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

From $10b^2=a^2+2ab$ you can see that $2\mid a^2$, hence $a=2m$ (for some $m$), from which you can conclude that $b^2=2m^2+2mb-4b^2$, which implies $2\mid b$, contradicting the (tacit) assumption $\gcd(a,b)=1$.
This is an interesting variant on the standard proof of irrationality in that it only invokes the implication $2\mid n^2\implies 2\mid n$ instead of the more general implication $p\mid n^2\implies p\mid n$ (with $p$ prime). Indeed, it gives an easy proof for the irrationality of $\sqrt{4k+3}$, whether $4k+3$ is prime or not:
$$\sqrt{4k+3}-1={a\over b}\implies(4k+3)b^2=a^2+2ab+b^2\implies a^2=2((2k+1)b^2-ab)\\
\implies a=2m\\
\implies2m^2=(2k+1)b^2-2mb\\
\implies b^2=2(m^2-kb^2+mb)\\
\implies 2\mid b
$$
which contradicts the assumption $\gcd(a,b)=1$. Thus $\sqrt{4k+3}-1$ is irrational, hence $\sqrt{4k+3}$ is irrational.
