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)

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

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

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

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

Answer 1

To prove that a rational number multiplied by an irrational number gives an irrational product, you can use the following approach:

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

Here is how this proof can be structured:

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

2. **Assume a rational number \( r \) and an irrational number \( x \)**:
- Let \( r = \frac{a}{b} \) (where \( a \) and \( b \) are integers, and \( b \neq 0 \)) and \( x \) be an irrational number.

3. **Consider the product \( r \cdot x \)**:
- The product \( r \cdot x = \frac{a}{b} \cdot x = \frac{a \cdot x}{b} \).

4. **Assume for contradiction that the product \( r \cdot x \) is rational**:
- If \( r \cdot x \) is rational, then \( \frac{a \cdot x}{b} \) must also be rational. This implies that \( a \cdot x = q \cdot b \) for some integer \( q \).

5. **From the above, isolate \( x \)**:
- Rearranging gives \( x = \frac{q \cdot b}{a} \).
- Since \( q \), \( b \), and \( a \) are integers (and \( a \neq 0 \)), \( \frac{q \cdot b}{a} \) is a rational number.

6. **Contradiction**:
- However, this contradicts the assumption that \( x \) is irrational.

Thus, we have shown that if \( r \cdot x \) were rational, it would force \( x \) to also be rational, which is a contradiction.

Therefore, it concludes that the product of a rational number and an irrational number must be irrational.

This method of direct proof demonstrates the truth of the claim and effectively disproves Ryiaz's assertion.