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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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    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.2011-05-01
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    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.) http://en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality2011-05-01
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    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.2011-05-01
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    @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.)2011-05-01
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    @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.2011-05-01
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    Actually I understand where you're coming from Martin... you should put yours as an answer!2011-05-01
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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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    why is b = 1 after g.c.d(a,b)?2011-05-01
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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