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

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 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.

Answer 1

The best response to Ryiaz's claim is:

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

Here's a brief outline of such a proof:

1. Definitions: Recall that a rational number can be expressed as a fraction of two integers, $r=\frac{a}{b}$, where $a$ and $b$ are integers and $b\ne 0$. An irrational number cannot be expressed as a fraction of two integers.

2. Assumption: Assume $r$ is a rational number and $x$ is an irrational number.

3. Multiplication Result: Consider the product $y=r\times x$.

4. Proof by Contradiction:

   • Suppose $y$ is rational. This means it can be expressed as $\frac{m}{n}$, where $m$ and $n$ are integers and $n\ne 0$.
   • Then we can rearrange the equation: $$x=\frac{y}{r}=\frac{\frac{m}{n}}{\frac{a}{b}}=\frac{mb}{na}$$
   • Since $mb$ and $na$ are both integers (since products and sums of integers are integers), this means that $x$ can be expressed as a fraction of integers, which contradicts the fact that $x$ is irrational.

5. Conclusion: Hence, our assumption that $y$ is rational must be false. Therefore, $r\times x$ must be irrational.

This proof shows that the product of a rational number and an irrational number is indeed irrational.