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Prove that $\sqrt{11}$ is irrational, subsequently prove that $n\sqrt{11}$ is also irrational for every $n \in \mathbb{N}$. You are allowed to use that if $p$ is prime, and $p | a^2$, then $p|a$.

Can't you also prove that $n\sqrt{11}$ is irrational simply by saying:

Assume $n\sqrt{11}$ is rational, then $n\sqrt{11} = \dfrac{p}{q}$, so $\sqrt{11} = \dfrac{p}{qn}$, which would mean that $\sqrt{11}$ is rational, which is false (assume we already proved that it's false).

So we know $n\sqrt{11}$ is irrational by contradiction.

Dolma
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2 Answers2

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I am providing you with a general proof that $\sqrt n$ is irrational if $n$ is not a perfect square which would be helpful to you.

Let $n$ be a whole number and not a perfect square.If possible,let $\sqrt n$ be a rational number $\frac pq$ with $gcd(p,q)=1$.

Now. $\frac pq=\sqrt n$

$\implies \frac {p^2}{q^2}= n$

$\implies {p^2}= n{q^2}$

$\implies n|p^2$

$\implies n|p$

Let $p=nm$ for $m\epsilon N$

Then, $p=nm$

$\implies p^2=n^2m^2$

$\implies nq^2=n^2m^2$ [$\because p^2=nq^2$]

$\implies n|q^2$

$\implies n|q$

$\therefore n|p$ and $n|q $

This contradicts our assumption that $gcd(p,q)=1$.This means that our supposition is wrong. Hence ,$\sqrt n$ can,t be a rational number.Thus $\sqrt n$ is irrational.

Remark - It follows from the above proof that $\sqrt 2,\sqrt 3,\sqrt 5,\sqrt 7,\sqrt 10,\sqrt 11$,etc are all irrational

As far as $n\sqrt 11$ is concerned ,your proof is correct.

Snehil Sinha
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I guess that your post also asked for a proof that $\sqrt{11}$ is irrational? Here it is.

Assume, for contradiction, that $\sqrt{11}$ is rational. Let $a$ and $b$ be relatively prime, positive integers. Then

\begin{align*} \sqrt{11} &= \frac{a}{b} \\ 11 &= \frac{a^2}{b^2}\\ 11 b^2 &= a^2 . \end{align*}

This means that $11 \vert a^2$ and so $11 \vert a$ as 11 is prime. So $a=11c$ and

\begin{align*} 11 b^2 &= (11c)^2 \\ b^2 &= 11c^2. \end{align*}

This means that $11 \vert b^2$ and so $11 \vert b$ as 11 is prime. But that both $11 \vert a$ and $11 \vert b$ contradicts $a$ and $b$ being relatively prime, so $\sqrt{11}$ is not rational, which means that it is irrational. Q.E.D.

user179189
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