Question

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)
Responses

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

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

Answer 1

To prove that a rational number multiplied by an irrational number gives an irrational product, the most effective approach is:

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

Here's a brief outline of that proof:

1. **Definitions**:
- A rational number can be expressed as \( r = \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \).
- An irrational number cannot be expressed as a fraction of integers.

2. **Proof by contradiction**:
- Assume \( r \) is a rational number and \( x \) is an irrational number. Assume for contradiction that the product \( rx \) is rational.
- Then, we could express \( rx \) as \( \frac{m}{n} \) where \( m \) and \( n \) are integers and \( n \neq 0 \).

3. **Solving for x**:
- Rearranging gives us \( x = \frac{rx}{r} \). Since \( r \neq 0 \), we have \( x = \frac{m}{n \cdot r} \).
- But this expression shows that \( x \) can be expressed as a ratio of integers (hence, rational), which contradicts our initial assumption that \( x \) is irrational.

4. **Conclusion**:
- Therefore, our assumption that the product \( rx \) is rational must be false, meaning that \( rx \) is irrational.

This proof demonstrates directly that the product of a rational number and an irrational number is indeed irrational.