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I have a question about example 1.1 on Baby Rudin (principles of mathematical analysis by Walter Rudin). This is the example that starts with $p^{2} = 2$ and tries to prove no rational number $p$ exists that satisfies the equation. He tries to prove this by contradiction by assuming there is a $p$ such that $p = m/n$. Shortly after this he has the equation $m^{2} = 2n^{2}$ and implies that $m^{2}$ is even. Okay this makes sense since any integer times an even integer is an even integer (parity property)-- in this case it is $2$ that is the even integer and $n^{2}$ which may be any integer -- either odd or even. He then goes on to say that since $m^{2}$ is even $m$ is also even and that $m^{2}$ is divisible by $4$. He then states: "It follows that the right side of $(m^{2} = 2n^{2}$) is divisible by $4$, so that $n^{2}$ is even, which implies that $n$ is even." However, I disagree with that last bit where he claims $n^{2}$ and $n$ are even because the fact is that $n^{2}$ may be odd and the right hand side $2n^{2}$ would still be even and divisible by $4$. This implies that $n$ may be odd and not just even, and I claim that since $n$ and $n^{2}$ are not necessarily even, his proof by contradiction is either incomplete or not correct. I am probably wrong, but his implication doesn't seem to be very sound. If someone could clarify that would be great.

CrypticParadigm
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    $m^2 = 2n^2$ implies $m$ is even so that $4q^2 = 2n^2$ for some integer $q$, so that $2q^2 = n^2$. It follows that $n^2$ is even so therefore $n$ is even. – MathematicsStudent1122 Sep 08 '17 at 01:10

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However, I disagree with that last bit where he claims $n^2$ and $n$ are even because the fact is that $n^2$ may be odd and the right hand side $2n^2$ would still be even and divisible by $4$.

If $n^2$ is odd, then $2n^2$ is not divisible by $4$.

It might be easier to see this part as:

  • $m^2=2n^2$.

  • So $m^2$ is even.

  • So $m$ is even.

  • So $m=2k$ for some $k$.

  • Then $2n^2=4k^2$.

  • This gives $n^2=2k^2$.

  • So $n$ is even (by the same logic we used to conclude that $m$ was even!).

Noah Schweber
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Because $m$ is even, $m^2 = 4k^2$ for some integer $k$. So $2n^2=4k^2$ which implies that $n^2=2k^2$, so $n^2$ is even, so $n$ is even.

littleO
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because the fact is that $n^2$ may be odd and the right hand side $2n^2$ would still be even and divisible by $4$.

No, it can't. If $n^2$ is odd then $2n^2$ can not be divisible by $4$.

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$m^2 = 2n^2$ implies $m^2$ is even. If $m$ is odd then $m^2$ is odd. That is not the case so $m$ is even.

So $m^2$ is divisible by $4$. So $2n^2$ is divisible by $4$. So $n^2$ is divisible by $2$. So $n^2$ is even.

Repeat. If $n$ is odd $n^2$ is odd. That is not so. So $n$ is even.

So if $(\frac mn)^2 = 2$ and $m,n$ are integers it must be that the are both even.

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Okay:

Look:

1) even * even = even.

Pf: $(2k)(2j) = 2(2kj)$.

2) odd * odd = odd.

Pf: $(2k + 1)(2j + 1) = 4kj +2k + 2j + 1 = 2(2kj +k + j) + 1$.

Therefore $even^2 = even*even = even$ and $odd^2 = odd*odd = odd$.

So:

$m^2 = 2n^2 \implies m^2$ is even $\implies m$ is even.

So $m = 2k$ for some $k$.

So $m^2 = (2k)^2 = 4k^2$.

So $4k^2 = m^2 = 2n^2$.

So $2k^2 = n^2$

So $n^2 = 2k^2 \implies n^2 $ is even $\implies n$ is even.

So $m^2 = 2n^2$ means both $m$ and $n$ are even.

fleablood
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