I have to prove the square root of 3 is irrational using proof by contradiction.
Say that is not irrational. Then it can be put into the form:
where p and q are integers ( ). For the sake of the argument, let this fraction be in lowest terms.
The we have that
Now, this implies the 3 divides evenly. But we assumed that the fraction is already in lowest terms, so 3 does not divide q. So the prime factorization of only contains a single 3, which is impossible.
Thus etc.
-Dan