Square Root of 2

This discovery greatly upset the Pythagoreans, since they revered the integers as the mystical foundation of the universe, and now apparently they were not even sufficient foundation for the numbers. Ironically, this unnerving discovery followed from applying their very own theorem–Pythagoras’ theorem–to the simplest possible right-angled triangle: half a square, a triangle with its two shorter sides both equal to one. This means its long side-the hypotenuse–has a length whose square is two. We shall now go through their argument showing that the length of this longest side cannot be written as a ratio of two integers, no matter how large you choose the integers to be. The basic strategy of the proof is to assume it can be written as a ratio of integers, then prove this leads to a contradiction. So, we assume we can write this number–the length of the longest side– as a ratio of two whole numbers, in other words a fraction m/n. This is the length whose square is 2, so m²/n² = 2, from which m² = 2n². Now all we have to do is to find two whole numbers such that the square of one is exactly twice the square of the other. How difficult can this be? To get some idea, let’s write down the squares of some numbers and look: 12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64, 92 = 81, 102 = 100, 112 = 121, 122 = 144, 132 = 169, 142 = 196, 152 = 225, 162 = 256, 172 = 289, '. On perusing this table, you will see we have some near misses. 32 is only one more than twice 22, 72 is only one less than twice 52, and 172 is only one more than twice 122. It’s difficult to believe that if we keep at it, we’re not going to find a direct hit eventually. In fact, though, it turns out this never happens, and that’s what the Pythagoreans proved. Here’s how they did it. First, assume we canceled any common factors between numerator and denominator. This means that m and n can’t both be even. Next, notice that the square of an even number is even. This is easy to check: if a is an even number, it can be written a = 2b, where b is another whole number. Therefore, a2 = 2x2xb2, so on fact a2 is not only even, it has 4 as a factor. On the other hand, the square of an odd number is always odd. If a number doesn’t have 2 as a factor, multiplying it by itself won’t give a number that has 2 as a factor. Now, back to the length of the square’s diagonal, m/n, with m² = 2n². Evidently, m² must be even, because it equals 2n², which has a factor 2. Therefore, from what we have just said above about squares of even and odd numbers, m must itself be even. This means, though, that m² must be divisible by 4. This means that 2n² must be divisible by 4, since m² = 2n² — but in this case, n² must be divisible by 2! It follows that n must itself be even — BUT we stated at the beginning that we had canceled any common factors between m and n. This would include any factor of 2, so they can’t both be even! Thus a watertight logical argument has led to a contradiction. The only possible conclusion is: the original assumption is incorrect. This means that the diagonal length of a square of side 1 cannot be written as the ratio of two integers, no matter how large we are willing to let them be. This was the first example of an irrational number—one that is not a ratio of integers. Legend has it that the Pythagoreans who made this discovery public died in a shipwreck.