Question

Ryiaz says it is impossible to prove that multiplying a rational number by an irrational number always results in an irrational product. How can you show that a proof is possible? (1 point)
You can prove it by choosing random numbers to plug into multiplication problems until
he's convinced.
Use proof by contradiction: assume the opposite is true and the product of a rational
•
number and an irrational number is always rational. This is a contradiction. According to the Closure Property, if the product is rational, both factors must be rational. So if one factor is irrational, the product must be irrational.

You can prove it directly 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 that the product of a rational number and an irrational number is always irrational.

Answer 1

The best approach to prove that multiplying a rational number by an irrational number always results in an irrational product is to use **proof by contradiction**. Here is how you can structure the proof:

1. **Definitions**:
- A rational number can be expressed in the form \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \).
- An irrational number cannot be expressed as a fraction of two integers.

2. **Assumption**:
Let's assume the opposite of what we want to prove. Specifically, assume that it is possible to multiply a rational number by an irrational number and get a rational product.

3. **Let’s illustrate this with examples**:
Let \( r \) be a rational number and \( x \) be an irrational number. By our assumption, their product is \( rx \), which we assume to be rational.

4. **Expressing the rational product**:
Since \( r \) is rational, we can express it as \( r = \frac{p}{q} \). According to our assumption:
\[
rx = \frac{p}{q} \times x \text{ is rational.}
\]

5. **Rearranging the equation**:
We can rearrange this to find \( x \):
\[
x = \frac{q(rx)}{p}.
\]
If \( rx \) is rational (as we assumed), then \( \frac{q(rx)}{p} \) is also rational since it is formed by multiplying/dividing rational numbers.

6. **Conclusion from the assumption**:
Since we found that \( x \) must be rational if our assumption is true (that \( rx \) is rational), this contradicts the fact that \( x \) is irrational.

7. Consequently, we conclude that our initial assumption must be wrong; thus, the product of a rational number and an irrational number must be irrational.

This proof shows that multiplying a rational number by an irrational number indeed results in an irrational product, thereby demonstrating that it is possible to prove this claim.