1.

Verify by the method of contradiction. `p:sqrt(7)`is irrational.

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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