Is the product of an irrational number with a rational number always an irrational number? ?

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. 

The full question isn’t shown. part of it is cut off

Dis the correct  answer