Question

yiaz 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)
Responses

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

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

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

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

Answer 1

The correct response is:

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

To elaborate on this, we can prove that the product of a non-zero rational number and an irrational number is indeed irrational.

Let's assume:
1. \( r \) is a non-zero rational number, which can be expressed as \( r = \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \).
2. \( s \) is an irrational number.

We want to show that the product \( r \times s = \frac{a}{b} \times s \) is irrational.

**Proof by Contradiction:**
Suppose \( r \times s \) is rational. This means we can express it as a fraction \( \frac{m}{n} \) where \( m \) and \( n \) are integers and \( n \neq 0 \).

Then we can rearrange:

\[ s = \frac{r \times s}{r} = \frac{\frac{m}{n}}{r} = \frac{m}{n} \times \frac{b}{a} = \frac{mb}{na} \]

Since \( \frac{mb}{na} \) is a ratio of integers (given that \( mb \) and \( na \) are integers), this means that \( s \) is rational.

However, this contradicts our original assumption that \( s \) is irrational. Therefore, our assumption that \( r \times s \) is rational must be false, meaning that \( r \times s \) must be irrational.

Hence, we have shown mathematically that a non-zero rational number multiplied by an irrational number results in an irrational number. This proves the claim.