Ryiaz claims it is impossible to prove that a rational number multiplied by an irrational number gives an irrational product. How can you prove this?(1 point)

Question

Responses

A) by using a proof by contradiction to show multiplication is a closed operation for rational numbers

B) by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer

C) He’s correct; it is impossible to prove in any way.

D) by choosing random numbers to plug into multiplication problems until he’s convinced

Answer 1

The correct response is:

B) by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.

To prove this, we can use the following reasoning:

Let \( r \) be a rational number, which can be expressed as \( r = \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \). Let \( x \) be an irrational number. We want to show that the product \( rx \) is irrational.

Assume for the sake of contradiction that \( rx \) is rational. Then we can express it in the form \( rx = \frac{m}{n} \) where \( m \) and \( n \) are integers and \( n \neq 0 \).

Now, if we rearrange this equation, we have: \[ x = \frac{rx}{r} = \frac{\frac{m}{n}}{\frac{a}{b}} = \frac{mb}{na}, \] where \( na \neq 0 \) since both \( n \) and \( a \) are integers with \( a \neq 0 \).

Since \( \frac{mb}{na} \) is a ratio of integers, \( x \) would be a rational number, contradicting our original assumption that \( x \) is irrational. Therefore, our assumption that \( rx \) is rational must be false, which implies that the product of a rational number and an irrational number is indeed irrational.