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.
To illustrate this, we can define a rational number \( r \) and an irrational number \( x \).
Let's say:
- \( r = \frac{p}{q} \) where \( p \) and \( q \) are integers and \( q \neq 0 \) (this makes \( r \) rational).
- \( x \) is an irrational number.
Now we can consider the product \( rx = \frac{p}{q} \cdot x \).
To prove that this product is irrational, we can use the fact that if \( rx \) were rational, it could be expressed in the form \( \frac{a}{b} \), where \( a \) and \( b \) are integers (and \( b \neq 0 \)).
Then we can manipulate it as follows:
- \( rx = \frac{p \cdot x}{q} \),
- If \( rx \) is rational, then \( p \cdot x \) must also be rational since \( q \) is a fixed rational number.
- Rearranging gives us \( x = \frac{rx \cdot q}{p} \).
Since both \( r \) and \( \frac{q}{p} \) are rational, their product \( rx \cdot q \) would also be rational. This implies that \( x \) must also be rational, which contradicts our original assumption that \( x \) is irrational.
Thus, we conclude that the product \( rx \) must be irrational. This direct proof effectively shows that a rational number multiplied by an irrational number results in an irrational product.
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