How to prove that $\sqrt 3$ is an irrational number? 
Possible Duplicate:
$a^{1/2}$ is either an integer or an irrational number 

I know how to prove $\sqrt 2$ is an irrational number. Who can tell me that why $\sqrt 3$ is a an irrational number?
 A: 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.
A: 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.
A: 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 $j = \lfloor \sqrt k \rfloor$.
A: 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.
A: Suppose $\sqrt{3} = a/b$ where $a$ and $b$ have no common factor (and note $b\neq 1$). Then $ 3 = a^2/b^2$, but $a^2$ and $b^2$ no common factors to cancel to produce an integer, so we have a contradiction.
A: A well-known variant of the usual proof may be clearer. If $\sqrt{3}=\frac{a}{b}$, then $a^2=3b^2$. Recall the Fundamental Theorem of Arithmetic and consider the exponent of $3$ in the prime factorization of both sides. On the left you have an even exponent. On the right you have an odd exponent, contradiction. This approach goes one to prove that $\sqrt m$ is rational iff $m$ is a square.
A: Alternatively, you can use the rational root test on the polynomial equation $x^2-3=0$ (whose solutions are $\pm \sqrt{3}$).  If $\frac{a}{b}$ is a solution to the equation (with $a,b\in \mathbb{Z}$ and $b\not=0$), then $b \vert 1$ and $a \vert 3$, hence $\frac{a}{b}\in \{\pm 1, \pm 3\}$.  However, it is straightforward to check that none of $\pm 1, \pm 3$ are solutions to $x^2-3=0$.  Therefore there are no such rational solutions and $\sqrt{3}$ is irrational.
In fact, in the above argument, if we replace 3 with an arbitrary prime $p\in \mathbb{N}$ and 2 with an arbitrary $m\in \mathbb{N}$, $m\geq 2$, the same argument shows that $\sqrt[m]{p}$ is irrational.
