I have to prove the square root of 3 is irrational using proof by contradiction.

2. Originally Posted by natewalker205
I have to prove the square root of 3 is irrational using proof by contradiction.
Say $p/q = \sqrt{3}$ where $p,q$ are reduced. Then $p^2 = 3q^2$ since RHS is divisible by 3 it means LHS is divisible by 3 but then $p^2$ makes LHS divisible by 3 and so $q$ needs to be divisible by 3. Thus, it is impossible, by infinite descent.

3. Originally Posted by natewalker205
I have to prove the square root of 3 is irrational using proof by contradiction.
Say that $\sqrt{3}$ is not irrational. Then it can be put into the form:
$\sqrt{3} = \frac{p}{q}$
where p and q are integers ( $q \neq 0$). For the sake of the argument, let this fraction be in lowest terms.

The we have that
$p^2 = 3q^2$

Now, this implies the 3 divides $p^2$ evenly. But we assumed that the fraction $\frac{p}{q}$ is already in lowest terms, so 3 does not divide q. So the prime factorization of $p^2$ only contains a single 3, which is impossible.

Thus etc.

-Dan