Let’s prove that is irrational. An irrational number is one that cannot be written in the form , where and are both integers; in other words there is no repeating pattern in it’s decimal.

First, let’s prove that if is a multiple of then so is . Assume that is not a multiple of . Then can be written as: where . Then . We can factor out a , so . This implies is not a multiple of since it’s an integer times *plus* a number that isn’t a multiple of (that’s the part). We have proved the contrapositive which implies our original proposition ( is the same as ).

Now let’s move on to the main proof. Assume, by way of contradiction that the square root of *is* rational. If the square root of is rational then we can write with and not sharing any common factors. We can do some math: which means . But this means is a multiple of since it’s an integer times . We just proved that if is a multiple of then so is . This means that is a multiple of 3. But if is a multiple of it can be written as . Let’s plug this in to get , simplifying we get that . Let’s divide through by to get . This means that is a multiple of since it’s times an integer. But again, if is a multiple of then so is . We started off by assuming that and had no common factors and we just showed that they share a common factor of . We’ve derived a contradiction to our proposition that is rational, therefore it must be irrational.