InterviewSolution
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InterviewSolution
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Prove that `sqrtp+sqrtq` is an irrational, where `p and q` are primes. |
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Answer» Let us assume that` sqrtp + sqrtq` is rational. Then , there exist co-primes a and b such that ` sqrtp + sqrtq = a/b` ` Rightarrow sqrtp = a/b -sqrtq` ` Rightarrow (sqrtp)^(2) = (a/b -sqrtq)^(2)` ` Rightarrow p = a^(2)/b^(2) - (2a)/bsqrtq + q` ` Rightarrow (2a)/(b) sqrtq = a^(2)/b^(2) + q -p` ` Rightarrow sqrtw = (a^(2) +b^(2)q - b^(2)p) xx b/(2a) = (a^(2)b+b^(3)(q-p))/(2a)` Since, a,b,p,q are intergers so `(a^(2)b+b^(3)(q-p))/(2a)` is rational Thus, ` sqrtq` is also rational. But, q being prime `sqrtq` is irrational. Since, a contradicition arises so our assumption is incorrect. Hence, `(sqrtp +sqrtq)` is irrational. |
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