Question #117109
Use the method of contradiction to prove that √5 is not a rational number.
1
Expert's answer
2020-05-25T21:11:29-0400

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 mn\dfrac{m}{n} where m,nm,n - are natural numbers.

Then 5=mm,5=m2n2,m2=5n2.\sqrt{5}=\dfrac{m}{m}, 5=\dfrac{m^2}{n^2}, m^2=5n^2.

m2m^2 is divisible by 5, so mm is divisible by 5. Then m2m^2 is divisible by 25. If so, then n2n^2 is divisible by 5, and therefore nn is divisible by 5. It turns out that the fraction mn\dfrac{m}{n} is short. We came to a contradiction of the original judgment (that the number is rational). Therefore, the number 5\sqrt{5} is irrational.

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


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