what is the proof that the square root of 6 is irrational?

1 ANSWERS

1. Guest1466
   

   It can be proved by contradiction. Suppose sqrt 6 is a rational number, then it can be stated as the ratio of m/n, with m and n bein integers with no common factors greater than 1. Square the both sides and m^2/n^2 = 6 and thus m^2 = 6 n^2. This means m^2 is an even number which results in m to be an even number: m = 2k
   
   m^2 = 4k^2 = 6 n^2 => 3 n^2 = 2k^2. As a result n^2 is an even number and from there we conclude that n has to be an even number. Thus m/n with m and n both bein even numbers has to have a common factor of 2 which contradicts the original assumption and proves that sqrt 6 cannot be rational.

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