How to show that $\sqrt{p\over p+1}$ is irrational?

Question

I want to prove that $\sqrt{p\over p+1}$ is irrational where $p$ is any prime number.

I have been thinking about this problem for several hours without any progress.

To show that $\sqrt{p\over p+1}$ is irrational, I assume that it is rational. So there are integers $m$ and $n$ (which have no common divisors) such that $\sqrt{p\over p+1}=\frac{m}{n}$. By squaring and rearranging, I get $pn^2=(p+1)m^2$ but I am not sure how to proceed from here. I can say that $(p+1)m^2$ is divisible by $p$ and then I do not know what to do!

@YvesDaoust, so $p \mid m^2$, therefore $p \mid m$. Similarly, $(p + 1) \mid n^2$. Then what? (Just trying to articulate what's going on in the OP's mind...) — Jose Arnaldo Bebita Dris, Mar 10 '17 at 10:08
$p$ and $p+1$ are definitely do not have any factors common other than 1 ...Moreover..No two consecutive integers are perfect squares.. — LM2357, Mar 10 '17 at 10:08
@JoseArnaldoBebitaDris No "Similarily". Ignore $(p+1)$. Work with $p$ and recognize the steps form the proof that $\sqrt2$ is irrational. — Arthur, Mar 10 '17 at 10:11
@user35508: Don't shoot me, but many mathematicians consider $0$ to be natural as well. — Mathematician 42, Mar 12 '17 at 15:16
@Mathematician..Oh yes , they do..But where I was taught ..We used the term Whole numbers to represent ${0}$ + $\mathbb{N}$ — LM2357, Mar 12 '17 at 15:18

score 2 · Accepted Answer · edited Mar 10 '17 at 10:31

2

Suppose

$$\sqrt\frac p{p+1}=\frac mn\;,\;\;m,n\in\Bbb Z\;,\;\;\frac mn\;\;\text{ is a reduced fraction} \implies pn^2=m^2(p+1)\implies p\,\mid\,m$$

since clearly $\;p\;$ divides the left side and $\;p+1\;$ cannot be divisible by $\;p\;$, but then

$$\;m=pk\implies m^2=p^2k^2\implies pn^2=m^2(p+1)=p^2k^2(p+1)\implies n^2=pk^2(p+1)$$

and thus also $\;p\,\mid n\;$ , contradiction.

edited Mar 10 '17 at 10:31

Nathanael Skrepek

2,558

answered Mar 10 '17 at 10:09

DonAntonio

211,718
17
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Thanks for your reply. I do not know the truth of some of the results that you used in your proof so I also need to prove them too for myself. – Mar 10 '17 at 10:18
Should not in your first line $pn=m^2(p+1)$ be $pn^2=m^2(p+1)$? – Mar 10 '17 at 10:25

Mathematician 42 · Answer 2 · 2017-03-10T10:29:12.363

Here's an alternative proof.

Consider the polynomial $(p+1)X^2-p$. By Eisenstein's criterium this polynomial is irreducible over $\mathbb{Q}$. Clearly $\sqrt{\frac{p}{p+1}}$ is a root of this polynomial. It follows that the extension degree $[\mathbb{Q}(\sqrt{\frac{p}{p+1}}):\mathbb{Q}]=2$. In particular this implies that $\sqrt{\frac{p}{p+1}}\notin \mathbb{Q}$.

Once you understand Eisenstein's criterium and field extensions you'll probably like this method. But it is of course overkill compared to the standard method.

How to show that $\sqrt{p\over p+1}$ is irrational?

2 Answers2