Question

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) Responses 1 He’s correct; it is impossible to prove in any way. 2 by choosing random numbers to plug into multiplication problems until he’s convinced 3 by choosing random numbers to plug into multiplication problems until he’s convinced 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 4 by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer

Answer 1

To prove that a rational number multiplied by an irrational number gives an irrational product, the best approach would be option 4: "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 how such a proof can be structured:

1. Definitions:

   • A rational number can be expressed as $r=\frac{a}{b}$, where $a$ and $b$ are integers, and $b\ne 0$.
   • An irrational number cannot be expressed as a fraction of integers.

2. Assumption:

   • Let $r$ be a rational number and $x$ be an irrational number.
   • We want to prove that the product $rx$ is irrational.

3. Proof by Contradiction:

   • Assume for the sake of contradiction that the product $rx$ is rational. Then, there exist integers $m$ and $n$ (with $n\ne 0$) such that $rx=\frac{m}{n}$.

4. Rearranging:

   • From the equation $rx=\frac{m}{n}$, we can express $x$ as: $$x=\frac{m}{n}\cdot \frac{b}{a}=\frac{mb}{na}$$
   • Since $a$, $b$, $m$, and $n$ are integers (and assuming $r\ne 0$), the right side is a fraction of integers, which means $x$ would be rational.

5. Contradiction:

   • This contradicts our original assumption that $x$ is irrational. Therefore, our assumption that $rx$ is rational must be false.

6. Conclusion:

   • We conclude that the product $rx$ must be irrational whenever $r$ is rational (and non-zero) and $x$ is irrational.

Thus, the statement that a rational number multiplied by an irrational number gives an irrational product is indeed provable and demonstrated through the proof above.

Answer 2

Use PROOF OF CONTRADICTION!