To prove that a rational number multiplied by an irrational number yields an irrational product, we can use a proof by contradiction. Here's how the proof would go:
-
Definition of Rational and Irrational Numbers:
- A rational number 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.
-
Assume the Opposite:
- Let \( r \) be a rational number, and let \( i \) be an irrational number. We will assume for the sake of contradiction that their product \( p = r \times i \) is a rational number.
-
Express the Rational Number:
- Let \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
-
Substitute and Rearrange:
- According to our assumption, \( p = r \times i = \frac{a}{b} \times i \).
- This implies \( p = \frac{a \times i}{b} \).
-
Solve for the Irrational Number:
- Rearranging this gives us \( a \times i = p \times b \).
- Here, \( p \times b \) is a product of two rational numbers (since \( p \) is assumed rational and \( b \) is rational), thus it is rational.
-
Contradiction:
- This leads to the conclusion that \( a \times i \) must also be rational since it equals \( p \times b \).
- However, since \( a \) is an integer and \( i \) is irrational, the product \( a \times i \) must be irrational. This contradicts our assumption that \( a \times i \) is rational.
-
Conclusion:
- Therefore, our initial assumption that the product \( p = r \times i \) is rational must be false. Hence, the product of a rational number and an irrational number is irrational.
In summary, the correct way to prove Ryaz wrong is by creating a proof that shows directly how multiplying a rational and an irrational number leads to an irrational answer.
1 answer