The Great great man of the east: Root 2 is Irrational & So am I… (Proof & a Little History)

Gelada_Baboon_facial_expression

The Pythagoreans would say that the diagonal of a square is incommensurable with its side. That is, if you draw a square, where each side is measured 1, then what is the length of its diagonal? The answer is $\sqrt{2}$ , an irrational number (one that cannot be written as a ratio, such as $\frac{m}{n}$ ).

There are many possible proofs that $\sqrt{2}$ is irrational, but I like this one because of its historical import. For instance, according to Boyer’s History of Mathematics:¹

“Aristotle refers to a proof of the incommensurability of the diagonal of a square with respect to a side, indicating that it was based on the distinction between odd and even.”

He’s referring to Aristotle’s Prior Analytics, 17.65b.16-21:²

For to put that which is not the cause as the cause, is just this: e.g. if a man, wishing to prove that the diagonal of the square is incommensurate with the side, should try to prove Zeno’s theorem that motion is impossible, and so established a reductio ad impossibile: for Zeno’s false theorem has no connection at all with the original assumption.

Proof that $\sqrt{2}$ is Irrational

I’ll start by assuming³ that $\sqrt{2}$ is rational and then proceed to show that this is impossible.⁴

If $\sqrt{2}$ is RATIONAL, then it can be written as a fraction, such as $\frac{m}{n}$ , where both $m$ and $n$ are elements of the Integers, $\mathbb{Z}$ , and $n \neq 0$ . Assume that I’ve also chosen both $m$ and $n$ to be in lowest terms — you can’t reduce the fraction down any further because they don’t share a common factor.

Given that

$\sqrt{2} = \frac{m}{n}$

then

$2 = (\frac{m}{n})^2 = \frac{m^2}{n^2}$

which implies that

$2n^2 = m^2$

This means that $m^2$ is and even integer by definition, and because of this, $m$ must also be an even integer (because the square of an odd integer is always odd). Therefore there exists an integer $r \in \mathbb{Z}$ such that $m = 2r$ .

We can then plug $2r$ into the above equation and get

$\begin{align} 2n^2 & = & (2r)^2 \\ 2n^2 & = & 4r^2 \\ n^2 & = & 2r^2 \\ \end{align}$

But that means that $n^2$ is also even, which (by the same property mentioned above) means that $n$ is even!

However, this presents a problem. If BOTH $m$ and $n$ are even, then the fraction $\frac{m}{n}$ cannot be in lowest terms — you can at least divide both the numerator and the denominator by 2, if not something else. But that’s a contradiction to our initial assumption that $\sqrt{2} = \frac{m}{n}$ .

Therefore, $\sqrt{2}$ is not a rational number, which is to say, it is irrational.

Now go lift something heavy,
Nick Horton

PS. The pic at the top is of a Gelada Baboon.

Boyer, Carl B. (1991). A History of Mathematics, second edition. ↩
McKeon, Richard (ed). (2001). The Basic Works of Aristotle. ↩
Remember that Assuming makes an Ass out of you and Uming. Poor Uming… ↩
For the Pythagorean version, see Thomas Health’s A History of Greek Mathematics, Vol. 1, pp. 91. ↩

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The Great great man of the east

Tuesday, September 1, 2015

Root 2 is Irrational & So am I… (Proof & a Little History)

Proof that $\sqrt{2}$ is Irrational

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Tuesday, September 1, 2015

Root 2 is Irrational & So am I… (Proof & a Little History)

Proof that is Irrational

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Proof that $\sqrt{2}$ is Irrational