Question

Ryiaz claims it is impossible to prove that a rational number multiplied by an irrational number gives an irrational productHow can you prove this? (1 point) by using a proof by contradiction to show multiplication is a closed operation for rational numbers by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer He's correct , it is impossible to prove in any way by choosing random numbers to plug into multiplication problems until he's convinced

Answer 1

To prove that multiplying a rational number by an irrational number yields an irrational product, we can use **proof by contradiction.**

Let’s denote the rational number as \( r \) and the irrational number as \( x \). We assume that when we multiply \( r \) and \( x \), the product \( r \cdot x \) is a rational number.

Since \( r \) is a rational number, we can express it as \( r = \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \).

Now, if we assume that \( r \cdot x \) is rational, we can denote it as \( q \), where \( q \) is a rational number. Therefore, we can express our equation as:

\[
r \cdot x = q
\]

This can be rearranged as:

\[
x = \frac{q}{r}
\]

Since \( r = \frac{a}{b} \), we find:

\[
x = \frac{q \cdot b}{a}
\]

Here, \( q \cdot b \) and \( a \) are both integers (since the product of rational numbers remains rational and the property of integers' closure under multiplication), meaning that \( \frac{q \cdot b}{a} \) is written as a fraction of integers (where \( a \neq 0 \)), hence \( x \) would be rational.

However, this leads to a contradiction because we started with the assumption that \( x \) is irrational.

Thus, our initial assumption that \( r \cdot x \) is rational must be false. Consequently, the product of a non-zero rational number and an irrational number must indeed be irrational.

Therefore, we have proven by contradiction that multiplying a rational number by an irrational number results in an irrational product.