steps on it. Suppose $\sqrt{4}=\frac{a}{b}$√4 = a/b in lowest terms. Squaring gives $a^2=4b^2$a² = 4b², so $a^2$a² is even and a is even. Write a = 2k: then $4k^2=4b^2$4k² = 4b², so $k^2=b^2$k² = b².

Here the argument stops. $k^2=b^2$k² = b² does not say that $b^2$b² is even, so nothing forces b to be even, and there is no contradiction. For $\sqrt{2}$√2, the step that made b even was $b^2=2k^2$b² = 2k², and the 2 in it came from the 2 under the root.

The diagonal of a square

The problem below uses one fact about any square: the square drawn on its diagonal has twice its area. Call the side s and the diagonal d. Then $d^2=2s^2$d² = 2s².

The drawing shows why. Four squares of side s make one big square. Each small square is cut in half by one of its diagonals, and those four diagonals make a tilted square with side d. The tilted square holds 4 half-squares, which is 2 whole small squares. So its area is $d^2=2s^2$d² = 2s².

This is Pythagoras' theorem for the case of a square: $d^2=s^2+s^2=2s^2$d² = s² + s² = 2s².

Each of the four small squares has side s and area $s^2$s². The tilted square has the diagonal d as its side, and it is made of four half-squares, so its area is $d^2=2s^2$d² = 2s².

The usual mistakes

Taking a calculator's answer as exact. A calculator shows only the first few decimal places of $\sqrt{2}$√2, such as 1.414. That is the fraction $\frac{1414}{1000}$1414/1000, but $1.414^2=1.999396$1.414² = 1.999396, which is not 2. Every decimal that stops is only an approximation to $\sqrt{2}$√2.

Getting the direction wrong. The proof needs "if $a^2$a² is even, then a is even", and that is justified by odd numbers having odd squares. "If a is even, then $a^2$a² is even" is also true, but it is not the step the proof uses.

Leaving out lowest terms. Without it, finding that a and b are both even is not a contradiction.