Prove that $\sqrt{11}-1$ is irrational by contradiction I am working on an assignment in discrete structures and I am blocked trying to prove that $\sqrt{11}-1$ is an irrational number using proof by contradiction and prime factorization.
I am perfectly fine doing it with only $\sqrt{11}$, but I am completely thrown off by the $-1$ when it comes to the prime factorization part.
My current solution looks like this :
$$ \sqrt {11} -1 = \frac {a}{b}$$
$$ \sqrt {11} = \frac {a}{b} + 1$$
$$ \sqrt {11} = \frac {a+b}{b}$$
$$ 11 = \left(\frac {a+b}{b}\right)^2$$
$$ 11 = \frac {(a+b)^2}{b^2}$$
$$ 11 = \frac {a^2 + 2ab + b^2}{b^2}$$
$$ 11 b^2 = a^2 + 2ab +b ^2$$
$$ 10b^2 = a^2 + 2ab $$
At that point, is it acceptable to conclude that a² is a multiple of 11 even though we have a trailing $2ab$?
The required method is then to conclude using prime factorization that $a = 11k$ and replace all that in the formula above to also prove $b$, however, I am again stuck with the ending $2ab$.
Would it instead be correct to prove that $\sqrt{11}$ is rational using the usual method and that, by extension, $\sqrt{11} - 1$ is also rational?
Thank you

No, you cannot conclude that $a^2$ is a multiple of $11$. You can instead rewrite
$$
a^2=10b^2-2ab=2(5b^2-ab)
$$
so $2\mid a$. Write $a=2c$, with $c$ integer. Then
$$
4c^2=2(5b^2-2bc)
$$
or $5b^2=2(c^2+bc)$. Since $2\nmid 5$, we conclude $2\mid b$.
This is a contradiction to $a$ and $b$ being coprime.
