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I was recently working on a question essentially worded in the following way:

Where does a proof of $\sqrt{4}$ being irrational fall apart when we try to apply the same principles used for proving that $\sqrt{2}$ is irrational.

I attempted by making the same (in this case, intuitively correct) assumptions that led to a contradiction in the case of $\sqrt{2}$:

  1. $\sqrt{4}$ is a rational number and can be written as $\dfrac{m}{n}$ where $n\neq0$

  2. $\dfrac{m}{n}$ is in lowest reduced terms; i.e. $m$ and $n$ are co-prime due to definition of rational numbers

Then I took the following steps:

$$m^2 = 4n^2$$

$$m^2 = 2(2n^2)$$

Thus, $m^2$ is even $\implies$ $m$ is even and can be written as $2k$.

$$m^2 = 4k^2 = 4n^2$$

$$k = n$$

Thus, $k$ is a factor of both $m$ and $n$, contradicting the second assumption that I made ($m$ and $n$ are co-prime).

Although I understand intuitively that this is not the case, doesn't this show that $\sqrt{4}$ is irrational?

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    $\begingroup$ What if $n = \pm 1$? $\endgroup$ – anomaly Aug 6 '19 at 17:16
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    $\begingroup$ Technically, you get $k^2=n^2.$ But in any event, $k$ being a factor of both just means $k=1.$ $\endgroup$ – Thomas Andrews Aug 6 '19 at 17:16
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    $\begingroup$ Since you know the square root of $4$ is $2$ you know that $m=2$ and $n=1$. Since $1$ divides every number there is no contradiction. $\endgroup$ – John Douma Aug 6 '19 at 17:25
  • $\begingroup$ Thus $\, m = 2k = 2n\ $ so $\,m/n = 2,\,$ no contradiction (of course). $\endgroup$ – Bill Dubuque Aug 6 '19 at 18:13
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You have proven that $n = k$ and $m = 2k$. In the case that $m$ and $n$ are coprime, set $k = 1$.

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All you can prove with this strategy is that $m=2n$, but the HCF is $1$ if we take $m=2,\,n=1$. By contrast, the analogous treatment of $\sqrt{2}$ shows $m,\,n$ must both be even.

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  • $\begingroup$ Can we do like this...we have $(m,n)= 1 $ and also we have $m=2n$ which means that $(2n,n)= n$ which contradicts the fact that $(m.n) = 1 $ or we can say that if k is even then so is n and so is m. so we have both m and n as even which contradicts (m,n)=1 $\endgroup$ – ReadThyOwnBook Aug 7 '19 at 23:57
  • $\begingroup$ @ReadThyOwnBook There's no contradiction because $n=1$. In the $\sqrt{2}$ case you have $m^2=2n^2\implies m=2k\implies n^2=2k^2\implies 2|n$. $\endgroup$ – J.G. Aug 8 '19 at 5:15
  • $\begingroup$ I don't quite see your point regarding this proof $\endgroup$ – ReadThyOwnBook Aug 8 '19 at 7:51
  • $\begingroup$ @ReadThyOwnBook My point about why you can't prove $\sqrt{4}$ irrational, or my point about how a contradiction shows $\sqrt{2}$ is irrational? $\endgroup$ – J.G. Aug 8 '19 at 10:57
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You proved that $k$ as a factor of both $n$ and $m$, since you can't say that $k=n$. But even if, if you set $k=1$ you do not get a contradiction from it.

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  • $\begingroup$ "Since you can't say that $k=n$". What do you mean by this? OP has shown that $k=\pm n$; the main flaw in their logic is that $k$ cannot be $1$. $\endgroup$ – Jam Aug 6 '19 at 19:06
  • $\begingroup$ Sorry typo. I wanted to say that $k=\pm 1$, still thinking of $m$ and $n$ as co-prime. Edited it so the "non-contradiction" becomes more clear. $\endgroup$ – sevenmaster Aug 6 '19 at 19:14

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