<s>[INST] 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
 [/INST] we prove $\sqrt11$ is irrational by contradiction.
let $\sqrt11$ is rational,$\exists a,b$ such that $(a,b)=1$ and 
$\sqrt{11}=\frac{a}{b}$ 
$\implies 11=\frac{a^2}{b^2}$
$a^2=11.b^2 \implies 11\mid a^2 $ and $11$ is prime 
therefore, $11\mid a, \exists a_1$ such that $a=11a_1$
$$a^2=(11a_1)^2=121a_1^2=11b^2$$
$$\implies b^2=11a_1^2 $$
$\implies 11\mid b$
it violated $(a,b)=1$.
by contradiction, $\sqrt{11}$ is irrational.
now I go to this problem,
 to prove $\sqrt{11}-1$ is irrational by contradiction,
assume that $\sqrt{11}-1$ is rational.
We know that $1$ is rational number. 
We also know that sum  of two  rational numbers is always rational number.
$[\sqrt{11}-1]+1$ is rational number.
but, $\sqrt{11}$ is irrational
by contradiction, $\sqrt{11}-1$ is irrational
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