Prove that $\sqrt{n^2 + 2}$ is irrational

Question

Suppose $n$ is a natural number. Prove that $\sqrt{n^2 + 2}$ is irrational.

From looking at the expression, it seems quite obvious to me that $\sqrt{n^2 + 2}$ will be irrational, since $n^2$ will be a natural number, and after adding $2$ to it, $n^2 + 2$ will no longer be a perfect square.

From that logic, I tried letting $n^2 + 2$ = ${a^2\over b^2}$ to find a contradiction, but I wasn't able to formulate a proof that worked. Any suggestions as to how I should approach this?

Try and prove that $n^2+2$ is never a sqaure and that the sqaure root of a non-sqaure is irrational — Joao, Dec 10 '14 at 06:19

score 3 · Answer 1 · answered Dec 10 '14 at 17:36

It can be easily proven that $\sqrt k$ is irrational if $k\in$ N is not a perfect square. At the same time, the difference between two consecutive perfect squares is $(n+1)^2-n^2=2n+1>2$ for $n>0$. So $n^2+2$ lies in between two consecutive perfect squares, meaning it is not one. Therefore, its radical is irrational.

score 2 · Answer 2 · 2014-12-10T06:46:02.623

2

Without loss of generality we can assume $x,y\in \mathbb{N}\ (?)$.

$$n^2+2=\dfrac{x^2}{y^2} \implies x^2=y^2\left(n^2+2\right)$$

For odd or even $n$, $x,y$ can be both odd, or exactly one of them should be even.

Case 1 ($n$ odd)

Case 1.1 (both odd)

$\left(x^2\equiv1\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 1\equiv3\pmod 4$

Case 1.2 ($x$ odd, $y$ even)

$\left(x^2\equiv1\pmod4, y^2\equiv0\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 1\equiv0\pmod 4$

Case 1.3 ($x$ even, $y$ odd)

$\left(x^2\equiv0\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 0\equiv3\pmod 4$

Case 2 ($n$ even)

Case 2.1 (both odd)

$\left(x^2\equiv1\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 1\equiv2\pmod 4$

Case 2.2 ($x$ odd, $y$ even)

$\left(x^2\equiv1\pmod4, y^2\equiv0\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 1\equiv0\pmod 4$

Case 2.3 ($x$ even, $y$ odd)

$\left(x^2\equiv0\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 0\equiv2\pmod 4$

edited Dec 10 '14 at 06:46

answered Dec 10 '14 at 06:25

Why the "?" is this really an answer? – John Smith Dec 10 '14 at 06:32
@JohnSmith: It is not an answer, just a hint. – Dec 10 '14 at 06:33
Your second line is wrong. The $2$ should be $2y^2$. – Dec 10 '14 at 06:33
@Bungo: It's now edited. Is it ok now? – Dec 10 '14 at 06:46
I think it's OK now. – Dec 10 '14 at 06:51
@Bungo: Thanks for noting my pathetic mistake. – Dec 10 '14 at 06:56
Would it suffice to say ${a^2\over b^2}$ is congruent to $0, 1 (mod 4)$ and $n^2 + 2$ is congruent to $2, 3 (mod 4)$? – user199081 Dec 10 '14 at 07:02
@user199081: No. Congruence is defined (so far as I know) for integers only. – Dec 10 '14 at 07:06
Since $n^2 + 2$ is a natural number, could I say $a^2$ is congruent to $0,1(mod 4)$ and $n^2 + 2$ is congruent to $2,3(mod 4)$? – user199081 Dec 10 '14 at 07:09
@user199081: Of course, you can claim that. I wanted to mean that it is incorrect to say $\frac{a^2}{b^2}$ congruent to $0,1$ unless $b=\pm1$. – Dec 10 '14 at 07:14

score 2 · Answer 3 · answered Dec 10 '14 at 06:27

When $n$ is even, show that $n^2+2$ will be $2$ times an odd number. Then show that such a number doesn't have a rational square root (you can modify the usual proof that $\sqrt{2}$ is irrational).

When $n$ is odd, show that $n^2+2\equiv 3$ (mod $4$). Then show that any integer squared is congruent to $0$ or $1$ mod $4$, so that $n^2+2$ is not a perfect square. As in Jaoa's comment, finish up by arguing the square root of a non-square is irrational, which again can be done by modifying the $\sqrt{2}$ proof.

user194150 · Answer 4 · 2014-12-10T06:43:48.627

$$\begin{array}{l} \sqrt {2 + n^2 } = \frac{p}{q};\quad and\quad p \wedge q = 1 \\ \sqrt {2 + n^2 } = \frac{p}{q} \Leftrightarrow 2 + n^2 = \left( {\frac{p}{q}} \right)^2 \\ \Leftrightarrow 2 = \left( {\frac{p}{q}} \right)^2 - n^2 \\ \Leftrightarrow 2 = \left( {\frac{p}{q} - n} \right)\left( {\frac{p}{q} + n} \right) \\ \Leftrightarrow 2q^2 = \left( {p - nq} \right)\left( {p - nq} \right) \cdots \left( * \right) \\ \left( * \right) \Rightarrow \left\{ \begin{array}{l} \exists m_0 \in\mathbb{ N} :p - nq = m_0 q \\ or \\ \exists m_1 \in \mathbb{N} :p + nq = m_1 q \\ \end{array} \right. \\ \Rightarrow \left\{ \begin{array}{l} \exists m_0 \in \mathbb{N} :p = \left( {n + m_0 } \right)q \\ or \\ \exists m_1 \in \mathbb{N} :p = \left( { - n + m_1 } \right)q \\ \end{array} \right. \\ \left\{ \begin{array}{l} p = \left( {n + m_0 } \right)q \Rightarrow q/p \\ or \\ p = \left( { - n + m_1 } \right)q \Rightarrow q/p \\ \end{array} \right. \\ \end{array}$$

contradiction with $\quad p \wedge q = 1$

score 1 · Answer 5 · edited Apr 13 '17 at 12:20

1

If $n\equiv0\pmod{2}$, then $n^2+2\equiv2\pmod{4}$.
If $n\equiv1\pmod{2}$, then $n^2+2\equiv3\pmod{4}$.

Thus, $n^2+2$ cannot be a perfect square since neither $2$ nor $3$ is a quadratic residue mod $4$.

As shown in this answer or using the Rational Root Theorem, if $n^2+2$ is not a perfect square, then $\sqrt{n^2+2}$ is irrational.

edited Apr 13 '17 at 12:20

Community

1

answered Aug 14 '15 at 09:04

robjohn

345,667

score 0 · Answer 6 · answered Oct 06 '23 at 22:32

$n=\sqrt(n^2)< \sqrt{n^2+2}<\sqrt{n^2+2n+1}=n+1$

So $\sqrt{n^2+2}$ is not an integer.

Assume $\sqrt{n^2+2}$ is rational $=p/q, [p,q\in N \land gcd(p,q)=1 ]$

Then $n^2+2=p^2/q^2$.

And $(n^2+2)q^2=p^2$

If $q=1$ then $\sqrt{n^2+2}=p$ is an integer. A contradiction.

If $q>1$, then $\exists $ prime $y | q\implies y | p \implies gcd(p,q)>1$. Another contradiction.

$q\in N \implies q\ge 1$.

Both cases are impossible so the assumption that $\sqrt{n^2+2}$ is rational is false.

Prove that $\sqrt{n^2 + 2}$ is irrational

6 Answers6