I have to prove the square root of 3 is irrational using proof by contradiction.
Say $\displaystyle p/q = \sqrt{3}$ where $\displaystyle p,q$ are reduced. Then $\displaystyle p^2 = 3q^2$ since RHS is divisible by 3 it means LHS is divisible by 3 but then $\displaystyle p^2$ makes LHS divisible by 3 and so $\displaystyle q$ needs to be divisible by 3. Thus, it is impossible, by infinite descent.
Say that $\displaystyle \sqrt{3}$ is not irrational. Then it can be put into the form:
$\displaystyle \sqrt{3} = \frac{p}{q}$
where p and q are integers ($\displaystyle q \neq 0$). For the sake of the argument, let this fraction be in lowest terms.
The we have that
$\displaystyle p^2 = 3q^2$
Now, this implies the 3 divides $\displaystyle p^2$ evenly. But we assumed that the fraction $\displaystyle \frac{p}{q}$ is already in lowest terms, so 3 does not divide q. So the prime factorization of $\displaystyle p^2$ only contains a single 3, which is impossible.
Thus etc.
-Dan