prove that there is no rational number $x$ that satisfies $x^2=2$

Nov 2015
246
4
hyderabad
I suck at proofs, I don't know why :(

I want to improve my mathematics skills, because I like mathematics a lot and want to pursue a career in it.

Please check the attached solution and let me know the changes/ way to solve it. Thanks a lot.

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Oct 2009
942
367
Not bad. You clearly don't "suck at proofs". But let me give a detailed deconstruction.

First, don't ever use the symbols $\exists$ or $\Rightarrow$ in proofs. Beginners do this a lot for some reason, but you'll never see these symbols in math papers. Your proof should consist of full sentences written with words. See https://sites.math.washington.edu/~lee/Writing/writing-proofs.pdf

Next, the contents:
1) You say, $q=\frac{p}{q}$, $q\neq 0$ and $\text{gcd}(p,q)=1$. Why does such $p$ and $q$ exist? You should refer to a previous theorem for this.
2) $p^2$ is an even number hence $p$ is an even number. You need to prove this, possibly in a lemma.
3) $\Rightarrow q$ is even. Why? I feel you're skipping a step here.
4) As $x$ is a rational $\Rightarrow\text{gcd}(p,q) = 1$. Why does the fact that $x$ is rational imply this? Isn't this just a hypothesis?
5) "which contradicts our assumption". What assumption? What contradicts what? It's not clear.
 
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skipjack

Forum Staff
Dec 2006
21,478
2,470
Did you want the proof to be from first principles?

The following proof by contradiction assumes some basic knowledge, such as "any rational can be written in lowest terms" and "the square of an odd integer is odd".

Suppose $x$ is a rational such that $x^2= 2$, then $x$ can be written in lowest terms as $p/q$,
where $p$ and $q$ are integers and $\gcd(p,q) = 1$.

As $x^2 = 2$ implies $(p/q)^2 = 2$, $p^2 = 2q^2$, which is even.

As the square of an odd integer is odd, $p$ is even, so $p = 2m$, where $m$ is an integer.

Hence $q^2 = 2q^2/2 = p^2/2 = (2m)^2/2 = 2m^2$, which is even.

As the square of an odd integer is odd, $q$ is even.

As $p$ and $q$ are both even, $\gcd(p,q) \geqslant 2$, which contradicts $\gcd(p,q) = 1$.

Hence the supposition that $x$ is a rational such that $x^2 = 2$ is false.
 
Nov 2015
246
4
hyderabad
Did you want the proof to be from first principles?

The following proof by contradiction assumes some basic knowledge, such as "any rational can be written in lowest terms" and "the square of an odd integer is odd".

Suppose $x$ is a rational such that $x^2= 2$, then $x$ can be written in lowest terms as $p/q$,
where $p$ and $q$ are integers and $\gcd(p,q) = 1$.

As $x^2 = 2$ implies $(p/q)^2 = 2$, $p^2 = 2q^2$, which is even.

As the square of an odd integer is odd, $p$ is even, so $p = 2m$, where $m$ is an integer.

Hence $q^2 = 2q^2/2 = p^2/2 = (2m)^2/2 = 2m^2$, which is even.

As the square of an odd integer is odd, $q$ is even.

As $p$ and $q$ are both even, $\gcd(p,q) \geqslant 2$, which contradicts $\gcd(p,q) = 1$.

Hence the supposition that $x$ is a rational such that $x^2 = 2$ is false.
Thanks for the proof :)