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?

  • 4
    $\begingroup$ What if $n = \pm 1$? $\endgroup$ – anomaly Aug 6 '19 at 17:16
  • 6
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

You have proven that $n = k$ and $m = 2k$. In the case that $m$ and $n$ are coprime, set $k = 1$.


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.

  • $\begingroup$ 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

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.

  • $\begingroup$ "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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.