Show that √3+√5 is an irrational number

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Firstly assumed that √3+√5 is a rational no. √3+√5=p/q (where p&q are any positive integer &co-prime &qis not equal to 0)p/q -√3=√5 (squaring both side)(p/q-√3)^2=(√5)^2p^2/q^2-2p√3/q+3= 5(Put 3 on RHS then ,we get)p^2/q^2-2p√3/q=2{Put (2p√3/q) on RHS then ,we get}(p^2/q^2)-2=-2p√3/q[(p^2)-(2q^2)]/q^2=2p√3/q(left only √3 on RHS,then solve ,we get)[p^2 – 2q^2]/2pq = √3. (We know that √3 are irrational no. )but {[p^2 – 2q^2]/2pq} is rationalThis implies that √3 is rational no.This implies that our contradict are wrong This implies that, √3 + √5 is irrational no.