Question

prove that √6 is an irrational number??? plz post correct explanation!!!!!!

Answer 1

Proof
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.