Question

Ryiaz says it is impossible to prove that multiplying a non-zero rational number by an irrational number always results in an irrational product. How can you show that a proof is possible?(1 point)
Responses

You can prove it by choosing random numbers to plug into multiplication problems until he’s convinced.
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 non-zero 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.
Use proof by contradiction: assume the opposite is true and the product of a non-zero 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.
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.
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 way to demonstrate that multiplying a non-zero rational number by an irrational number always results in an irrational product is through **proof by contradiction**.

Here’s the reasoning step by step:

1. **Assume the Opposite**: Let's assume that the product of a non-zero rational number \( r \) and an irrational number \( x \) is a rational number. So, we have:
\[
r \cdot x = q
\]
where \( q \) is a rational number.

2. **Express \( x \)**: Since \( r \) is non-zero and rational, we can manipulate the equation to express \( x \) in terms of \( q \):
\[
x = \frac{q}{r}
\]
Here, \( \frac{q}{r} \) is the quotient of two rational numbers.

3. **What is \( \frac{q}{r} \)?**: According to the properties of rational numbers, the quotient of two rational numbers \( q \) and \( r \) (where \( r \neq 0 \)) is also a rational number.

4. **Contradiction**: This implies that \( x \) must be rational, contradicting our initial condition that \( x \) is irrational.

5. **Conclusion**: Since our assumption led to a contradiction, we conclude that our initial assumption is false. Therefore, it must be true that multiplying a non-zero rational number \( r \) by an irrational number \( x \) always results in an irrational product.

Thus, you can indeed prove that the product of a non-zero rational number and an irrational number is always irrational.