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
 A: 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.
A: 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
A: Continuing where you stopped:
$$
10b^2 = a^2 + 2ab
$$
Write this as
$$
b(10b-2a) = a^2
$$
Therefore, $b$ divides $a^2$. Since $\gcd(a,b)=1$, the only possibility is $b=1$. But then $\sqrt{11}-1=a$ is an integer, which implies $\sqrt{11}$ is an integer, which it clearly isn't because $3^2 < 11 < 4^2$.
A: 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.
A: 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.
A: Alternative proof: $\sqrt{11} - 1$ is a root of $x^2 + 2x - 10 = (x + 1)^2 - 11$. Using that $p/q\in\Bbb Q$ (irreductible fraction) is a root of $P(x)\in\Bbb Z[x] = a_nx^n + \cdots + a_1x + a_0\implies p\vert a_0,q\vert a_n$ (the rational root theorem), you can check that all the possible $p/q$ aren't roots of $x^2 + 2x - 10$.
