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)

Question

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 option for showing that multiplying a rational number by an irrational number always results in an irrational product is:

Use proof by contradiction: assume the opposite is true and the product of a rational number and an irrational number is rational. Then, derive a contradiction from this assumption.

Here's how the proof would go:

1. Assume we have a rational number \( r \) and an irrational number \( x \).
2. By definition, a rational number can be expressed as \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers, and \( b \neq 0 \).
3. Now, assume that the product \( r \cdot x \) is rational.
4. Then, we can express this product as a rational number \( q = r \cdot x \).
5. Rearranging gives us \( x = \frac{q}{r} \).
6. Since \( q \) is rational (by our assumption) and \( r \) is rational (by our definition), the division of two rational numbers \( \frac{q}{r} \) is also rational.
7. This means that \( x \) would have to be rational, which contradicts the fact that \( x \) is an irrational number.

Since we reached a contradiction from our initial assumption, we conclude that multiplying a rational number by an irrational number must indeed result in an irrational product. Thus, the statement is proven.