# If $n$ is any positive integer, prove that $\sqrt{4n-2}$ is irrational

If $n$ is any positive integer, prove that $\sqrt{4n-2}$ is irrational.

I've tried proving by contradiction but I'm stuck, here is my work so far:

Suppose that $\sqrt{4n-2}$ is rational. Then we have $\sqrt{4n-2}$ = $\frac{p}{q}$, where $p,q \in \mathbb{Z}$ and $q \neq 0$.

From $\sqrt{4n-2}$ = $\frac{p}{q}$, I just rearrange it to:

$n=\frac{p^2+2q^2}{4q^2}$. I'm having troubles from here, $n$ is obviously positive but I need to prove that it isn't an integer.

Any corrections, advice on my progress and what I should do next?

• This is my first time (trying) to use the math latex syntax... I'm not sure what not equals is, can someone correct that? Thanks. – meiryo May 1 '11 at 12:39
• You could simply notice that your number is 2*(2n-1), i.e. 2-times odd number and then proceed similarly as in proving that $\sqrt{2}$ is irrational. (Have a look on the exponent of 2 in the prime factorizations of p and q.) en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality – Martin Sleziak May 1 '11 at 12:49
• If you inspect the proof for the irrationality of $\sqrt{2}$ carefully, you should be able to deduce a necessary and sufficient condition in terms of the prime decomposition of a natural number $\alpha$ for it to have a rational square (or any $k$-th, for that matter) root. Can you see what that condition is? This particular exercise will follow as a special case. – Alex B. May 1 '11 at 12:52
• @meiryo: The case $n=1$ is the irrationality of $\sqrt{2}$, which presumably was done in detail. Your rearranging could be useful, but probably the simpler $(4n-2)q^2=p^2$ will do. (To recall a proof you may have seen, easily from this, $p$ is even. But then easily $q$ is even, and so on.) – André Nicolas May 1 '11 at 13:07
• @Martin so I should forget about proving it via contradiction? I'm not quite sure why we have to prove $\sqrt{2}$ is irrational in order to prove the rest is irrational. – meiryo May 1 '11 at 13:15

The number $\sqrt{4n-2}$ is rational iff $4n-2 = a^2$ reduction mod 4 shows that this is impossible.

Here is a proof of the general fact that $\sqrt{k}$ is irrational unless $k$ is a square: Suppose $\frac{u}{v}$ is a solution to $x^2 - k = 0$, then it is an integer $i$ by Gauss lemma, but then $k = i^2$.

• Alternatively, one could apply the rational root theorem to $x^2-k$... – J. M. is a poor mathematician May 1 '11 at 13:29

$4n-2 = (a/b)^2$ so $b$ divides $a$.

But $\operatorname{gcd}(a,b) = 1$ so $b = 1$.

So now $2$ divides $a$ so write $a = 2k$ then by substitution, we get that $2n-1 = 2k^2$

Left side is odd but the right side is even. Contradiction!

• why is b = 1 after g.c.d(a,b)? – meiryo May 1 '11 at 13:26
• @meiryo $(4n -2)$ is an integer therefore $(a/b)$ is also an integer. If $b \neq 1$ with $\gcd(a,b) = 1$ then $(a/b)$ fails to be an integer. – TRUSKI Oct 30 '17 at 9:25

$$\sqrt{4n - 2} = \sqrt{2(2n - 1)} = \sqrt{2}\sqrt{2n - 1}$$ Now for some natural number $p$ and some natural number $q$, let: \begin{align} \sqrt{2} &= \{p/q : p/q \text{ is simplified to lowest terms}\} \\ \implies 2 &= (p/q)^2 \\ &= p^2/q^2 \\ \implies 2q^2 &= p^2 \\ \implies p &= 2r \text{ for some natural number r (simply, r \in \mathbb{N})} \\ \implies 2q^2 &= (2r)^2 \\ &= 4r^2 \\ \implies q^2 &= 2r^2 \end{align} However if both $q^2$ and $p^2$ are divisible by $2$, they are both even; contrary (a contradiction) to our original assumption since we already established that $p/q$ was simplified to lowest terms. \begin{align} \therefore \sqrt{2} &\neq p/q \\ \therefore \sqrt{4n - 2} &\neq p/q \end{align}

Suppose $n$ is a natural number. If $n$ is even, then $n^2 = (2k)^2 = 4k^2$ is divisible by 4. If $n$ is odd, then $n^2 = (2k-1)^2 = 4k^2-4k+1 = 1\ \textrm{mod}\ 4$.

Hence, if $n$ is a natural number, then $n^2$ is either 0 or 1 mod 4.

Thus, if $m$ is a natural number and is 2 or 3 mod 4, then $\sqrt{m}$ is not a natural number, and hence must be irrational, since $\sqrt{m}$ is either irrational or natural.

Hence, as a corollary, since $4n-2$ is 2 mod 4, then $\sqrt{4n-2}$ is irrational.

(This shows that $\sqrt{4n-3}$ is also irrational.)

• What does this add to @quanta's solution? // Re your last parenthesis, try $n=3$ and have a surprise. – Did Dec 14 '11 at 14:36