Question

find the answer to start fraction square root of 196 end square root over seven end fraction times square root of 108 end square root and select the correct option below. Also, is the product of a nonzero rational number and an irrational number classified as rational or irrational?

Answer 1

1. From your description, I can infer that the multiplication is:

`( √(196) )/(7) * √(108)`

The first thing we are going to do is simplify the radicands 196 ans 108 (picture 1):

`196=2^2*7^2` and
`108=2^2*3^3`
Knowing this, we can rewrite our radicals as follows:

`( √(196) )/(7) * √(108)= ( √(2^2*7^2) )/(7) * √(2^2*3^3)`

Remember that
`\sqrt[n]{x^n} =x`; in other words if the radicand is raised to the same power as the index of the radical, we can take the radicand out. Since 2 and 7 are raised to the power 2 and the index of the radical is also 2 (square root), we can take out 2 and 7:

`( √(2^2*7^2) )/(7) * √(2^2*3^3)= (2*7)/(7) *2 √(3^3)`

Look! we have the same numerator and denominator in our fraction, so we can cancel them both:

`(2*7)/(7) *2 √(3^3)=2*2 √(3^3) =4 √(3^3)`

Notice that we can write
`3^3` as
`3^2*3`, so we can rewrite our expression one last time:

`4 √(3^3) =4 √(3^2*3) =4*3 √(3) =12 √(3)`

We can conclude that the correct option is:
`12 √(3)`

2. The product of a nonzero rational number and an irrational number is always an irrational number.

Proof by contradiction:
Lets assume that the product of an irrational number and a rational non-zero number is always rational.
Let
`x` be and irrational number and let
`(a)/(b)` and
`(c)/(d)` be two rational numbers with
`a`,
`b`,
`c`, and
`d` are non-zero integers.

`x* (a)/(b) = (c)/(d)`

`x= (c)/(d) * (b)/(a)`

`x=(cb)/(da)`
Since integers are closed under multiplication,
`(cb)/(da)` is a rational number. Since
`x` is an irrational number and
`x=(cb)/(da)`, we have a logical contradiction, so we can conclude that the product of an irrational number and a rational non-zero number is always an irrational number.