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rove that V5 is irrational |
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Answer» Let us assume, to the contrary, that √5 is rational. i.e, we can find integer a and b such that √5= a/b.suppose a and b have a common factor other than 1,then we can divide by the common factor and assume that a and b are coprime. So, b√5=a.Squaring on both side, and rearranging, we get 5b^2=a^2. Therefore , a^2 is divisible by 5, and by Theorem 1.3, it follows that a is also divisible by 3.So, we can write a =3c for some integer c. Substituting for a, we get 5b^2=25c^2, i.e , b^2 =3c^2.This means that b^2 is divisible by 5 and so b is also divisible by 5. Therefore, a and b have at least 5 as a common factor. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that √5 is rational. So, we conclude that √5 is irrational. |
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