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The more known proof of square root of $2$ is by contradiction when we assume it can be expressed as an irreducible fraction and later finding that it isn't irreducible, but... if we assume the same conditions for example: square root of $9$ we find that it isn't irreducible either. So, what's the trouble here?

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marked as duplicate by Peter Taylor, heropup, 6005, Qwerty, Henrik Aug 16 '16 at 21:47

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    $\begingroup$ @ZackNi I don't really see how your comment is relevant to this question. $\endgroup$ – Noah Schweber Aug 16 '16 at 0:45
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    $\begingroup$ The proof of the irrationality of $\surd 2$ hinges on the prime factorisation of $2$ having odd indices (aka $2^{\bf 1}$). The prime factorisation of integers such as $4$ or $9$ do not have odd indices; they are $2^{\bf 2}, 3^{\bf 2}$ respectively. $\endgroup$ – Graham Kemp Aug 16 '16 at 0:49
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    $\begingroup$ When you have $9a^2 = b^2$ and you conclude $9|b^2$ you don't have to conclude 9|b because you can have just 3|b. Which will not lead to us requiring 9 divides a. There is no contradiction. $\endgroup$ – fleablood Aug 16 '16 at 1:00
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    $\begingroup$ I like to explain it in terms of operations. Square root is by definition the inverse operation of squaring, i.e. multiplying a number by itself. We know that the result of multiplication of two natural numbers is always a natural number too. Thus, we necessarily have a set of 'perfect squares', i.e. natural numbers which are the results of squaring some other natural number. So, when we take the square root of a perfect square we have to obtain a natural number too. This explanation might not be rigorous, but it's how I understand all of this myself $\endgroup$ – Yuriy S Aug 16 '16 at 2:21
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    $\begingroup$ Have you tried to actually follow the same proof in the case of $9$? (I'm not being patronizing here, just wondering.) $\endgroup$ – Asaf Karagila Aug 16 '16 at 5:40
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Let's look at the proof that $\sqrt{2}$ is irrational:

To show $\sqrt{2}$ is irrational, we argue by contradiction. Suppose $\sqrt{2}={p\over q}$, where $p, q$ are natural numbers. Without loss of generality, ${p\over q}$ is in lowest terms. So ${p^2\over q^2}=2$, and so $p^2=2q^2$. $\color{red}{\mbox{This means that}}$ $2$ divides $p$, that is, $p=2k$. So $2q^2=4k^2$, and $q^2=2k^2$. $\color{red}{\mbox{Again, this means}}$ that $2$ divides $q$. But then $p$ and $q$ have a common factor, contradiction.


Where this breaks down for $\sqrt{4}$: Look at the red phrases. They depend on the following fact: $$\mbox{If $2$ divides $ab$, then either $2$ divides $a$ or $2$ divides $b$}.$$ This fact requires proof, and relies on the fact that $2$ is prime. $4$, by contrast, is not prime, and indeed the fact fails for $4$: $4$ divides $2\cdot 2$, but $4$ does not divide $2$.

How it generalizes: It's a good exercise to show that the usual argument does work for any number $n$ such that some prime $p$ divides $n$ an odd number of times - that is, $p^{2i+1}$ divides $n$ but $p^{2i+2}$ doesn't, for some $i$. For example, the following numbers all have this property:

  • $12$ ($p=3, i=0$)

  • $27$ ($p=3, i=1$)

  • $24$ ($p=2, i=1$ or $p=3, i=0$)

etc. Any such number has an irrational square root. By contrast, if every prime dividing $n$ divides $n$ an even number of times, then $\sqrt{n}$ is rational (exercise!). So this is a complete characterization.

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  • $\begingroup$ Euclid's lemma says that if $p$ is prime and $a,b\in\mathbb Z$ and $p\mid ab$, then either $p\mid a$ or $p\mid b$, or more generally if $n,a,b\in\mathbb Z$ and $\gcd(n,a)=1$ and $n\mid ab$, then $n\mid b$. $\endgroup$ – user236182 Aug 16 '16 at 16:29
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The standard proof that $\sqrt{2}$ is irrational is of course as follows: Assume for contradiction there exist $a, b$ positive whole numbers that share no factors (besides $1$) and $a / b = \sqrt{2}$. Then $a^2 = 2 b^{2}$. Now the observation here is that $2 b^{2}$ is divisible by $2$, a prime number, so $a$ must also be divisible by $2$.

Now, if instead we considered $9$, then we'd be looking at $a^2 = 9b^2$. We cannot conclude from here that $a$ is divisible by $9$, so we are dead in our tracks.

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$$\sqrt{9}=\frac{a}{b}$$

where $\gcd(a,b)=1$. $$\sqrt{9}b=a$$ Squaring both sides, $$9b^2=a^2$$

We know that $a$ must be a multiple of $3$ (notice I am saying $a$ is a multiple of $3$ rather than $9$ and compare it with the proof for irrationality of $\sqrt{2}$.)

So $a=3k$ and hence $b=k$. $k$ takes value $1$ in this case.

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  • $\begingroup$ "$k$ takes value $1$ in this case": of course, that is one value that it can take, but it doesn't necessarily do so. (Also, of course we could as well have $b = -k$.) $\endgroup$ – LSpice Aug 16 '16 at 20:03
  • $\begingroup$ Let me impose one more condition, $a$,$b>0$. $\endgroup$ – Siong Thye Goh Aug 16 '16 at 20:57
  • $\begingroup$ Ah, I read too quickly, and missed the (crucial) bit that $\gcd(a, b) = 1$. Of course then (with your added positivity condition) you are right that $k = 1$. $\endgroup$ – LSpice Aug 16 '16 at 21:03
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See this proof that if $n$ is not a perfect square then $\sqrt{n}$ is irrational:

Follow-up Question: Proof of Irrationality of $\sqrt{3}$

The proof starts by saying that if $n$ is not a perfect square then there is a $k$ such that $k^2 < n < (k+1)^2$. The proof breaks down if $k^2 = n$.

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Because the proof relies on that the root isn’t an integer

$\gcd(p,q)=1, \frac{p}{q}=sqrt(9)\implies p^2=9*q^2 \implies 9|p^2$. Normally we would say now that this means $9|p$ as we would with $2$, but that neglects the case where $q=1$, i.e. $9=p^2$, but in this case, that isn’t true.

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    $\begingroup$ This isn't really true - nowhere in the standard proof do we assume that $q\not=1$. Instead, what's going on is that we use the fact that $2$ is prime - which means that if $2$ divides a product $a\cdot b$, then $2$ divides one of the factors. In particular, if $2$ divides $a^2$ then $2$ divides $a$. This fails for nonprimes in general - e.g. $9$ divides $6^2$ but does not divide $6$. $\endgroup$ – Noah Schweber Aug 16 '16 at 0:52
  • $\begingroup$ Right, but showing that 2 is prime usually is done by showing that its irreducible, i.e. doesn't have any integer factors (aside from 1 and itself), which would imply a that its square root isn't an integer. $\endgroup$ – rikhavshah Aug 16 '16 at 0:56

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