Proof that Square Root of 6 is Irrational


We have learned that \sqrt{3} is irrational and thousands of sites on the internet will show you the proof that \sqrt{2} is irrational. In this post, we show that \sqrt{6} is irrational. This is a prerequisite for the next post which will show that \sqrt{2} + \sqrt{3} is irrational.

You will probably argue that since \sqrt{2} is irrational and \sqrt{3} is irrational, then their product should be irrational. This is not always the case. The product of two irrational numbers is not always irrational.  For instance, \sqrt{2} is irrational but (\sqrt{2})(\sqrt{2}) = 2 is rational. And now, we proceed the the theorem and its proof.


\sqrt{6} is irrational.


To prove the theorem above, we use proof by contradiction.

Assuming that \sqrt{6} is rational. Then, \sqrt{6} = \frac{a}{b} where  a and b are integers and \frac{a}{b} is in lowest terms. This means that a and b cannot be both even (Why?) (*)

Squaring both sides, we have \displaystyle 6 = \frac{a^2}{b^2}.

Multiplying both sides by b^2, we have 6b^2 = a^2. It follows that a^2 is even and  a is even. (**)

If a is even, then it can be expressed as 2c where c is an integer. Substituting to the equation above, we have 6b^2 = (2c)^2 which simplifies to

6b^2 = 4c^2.

Dividing both sides by 2 gives 3b^2 = 2c^2. This implies that 3b^2 is even which means that b is even (Why?). (***)

From *, we assumed that a and b cannot be both even. But from ** and *** a and b are even. A contradiction!

Therefore, \sqrt{6} is irrational.

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