Prove by contradiction that 3 root2 - 4 root5 is irrational, given that root2 is irrational.?

Answer 1

by contradiction: I suppose 3sqrt(2)-4sqrt(5)=r (where r is rational)

thus 4sqrt(5)=3sqrt(2)-r
we squarre: 80=18-6r*sqrt(2)+r^2
so sqrt(2)=(r^2-62)/(6r)

impossible because sqrt(2) is irrationnal !

Answer 2

Let's prove a general result: If a and b are rational and r1 and r2 are non negative rationals, then

S = sqr(r1) + b sqrt(r2) is rational if, and only if, so are sqrt(r1) and sqrt(r2)

Proof:
If sqrt(r1) and sqrt(r2) are rational, then it's immediate that S is rational.

For the converse, suppose S is rational and admit that at least one of the numbers sqrt(r1) and sqrt(r2), say sqrt(r1), without loss of generality, is irrational. We have that

b sqrt(r2) = S - a sqrt(r1). Squaring,

b^2 r2 = S^2 - 2aS sqrt(r1) + a^2 r1 (1)

Since a, b, r1 and S are rational, it follows so are b^2 r2 and S^2 + a^2 r1. And since sqrt(r1) is irrational, it follows so is 2aS sqrt(r1) and, therefore S^2 - 2aS sqrt(r1) + a^2 r1. Hence, (1) says a rational number equals an irrational one - a contradiction that proves that if S is rational then so are sqrt(r1) and sqrt(r2).

Back to your case, it's a particular case of what we've proved, with a = 3, r1 = 2, b = -4 and r2 = 5. Since 2 and 5 are not perfect squares, their square roots are irrational. So, it follows from what we proved that 3 root2 - 4 root5 is irrational.