Proof: $\sqrt {a^2-b^2}$ irrational

Question

I'm trying to prove that the following expression is irrational: $$ \sqrt{x^2+1}$$

$$ \sqrt{x^2+1}=\frac a b \\ x^2+1=\left(\frac a b\right)^2=\frac {a^2} {b^2 } \\ x^2=\frac {a^2} {b^2 } -1 \\ x=\sqrt{\frac {a^2} {b^2 } -1}=\sqrt{\frac{a^2-b^2}{b^2} } \\ \frac {\sqrt{a^2-b^2}} {\sqrt{b^2 }} \\ x=\frac {\sqrt{a^2-b^2} }b $$

Here is where I get stuck. I know that to prove that it is irrational I must prove that $$\sqrt {a^2-b^2} $$ is not a rational expression.

Any help would be appreciated.

Answer 1

What about $$a=5,b=4,c=3$$ I think the assumption that this is irrational is wrong, or maybe just needs more context

Answer 2

You are on a wrong track.

If you know an algebraic proof that $\sqrt{2}$ is irrational you can generalize it to show that the square root of an integer is rational if and only if it is the square of an integer*. Since adding $1$ to a square (other than $0$) can't produce a square you're done.

*Hint. Write $a/b$ in lowest terms. Then show that every prime that divides $a$ must divide $n^2 + 1$ twice.