Using the fact that 2 is prime, show that there do not exist integers p and q such that $p^2=2q^2$

Question

Using the fact that 2 is prime, show that there do not exist integers p and q such that $p^2=2q^2$. Demonstrate that therefore $\sqrt2$ cannot be a rational number.

Second attempt: Suppose $p^2=2q^2$, then: $$p^2=2q^2$$ $$1=2\frac{q^2}{p^2}$$ $$1=2(\frac{q}{p})^2$$ so $$(\frac{q}{p})^2=1/2$$ $$\frac{q}{p}=\sqrt{1/2}$$ $$\frac{q}{p}=\sqrt{1}/\sqrt{2}$$ $$\frac{q}{p}=\frac{1}{\sqrt{2}}$$ so $q=1$ and $p=\sqrt{2}$, which is not an integer So there are no such integers p and q such that:

$$\sqrt{2}=\frac{p}{q}$$

So $\sqrt 2$ cannot be a rational number.

First attempt: I don't think my answer is right because it doesn't really use the fact that 2 is prime. My attempt: Let p and q be integers. Then, if p=q $p^2=q^2\ne2q^2$. So either p>q or q>p. $$p^2=2q^2$$ $$pp=2qq$$ $$\frac{p}{q}=2(\frac{q}{p})$$ Since $p\neq$, then either $\frac{p}{q}$ or $\frac{q}{p}$ is rational, and the other one is irrational, so they cannot equal each other.

Suppose $p=\sqrt2$, then $$p^2=2q^2$$ $$(\sqrt 2)^2=2q^2$$

That's as far as I could get (not even sure if I was headed in the right direction)

Answer 1

There are various proofs of this result. Since you are asked to use the fact that $2$ is prime, possibly the following one is intended.

Suppose that $p^2=2q^2$, and factorise each side into primes. Since $p^2$ is a square, the number of factors of $2$ on the LHS is even. Similarly, the number of factors of $2$ in $q^2$ is even; but the extra $2$ makes the number of factors of $2$ on the RHS odd. Therefore LHS cannot equal RHS.

Answer 2

This might count under your rules... or not, I leave it with you

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First, find the largest $k$ so that $2^k$ divides both $p$ and $q$. $k$ will be zero if one of them is odd.

Now define $r=\frac{p}{2^k}$ and $s=\frac{q}{2^k}$. Note that we are sure that at least one of $r$ and $s$ is odd, or we could have increased $k$. Then

$$\begin{align} (r2^k)^2 &=2(s2^k)^2\\ r^2(2^k)^2 &=2s^2(2^k)^2\\ r^2 &=2s^2\\ \end{align}$$

Now we see that $r^2$ is even, and since $2$ is prime $r$ must also be even (and $s$ must be odd from our previous work). Define $t=\frac{r}{2}$, so $r=2t$. Then

$$\begin{align} (2t)^2 &=2s^2\\ 4t^2 &=2s^2\\ 2t^2 &=s^2\\ \end{align}$$

But now we find that $s^2$ is even, which implies that $s$ is even as $2$ is a prime, which contradicts our construction of $r$ and $s$. So no such integers $p$ and $q$ can exist with the given property.

Answer 3

By contradiction.If $\sqrt 2$ is rational there is a LEAST positive integer $p$ for which $p^2=2q^2$ for some positive integer $q.$ This implies that $p$ is even because $p^2=2q^2$ is even. (In general if $x$ is an integer and $x^2$ is divisible by the prime $y$, then $x$ is divisible by $y$.) So let $p=2p'.$ Then $p'$ is a positive integer. We have $4p'^2=p^2=2q^2,$ therefore $2p'^2=q^2.$ So $q$ is even. Let $q=2q'.$ Then $q'$ is a positive integer. We have $2p'^2=q^2=4q'^2$. Therefore $p'^2=2q'^2.$ But $p'=p/2<p$ and this means $p$ is NOT the least,a contradiction...... Another way of saying this is that if $S=\{p\in Z^+ :\exists q\in Z^+ (p^2=2q^2)\}$ is not the empty set, then for any $p\in S$ we have $S\supset T=\{p,p/2,p/4,p/8,...\}$ which makes $T$ an infinite strictly descending sequence of positive integers, which is impossible.