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Verify by the method of contradiction. `p:sqrt(7)`is irrational. |
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Answer» Let us assume `p:sqrt7` is a rational. `=>sqrt7 = a/b` It means, ,`a` and `b` have no common factor. Also, `a^2/b^2 = 7=> a^2 = 7b^2->(1)` Above equation means, `a^2 ` has a factor of `7`. So, `a` will also have a factor of `7`.We can also write it as, `a = 7c->(2)` where `c` is another real number. From (1) and (2), `49c^2 = 7b^2=> b^2 = 7c^2` Above means, that `b` is also having a factor of `7`. But, earlier we proved `a` and `b` have no common factor. So, our assumption is incorrect that `sqrt7` is a rational number. It means, `sqrt7` is an irrational number. |
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