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)

Question

Responses

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

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

GPT-4o mini · Answer

The correct response is: **by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.** To provide a clearer explanation, we can demonstrate that multiplying a rational number by an irrational number usually results in an irrational number. Here’s a proof: 1. Let $ r $ be a rational number, which can be expressed as $ r = \frac{a}{b} $, where $ a $ and $ b $ are integers and $ b \neq 0 $. 2. Let $ x $ be an irrational number. 3. We want to show that the product $ r \cdot x $ is irrational. Assume for the sake of contradiction that $ r \cdot x $ is rational. Then we could write: $$ r \cdot x = \frac{m}{n} $$ for some integers $ m $ and $ n $ (where $ n \neq 0 $). Rearranging gives: $$ x = \frac{r \cdot x}{r} = \frac{m}{n \cdot b} = \frac{(a \cdot m)}{(b \cdot n)} $$ Since $ a $, $ b $, $ m $, and $ n $ are all integers, the right side of the equation is a ratio of integers, which means $ x $ would be rational. This contradicts our initial statement that $ x $ is irrational. Therefore, $ r \cdot x $ must be irrational. This proof demonstrates that multiplying a non-zero rational number by an irrational number results in an irrational product, directly disproving Ryiaz's claim.