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

Answer 1

The best way to prove that a rational number multiplied by an irrational number yields an irrational product 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 how such a proof could be structured:

1. Definitions:

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

2. Proof:

   • Let \( r \) be a rational number, such that \( r = \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \).
   • Let \( x \) be an irrational number.
   • Assume for the sake of contradiction that the product \( rx \) is rational.
   • If \( rx \) is rational, then we can express it as \( \frac{m}{n} \), where \( m \) and \( n \) are integers and \( n \neq 0 \).
   • Rearranging gives \( x = \frac{mx}{r} = \frac{m}{r} \), where \( r \neq 0 \), which means \( x \) can be expressed as a ratio of integers, which contradicts the assumption that \( x \) is irrational.

Thus, our initial assumption that \( rx \) is rational must be false, implying that \( rx \) must be irrational.

Hence, we have proven that the product of a rational number and an irrational number is indeed irrational.