Prove that $\sqrt{11}-1$ is irrational by contradiction

Question

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

Answer 1

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.

Answer 2

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

Answer 3

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 $3^2 < 11 < 4^2$.

Answer 4

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.

Answer 5

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 $x^2 + 2x - 10$.

Answer 6

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.