Show3√2-5√7 is irrational

-

We can solve the by the method of contradiction. Let us assume that 3✓2-5✓7 is a rational number. A rational number is always in the form of p/q where p and q are intergers, p and q are co prime and q not equal to zero.  Therefore, 3✓2-5✓7 = p/q Squaring both sides (3✓2-5✓7)^2 =(p/q)^2 193 - 30✓14= p^2/q^2 -30✓14= p^2/q^2 -193 -30✓14=p^2-193q^2/q^2 ✓14=p^2-193q^2/-30q^2 Notice that  LHS is  irrational(14 is not a perfect square) but RHS is rational. Hence this situation is not possible. Hence proved.