# Courant's proof of irrational number

In Courant's Differential and Integral Calculus he proves that if a right-angle triangle has sides of unit 1 length then, using Pythagoras, we have $h^2 = 1^2 + 1^2 = 2$. Now, if $h$, the hypotenuse, is a rational number, it can be represented as $p/q$. And manipulation of Pythagoras gives $p^2 = 2q^2$. And since $p^2$ is an even number (it equals $2q^2$ and any number times 2 is even) . . . and here's where I get stuck. Courant says since $p^2$ is even, $p$ itself must be even, say $p = 2p\prime$. Substituting this expression for $p$ gives us $4p\prime^2 = 2q^2$, or $q^2 = 2p\prime^2$, consequently $q^2$ is even, and so (again, my number theory breakdown!) so $q$ is even. Hence, $p$ and $q$ both have the factor 2 . . . which contradicts our premise that a good rational number has no "hidden" common factor. Thus the hypotenuse $h$ cannot be represented by a fraction $p/q$. . . And so my confusion, Why can we say, apparently, when a square is even, it's root must be even?

• This proof is quite a bit older than Courant. It's attributed to Pythagoras, or to a student of his. Commented Feb 15, 2015 at 14:05
• Have you understood the problem? If so, can you mark the right answer? Commented Feb 17, 2015 at 8:21

• more generally , for any prime p , if $n^2$ divide p then n divides p Commented Feb 15, 2015 at 5:30
Squares of even numbers are even (and in fact divisible by $4$), since $(2n)^2 = 4n^2$.