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Everyone with any basic knowledge of number theory knows the classic proof of the irrationality of $\sqrt{2}$. Curious about generalizations using elementary methods, I looked up the irrationality of $\sqrt{3}$, and found the following:

Say $ \sqrt{3} $ is rational. Then $\sqrt{3}$ can be represented as $\frac{a}{b}$, where a and b have no common factors.

So $3 = \frac{a^2}{b^2}$ and $3b^2 = a^2$. Now $a^2$ must be divisible by $3$, but then so must $a $ (fundamental theorem of arithmetic). So we have $3b^2 = (3k)^2$ and $3b^2 = 9k^2$ or even $b^2 = 3k^2 $ and now we have a contradiction.

Such a proof follows the same basic logic as the proof for $\sqrt{2}$, except for using the fundamental theorem of arithmetic to replace and generalize the trivial fact that $n$ is even if $n^2$ is even. However, when I apply this proof format to $\sqrt{4} $ (which is clearly an integer and thus rational) I get the following:

Say $ \sqrt{4} $ is rational. Then $\sqrt{4}$ can be represented as $\frac{a}{b}$, where a and b have no common factors.

So $4 = \frac{a^2}{b^2}$ and $4b^2 = a^2$. Now $a^2$ must be divisible by $4$, but then so must $a $ (fundamental theorem of arithmetic). So we have $4b^2 = (4k)^2$ and $4b^2 = 16k^2$ or even $b^2 = 4k^2 $, which implies that $b=4n$ by the fundamental theorem. Now we have a contradiction (since can note that both $a$ and $b$ are divisible by $4$ and we assumed they were coprime)

This proof is clearly false, yet I fail to see where it differs. Where does it do so?

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    $\begingroup$ Whereas $2$ and $3$ are primes, $4$ is not a prime. So the claim that $4 \mid a^2$ (i.e. $4$ divides $a^2$) does not imply $4 \mid a$. As a counterexample, $4$ divides $a^2 = 36$, but $4$ does not divide $a = 6$. $\endgroup$ Commented Apr 18, 2016 at 21:04
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    $\begingroup$ Perhaps more obviously (if we were to take the obvious approach and set $a=2$, $b=1$), $4$ divides $2^2$ but does not divide $2$. $\endgroup$ Commented Apr 18, 2016 at 21:07
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    $\begingroup$ The essential property here that $2$ and $3$ possess but $4$ does not is squarefreeness: ihey are composed of distinct prime factors. If $d$ is divisible by any prime squared then it is false that "$d$ divides $a^2$" implies "$d$ divides $a$". $\endgroup$ Commented Apr 18, 2016 at 21:51
  • $\begingroup$ Note regarding above comments: primality and squarefreeness are not sufficient and necessary conditions; not being a perfect square is. The argument that the square root of an integer is irrational applies to any non-perfect square, whether composite or containing a nontrivial square factor. $\endgroup$ Commented Apr 18, 2016 at 21:57
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    $\begingroup$ There is a foolproof method of finding at what step a false proof like this fails, which is indicated in Fimpellizieri's comment: you substitute what you know to be a counterexample to the final result, and see at what point you've written down a false statement about your counterexample. You can always do this; it requires essentially no cleverness. $\endgroup$ Commented Apr 19, 2016 at 19:11

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Just because $a^2$ is divisible by $4$, that doesn't mean $a$ is.

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Your error is stating that if $a^2$ is divisible by 4 so must $a$ be. The fundimental theorem states if a prime $p $ divides $ab $ then $p $ must divide $a $ or $p$ must divide $b $. That is true because $p $ is indivisable.

But if $p $ is composite it doesn't hold. $p$ could equal $jk $ and $j$ could divide $a $ and $k $ divide $b $. Example: 3 divides 4 times 9 so 3 either divides 4 or 3 divides 9 because 3 is prime. But 6 divides 4 times 9 but 6 neither divides 4 nor 9 but instead 3 divides 9 while 2 divides 4 so 6=2 times 3 divide 4 times 9.

So 4 divides $a^2$ means $2*2$ divides $a*a$ so 2 divides $a $ is all you can conclude with certainty. ... because 4 is not prime.

Actually because 4 is not square free. All of its prime factors must divide into $a$ but the square powers can be distributed among (and are) distributed among the square powers of $a$.

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Now If you want to prove that square root of a non perfect square is irrational.

Then you can do this what I wrote below as a proof.

Let $p$ is a non perfect square. We are required to prove $\sqrt{p}$ is irrational. Let $\sqrt{p}$

is rational and of the form $\frac{a}{b}$ where $a\;\;and\;\;b$ are coprime and $b\neq{0}$. Now, we have

$$\sqrt{p}=\frac{a}{b}$$ By squaring both sides we have $$p=\frac{a^2}{b^2}$$ $$\implies{a^2}=p{b^2}$$ We know that $a\;and\;b$ are coprime to each other and $g.c.d(a,b)=1$. So we have $$ax+by=1$$ for some non zero integers $x$ and $y$.Then multiply $\sqrt{p}$ on both sides.Then we get $$\sqrt{p}ax+\sqrt{p}by=\sqrt{p}$$See that $a=\sqrt{p}b$. So we have $$pbx+ay=\sqrt{p}$$ Now see that the L.H.S belongs to integer set whereas the R.H.S belongs to $\mathbb{R}\backslash \mathbb{Z}$.

Hence it is a contradiction.So $\sqrt{p}$ is irrational. May be it will help you.

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  • $\begingroup$ It is a nice proof but it really doesn't do anything to answer the OP. The question is not "how cause I prove this square root is irrational?" but "I know this proof is wrong somehow, but where is the error?" $\endgroup$ Commented Apr 19, 2016 at 20:02
  • $\begingroup$ That's because it doesn't mean if 4|a^2,then 4|a.A counter example is a=36.4|36 but 4 doesn't divide 6. $\endgroup$ Commented Apr 20, 2016 at 10:04
  • $\begingroup$ Yes, that was given in the other answers and comments. I just find it odd that you submitted an answer to a different question than the one asked here. $\endgroup$ Commented Apr 20, 2016 at 10:25

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