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I have a proof for it. I just want to make sure it's correct.

Assume, for the sake of contradiction, that $\sqrt{2}$ is rational.

Then, $\sqrt{2} = \frac{a}{b}$ (1) , where a and b are integers and gcd(a,b)=1.

Note that $\sqrt{2}$ is non-integer. Thus, b does not divide a

Squaring both sides of equation (1),

2 = $\frac{a^2}{b^2}$. This means that $b^2|a^2$

Let p be a prime where p|$a^2$. Then $p|a$

Now assume that p| $b^2$. Then p|b. However, this is a contradiction because $gcd(a,b)= 1$

Therefore, $gcd(a^2, b^2)=1$. However, this contradicts the earlier conclusion that $b^2|a^2$

Hence, $\sqrt{2}$ cannot be rational.

Therefore, $\sqrt{2}$ irrational.

Is there something wrong with this proof?

Mohamed Lotfi
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    You should make further clear why "$b^2$ divides $a^2$" is (or leads to) a contradiction. For formatting place the math expressions between dollarsigns '$'. – drhab Nov 10 '19 at 13:13
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    It is not that it is wrong wrong, but that it is incomplete. Why is it impossible for $b^2$ to divide $a^2$ while $b$ doesn't divide $a$? – conditionalMethod Nov 10 '19 at 13:15
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    "$b^2\mid a^2$ is a contradiction". You did not show that this is impossible. – Peter Nov 10 '19 at 13:15
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    just mention that the greatest common divisor of $a$ and $b$ is $1$ and then you're done. – thesmallprint Nov 10 '19 at 13:18
  • @Peter Is it complete now? – Mohamed Lotfi Nov 10 '19 at 13:30
  • @MohamedLotfi Almost. Just mention also the case $b=1$ in which we have no prime factor of $b$. If we have $b>1$, your argument applies since then $b$ must have at least one prime factor and this must also be a prime factor of $a^2$ and hence of $a$. – Peter Nov 10 '19 at 13:54
  • @Peter The argument is not correct because it does not correctly prove there is a prime $,p\mid a,b,,$ cf. "Now assume $p\mid b^2\ldots$" (my emph.) – Bill Dubuque Nov 10 '19 at 20:00
  • @BillDubuque We have established $b^2\mid a^2$ and if $b>1$ holds , then for some prime $p$ , we must have $p\mid b$ , for this prime we also have $p\mid a^2$ and hence $p\mid a$. So where is the flaw in this argument ? – Peter Nov 11 '19 at 20:48
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    @Peter But that is not what is written above. Rather it says: "let p be a prime where p|a^2 ... Now assume p|b^2 ..." Read it carefully. – Bill Dubuque Nov 11 '19 at 21:01
  • @BillDubuque OK, that's a point. Thank you for pointing it out. – Peter Nov 11 '19 at 21:02

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