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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?

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    For not equal, both \neq and \not = work when enclosed in dollar signs. \not will put a slash through lots of things, so you can have \not \ge for example. To check whether it works on this page, specifically, you can just type it into an answer box.2011-05-01

4 Answers 4

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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 $k = i^2$.

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    Alternatively, one could apply the rational root theorem to $x^2-k$...2011-05-01
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$4n-2 = (a/b)^2$ 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!

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    @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.2017-10-30
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$\sqrt{4n - 2} = \sqrt{2(2n - 1)} = \sqrt{2}\sqrt{2n - 1}$ 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. $\begin{align} \therefore \sqrt{2} &\neq p/q \\ \therefore \sqrt{4n - 2} &\neq p/q \end{align}$

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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.)

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    What does this add to @quanta's solution? // Re your last parenthesis, try $n=3$ and have a surprise.2011-12-14