Answer to Question #117109 in Combinatorics | Number Theory for Priya

Question

Use the method of contradiction to prove that √5 is not a rational number.

Answer 1

Solution.

Assume that √5 is a rational number. Suppose there exists a rational number whose square is 5. This number can be given as an unbreakable fraction $\dfrac{m}{n}$ where $m,n -$ are natural numbers.

Then $\sqrt{5}=\dfrac{m}{m}, 5=\dfrac{m^2}{n^2}, m^2=5n^2.$

$m^2$ is divisible by 5, so $m$ is divisible by 5. Then $m^2$ is divisible by 25. If so, then $n^2$ is divisible by 5, and therefore $n$ is divisible by 5. It turns out that the fraction $\dfrac{m}{n}$ is short. We came to a contradiction of the original judgment (that the number is rational). Therefore, the number $\sqrt{5}$ is irrational.

Answer: the number  $\sqrt{5}$ is irrational.