So I was reading Spivak's Calculus for fun and I found on the bottom of page 25, 4th edition:(adsbygoogle = window.adsbygoogle || []).push({});

Suppose that [itex]\sqrt{2}[/itex] were rational; that is, suppose there were natural numbers p and q such that [itex]\frac{p^{2}}{q^{2}} = 2[/itex].

We can assume that p and q have no common divisor (since all common divisors could be divided out to begin with). Now we have

p[itex]^{2}[/itex] = 2q[itex]^{2}[/itex].

This shows that p[itex]^{2}[/itex] is even, and consequently p must be even; that is p = 2k for some natural number k. Then

p[itex]^{2}[/itex] = 4k[itex]^{2}[/itex]=2q[itex]^{2}[/itex],

so 2k[itex]^{2}[/itex] = q[itex]^{2}[/itex].

This shows that q[itex]^{2}[/itex] is even, and consequently q is even. Thus both p and q are even, contradicting the fact that p and q have no common divisor. This contradiction completes the proof.

Am I missing something or is this proof not complete and sloppy?! Spivak didn't even consider the possibility that q is odd and p is even; in which case, we can write p as 2k and q as 2m+1 and THEN check the equivalency. In this case we find that two sides won't be equivalent when we substitute 2k and 2m+1 in the given equation and that would complete the proof.

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# Sloppy Proof in Spivak's Calculus?

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