If a, b are non-zero rational numbers, and c is a non-zero irrational number, what can we say about (a/b)^2

We can say that (a/b)^2 is a rational number

Proof:

Rational means the number is a ratio of two whole numbers

If 'a' is rational, then a = p/q for some whole numbers p,q (eg: p = 2, q = 3 so p/q = 2/3)

The same applies for b, which I'll let b = r/s, with r & s being whole numbers.

To avoid division by zero, I'm going to make p,q,r,s nonzero which leads to 'a' and 'b' being nonzero.

Now let's divide 'a' over b to get

a/b = (p/q) divided by (r/s)

a/b = (p/q) * (s/r)   flip the second fraction; multiply

a/b = (ps)/(qr)

The new numerator is p*s which is a whole number (because the product of two whole numbers is also a whole number). The same goes for q*r in the denominator. So a/b is rational if both 'a' and 'b' are rational together.

We can extend this to (a/b)^2 as well

(a/b)^2 = [ (ps)/(qr) ] ^2

(a/b)^2 = [ (ps)^2 ]/[ (qr)^2 ]

(a/b)^2 = (p^2*s^2)/(q^2*r^2)

p is a whole number, and so is p^2. Same for s and s^2. Overall, the final numerator p^2*s^2 is also a whole number. Similarly, q^2*r^2 is a whole number.

So we have yet another ratio of two whole numbers proving that (a/b)^2 is a rational number.

step-by-step explanation: