# Proof square root of 4 is not irrational.

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?

• What if $n = \pm 1$? Aug 6, 2019 at 17:16
• Technically, you get $k^2=n^2.$ But in any event, $k$ being a factor of both just means $k=1.$ Aug 6, 2019 at 17:16
• 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. Aug 6, 2019 at 17:25
• Thus $\, m = 2k = 2n\$ so $\,m/n = 2,\,$ no contradiction (of course). Aug 6, 2019 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.
• 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 Aug 7, 2019 at 23:57
• @ReadThyOwnBook There's no contradiction because $n=1$. In the $\sqrt{2}$ case you have $m^2=2n^2\implies m=2k\implies n^2=2k^2\implies 2|n$.
• @ReadThyOwnBook My point about why you can't prove $\sqrt{4}$ irrational, or my point about how a contradiction shows $\sqrt{2}$ is irrational?