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 question,
$$\sqrt {11} -1 = \frac {a}{b}$$ where $(a,b)=1$.
From last  step,
$10b^2 = a^2 + 2ab$
$LHS $ is multiple of  $5$. then $RHS$ is also multiple of $5$.
in $RHS,a^2 + 2ab$  exactly one  of $a$ and $b$ is multiple.
if  both $a$ and $b$  are multiple of $5$ , it violated $(a,b)=1$.
if $b$ is  multiple of $5$. let $b=5b_1 \implies 250b_{1} ^2 =a^2+50ab_1 \implies 5 \mid a $.
but, it violated $(a,b)=1$.
therefore, $a$ is  multiple of $5$. let $a=5a_1$.
$10b^2= 25 a_1^2+2ab$
$10b^2-25 a_1^2= 10a_1b$
$/5 \implies 2b^2-5a_1^2=2a_1b$
Therefore $2 \ mid a_1 , \exists a_2$ such that $a_1=2a_2$
$ 2b^2-10a_2^2 =4a_2b$
$\implies b^ 2 -2a_2b=  5a_2^2 $
$\implies 5\mid a_2 , \ exists a_3$ such that a_2=5a_3
$ b^2  -10a_3b =25a_3^2$
$ b^2  =25a_3^2 +10a_3b$
Therefore $ 5 \mid b$
Which contradict our condition.
