Yes the product of any rational number with an irrational number is always irrational. For example, 2/3 times pi is going to be an irrational result. In this case simply (2/3)pi or (2pi)/3
So the template is
(rational)*(irrational) = irrational
-------------------------------------------------------------------------------------
Here's proof of the claim. Let's make p, q, r, s to be integers. Let's also make m be some irrational number
For now let's assume that multiplying a rational number with an irrational number leads to a rational result. The aim is to use contradiction to prove the opposite.
p/q is rational by definition (make q nonzero)
so is r/s
m is irrational
So the claim so far is
(p/q)*m = r/s
Now multiply both sides by the reciprocal of p/q. That would be q/p
(p/q)*m = r/s
(q/p)*(p/q)*m = (q/p)*(r/s)
m = (q*r)/(p*s)
The right side of the last equation is a rational number since q*r and p*s are integers (p and s are nonzero)
But m was originally irrational. So this is where the contradiction kicks in. This invalidates the initial claim "the product of a rational and irrational number is rational" proving the claim made by your textbook.
So this completes the proof.
