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Question
If p, q are prime positive integers, prove that \[\sqrt{p} + \sqrt{q}\] is an irrational number.
Solution
Let us assume that `sqrtp+sqrtq` is rational. Then, there exist positive co primes a and b such that
`sqrtp +sqrtq=a/b`
`sqrtp=a/b-sqrtq`
`(sqrtp)^2= (a/b-sqrtq)^2`
`p= (a/b)^2-(2asqrtq)/b+q`
`p-q=(a/b)^2-(2asqrtq)/b`
`p-q=(a/b)^2-(2asqrtq)/b`
`(a/b)-(p-q)= (2asqrtq)/b`
`(a^2-b^2(p-q))/b^2 = (2asqrtq)/b`
`(a^2-b^2(p-q))/b^2(b/(2a))=sqrta`
`sqrtq=((a^2-b^2(p-q))/(2ab))`
Here we see that `sqrtq` is a rational number which is a contradiction as we know that `sqrtq` is an irrational number
Hence `sqrtp+sqrtq` is irrational
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