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 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
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

He’s correct; it is impossible to prove in any way.
He’s correct; it is impossible to prove in any way.

Answer 1

To prove that a rational number multiplied by an irrational number results in an irrational number, the best approach 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 that proof:

1. Definitions:

   • A rational number can be defined as any number that can be expressed as the fraction \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b \neq 0\).
   • An irrational number cannot be expressed as a fraction of two integers.

2. Assume:

   • Let \(r\) be a rational number and \(i\) be an irrational number.
   • Suppose \(r \times i\) is rational. By definition of rationality, this means that \(r \times i = \frac{m}{n}\) for some integers \(m\) and \(n\), where \(n \neq 0\).

3. Rearranging:

   • We can manipulate the assumption: \(i = \frac{m}{n \cdot r}\). Since \(r\) is rational (and not zero), \((n \cdot r)\) is also a non-zero rational number. Thus, \(i\) can be expressed as a fraction of integers—contradicting the fact that \(i\) is irrational.

4. Conclusion:

   • Therefore, our original assumption that \(r \times i\) is rational must be false. Hence, \(r \times i\) must be irrational.

This proof shows that multiplying a rational number by an irrational number always yields an irrational result.