Prove or disprove that $\sqrt{2+\frac{1}{n}}$ is irrational for $n \in \mathbb{Z}^+$

I have good reason to suspect that $\sqrt{2+\frac{1}{n}}$ is irrational for all $n \in \mathbb{Z}^+$ but a proof of this eludes me. I've tried proof by contradiction have had no success. I've also tried induction where clearly the base case is true, i.e. $\sqrt{2+1}=\sqrt{3}$ is irrational, but I haven't been able to show the induction step.

• To prove something is irrational, one must (almost) always use a contradiction proof. – Michael Burr Jun 30 '15 at 2:29
• Are you serious ? Elementary scrutiny would have been to try a few perfect squares, starting with $n=4$ ! – Yves Daoust Jul 20 '15 at 10:41

In fact, there are infinitely many $n$ with $\sqrt{2+\frac1n}$ rational.

You need $n$ and $2n+1$ to both be perfect squares, say $y^2$ and $x^2$. Then you are trying to solve:

$$x^2-2y^2=1\tag{1}.$$

Then $n=y^2$.

There are infinitely many solutions to this equation, starting with $(x_1,y_1)=(3,2)$ and $(x_{k+1},y_{k+1})=(3x_k+4y_k,2x_k+3y_k)$. So the first is $n=4$, next $n=144$. Next: $n=4900$.

You can actually get the closed form:

$$n =\left\lfloor\frac{(17+12\sqrt 2)^{k}}{8}\right\rfloor.$$

The equation (1) is called Pell's Equation for $2$.

Indeed, if $D$ is a positive integer, then:

1. If $D$ is not s perfect square then there are infinitely many positive $n$ such that $\sqrt{D+\frac1 n}$ is rational;
2. If $D$ is a perfect square, then there is no positive $n$ such that $\sqrt{D+\frac1n}$

This all follows from elementary results about Pell's equation.

Disproof: $n=4 \rightarrow 2+1/4= 2.25 ; \sqrt{2.25} =1.5=3/2$

• Better to just keep fractions - decimal notation often gives the impression of being approximations. – Thomas Andrews Jun 30 '15 at 3:07

Let's start a proof by contradiction to try to show that no solutions exist and see where the proof goes wrong. Suppose that $\sqrt{2+\frac{1}{n}}$ is rational and can be written as $\frac{p}{q}$ where $p$ and $q$ share no factors. Then, we know that $$2+\frac{1}{n}=\frac{p^2}{q^2}.$$ By clearing $q^2$ from both sides, we have that $$2q^2+\frac{q^2}{n}=p^2.$$ Therefore, $\frac{q^2}{n}$ must be an integer, so $n\mid q^2$.

Now, if $\frac{q^2}{n}\not=1$, then there is some prime $s$ dividing $\frac{q^2}{n}$. Therefore, this prime divides $q$. Then, it follows that this prime divides the LHS so it divides $p^2$, but then $p$ and $q$ share a factor (a contradiction).

Therefore, it must be that $q^2=n$. Therefore, the equality becomes: $$2n+1=p^2$$ (which is the same as $2q^2+1=p^2$).

We now begin the construction: Let us assume that $n$ is a square integer such that $2n+1$ is a square. Let $p=\sqrt{2n+1}$ and $q=\sqrt{n}$. Then $\sqrt{2+\frac{1}{n}}=\frac{p}{q}$ and this is the only way to have a counterexample (as seen in the provided disproof).

• Wow, you do all the work, and I get all the points -- my idea of teamwork ;). – Gary. Jun 30 '15 at 2:40
• Cheers! @Gary. But, at the end of the day, one must come up with an example that satisfies the conditions I provided (which you did). – Michael Burr Jun 30 '15 at 2:42
• Isn't it easier to just show that if $p,q$ relatively prime, then $p^2$ and $q^2$ are relatively prime. And $2n+1$ and $n$ are relatively prime, so $p^2=2n+1$ and $q^2=n$. You just need to know that every rational number has a unique representation as a quotient of relatively prime numbers. – Thomas Andrews Jun 30 '15 at 2:46
• @ThomasAndrews Nice observation. – Michael Burr Jun 30 '15 at 2:50