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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?

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

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The square of an odd number is odd. The square of an even number is even.
So, if the square is even, the square-root must be even.

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  • $\begingroup$ more generally , for any prime p , if $n^2$ divide p then n divides p $\endgroup$
    – VigneshM
    Commented Feb 15, 2015 at 5:30
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    $\begingroup$ To do this, you have to prove (1) that all integers are either even or odd and (2) no integer is both even and odd. To make it more fun, start with the Peano postulates and define even and odd. $\endgroup$ Commented Feb 15, 2015 at 5:40
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Squares of even numbers are even (and in fact divisible by $4$), since $(2n)^2 = 4n^2$.

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