4

I start the proof by assuming there is $m/n \in Q$, expressed in lowest terms, such that $$(m/n)^2=2$$

My textbook tells me to derive a contradiction by showing $m,n$ are both even. Here's how I did it.

$(m/n)^2=2$

$m^2/n^2=2$

$m^2=2n^2$

Hence, $m^2$ is even and $m$ must be even. Then, $m^2=2k$ for some integer $k.$ Plugging $m^2$ back to the equation,

$(2k)^2/n^2=2$

$4k^2=2n^2$

$n^2=2k^2$

Hence, $n^2$ is even so as $n.$ Therefore, it contradicts $m/n$ are in lowest terms.

Is this right way to prove this? I feel I am doing wrong because I am plugging back the equation I got from the equation to the same equation.

TonyK
  • 64,559
user3000482
  • 1,516
  • 3
    It is the standard way to do it. – Beni Bogosel Dec 22 '16 at 20:56
  • 4
    It is correct, just typo, the $m^2=2k$ should be $m=2k$. – Sil Dec 22 '16 at 20:57
  • 2
    Why are there close votes for this question? Yes, it's a duplicate, but with the [proof-verification] tag it is obvious that the user is asking for feedback on his/her own proof, not a pointer to someone else's proof. Closing as a duplicate defeats the entire purpose of this tag. – Gyu Eun Lee Dec 22 '16 at 22:31
  • The statement in the title is just wrong; there *is so* a rational number such that the square of it cannot be $2.$ For instance, $1$ is such a number; $1$ is rational and $1^2\ne2.$ – bof Dec 22 '16 at 22:52

1 Answers1

2

Yes, that's a classical proof. It can be recast into the following proof comparing parity of powers of $2$ in factorizations (a special case of uniqueness of prime factorizations). Suppose $\, A^{\large 2} = 2B^{\large 2}$

It is straightforward to prove that every natural can be written uniquely in the form $\, 2^{\large k} n\,$ for odd $\,n.\,$ So we have $\, A= 2^{\large J} a,\ B = 2^{\large K} b\,$ for $\,a,b\,$ odd. Therefore $\, A^{\large 2}\! = 2 B^{\large 2}$ $\Rightarrow$ $\,2^{\color{#c00}{\large 2J}} a^{\large 2} = 2^{\color{#0a0}{\large 1+2K}} b^{\large 2}.\ $ But $\,a,b\,$ odd $\,\color{#f2f}\Rightarrow$ $\,a^{\large 2},b^{\large 2}\,$ odd too, which contradicts uniqueness (LHS has $\rm\color{#c00}{even}$ vs. RHS $\rm\color{#0a0}{odd}$ power of $2).$

To generalize this proof from $\,p = 2\,$ to an arbitrary prime $\,p\,$ would require generalizing the $\,\rm odd^{\large 2} = odd\,$ step that we used above, $ $ i.e. we would need a proof of the following inference $\ p\nmid a,b\,\color{#f2f}\Rightarrow\, p\nmid a^{\large 2},b^{\large 2}.\ $ Just as for the case $\,p=2,\,$ for any fixed prime $\,p\,$ we could prove it by a brute-force case analysis. But to prove the result for all primes $\,p\,$ we need something much deeper, e.g. Euclid's lemma $\ p\mid ab\,\Rightarrow\, p\mid a\,$ or $\,p\mid b\ $ (or closely related results).

Bill Dubuque
  • 272,048