How to prove that $\sqrt 3$ is an irrational number? [duplicate]

Solution 1:

If you follow through the usual proof for $\sqrt{2}$ substituting $3$ for $2$, it goes through just fine. Let $\sqrt{3}=\frac{p}{q}, p,q $ relatively prime. $3=\frac{p^2}{q^2}$, so $3$ divides $p$ and so on.

Solution 2:

The proof is very similar to the irrationality of square root of two.

Let $\sqrt{3} = \frac a b$, where a and b have no common factors besides $1$

As $3b^2 = a^2$ so $a^2$ is a multiple of $3$, and hence $a$ should be a multiple of $3$. Let $a = 3k$, then $b^2 = 3k^2$, and $b$ must also be a multiple of three. You will arrive at a contradiction to the earlier assumption that $a$ and $b$ have no common factors.

Solution 3:

Another variation on a theme:

If $\sqrt 3 = m/n$, where $n$ is as small as possible, then $$ \frac{m}{n} = \sqrt 3 \frac{\sqrt 3 - 1}{\sqrt 3 - 1} = \frac{3-\sqrt 3}{\sqrt 3 - 1} = \frac{3-m/n}{m/n-1} = \frac{3 n - m}{m-n}$$ and the right side has a smaller denominator, since $m < 2n$ (i.e., $\sqrt 3 < 2$).

This can be used to show (IIRC) that $\sqrt k$ is irrational for any non-square k by multiplying $\sqrt k$ by $\frac{\sqrt k - j}{\sqrt k - j}$ where $j = \lfloor \sqrt k \rfloor$.

Solution 4:

A continued fraction proof of the irrationality of $x = \sqrt{3} - 1$, from which the irrationality of $\sqrt{3}$ follows. (A continued fraction proof of $\sqrt{2}$ can be found here: How can you prove that the square root of two is irrational? )

Notice that $x = \sqrt{3} - 1$ is a root of the equation $x^2 + 2x - 2 = 0$

This can be re-written as

$$x(3+x) = 2+x$$

$$x = \frac{2+x}{3+x} = \cfrac{1}{1 + \cfrac{1}{2 + x}}$$

And thus

$$x = [1,2,1,2,\dots]$$

and so is irrational.