A Quick Little Proof

Let’s prove that \sqrt{3} is irrational. An irrational number is one that cannot be written in the form \frac{a}{b} , where a and b are both integers; in other words there is no repeating pattern in it’s decimal.

First, let’s prove that if n^2 is a multiple of 3 then so is n . Assume that n is not a multiple of 3 . Then n can be written as: n = 3m + \ell where \ell \in \{1,2\} . Then n^2 = 9m^2 + 6m\ell + \ell^2 . We can factor out a 3 , so n^2 = 3(3m^2 + 2m\ell) + \ell^2 . This implies n^2 is not a multiple of 3 since it’s an integer times 3 plus a number that isn’t a multiple of 3 (that’s the \ell part). We have proved the contrapositive which implies our original proposition (P \Rightarrow Q is the same as \neg Q \Rightarrow  \neg P ).

Now let’s move on to the main proof. Assume, by way of contradiction that the square root of 3  is rational. If the square root of 3 is rational then we can write \sqrt{3} = \frac{a}{b} with a and b not sharing any common factors. We can do some math: 3 = \frac{a^2}{b^2} which means 3b^2 = a^2 . But this means a^2 is a multiple of 3 since it’s an integer times 3 . We just proved that if n^2 is a multiple of 3 then so is n . This means that a is a multiple of 3. But if a is a multiple of 3 it can be written as 3m . Let’s plug this in to get 3b^2 = (3m)^2 , simplifying we get that 3b^2 = 9m^2 . Let’s divide through by 3 to get b^2 = 3m^2 . This means that b^2 is a multiple of 3 since it’s 3 times an integer. But again, if b^2 is a multiple of 3 then so is b . We started off by assuming that a and b had no common factors and we just showed that they share a common factor of 3 . We’ve derived a contradiction to our proposition that \sqrt{3} is rational, therefore it must be irrational.

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