# Proving the irrationality of $\sqrt{5} = x \sqrt{7} + y$

I need to prove that there are no rational numbers $x, y$ that

$$\sqrt{5} = x \sqrt{7} + y$$

We know that square root of prime is irrational so $y = 5 - 7x$ so the only number for it to be rational is if $y = 0$ so $x = \frac{\sqrt{7}} {\sqrt {5}}$ but that is irrational or when $x = 0$, $y = \sqrt{5}$ which is irrational.

$x = \sqrt{5} - y \sqrt {7}$, so I'm stuck here.

Or I need to try to prove each at turn, like

1 case : let $x =$ blah blah and do it as irrational prime proof

2 case : let $y =$ ???

Help would be appreciated.

## 2 Answers

Assume we have rationals $x,y$ such that $\sqrt{5} = x\sqrt{7} + y$.

Square both sides, we get $5 = 7x^2 + y^2 + 2xy\sqrt{7}$.

So $\sqrt{7} = {{5 - 7x^2 - y^2} \over {2xy}}$ which is rational, contradiction.

• why RHS is rational? – talmudist Oct 31 '15 at 17:57
• Because $\mathbb{Q}$ is a field, so closed under +,- and division. – Henno Brandsma Oct 31 '15 at 17:58
• Because $x,y$ are rational, and the rational numbers are closed under the operations of arithmetic. – Lee Mosher Oct 31 '15 at 17:58
• aaa i get it now , thank you very much , i was doing it this way but then got stuck – talmudist Oct 31 '15 at 18:00
• addition: check that $xy \neq 0$ to justify the division. – Henno Brandsma Oct 31 '15 at 18:00

If $x,y\in Q$, then $\sqrt 5= x\sqrt 7+y\implies \sqrt 5-y=x\sqrt 7\implies 5-2 y\sqrt 5 +y^2=7x^2\implies$ $$(-2 y)\sqrt 5=7x^2-5-y^2\in Q\implies y=0\implies$$ $$\sqrt 5=x\sqrt 7\implies x=\sqrt {5/7}\not \in Q$$contradicting $x\in Q$.

• Why did let y=0 ? – talmudist Nov 1 '15 at 10:28
• We got $y=0$ because we got $-2 y\sqrt 5\in Q$. That would make $\sqrt 5 \in Q$ if $y$ were non-zero. – DanielWainfleet Nov 1 '15 at 15:36