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.