To prove that a rational number multiplied by an irrational number gives an irrational product, you can use the following approach:
By creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.
Here is how this proof can be structured:
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Definition of rational and irrational numbers:
- A rational number can be expressed as \( \frac{a}{b} \), where \( a \) and \( b \) are integers, and \( b \neq 0 \).
- An irrational number cannot be expressed as a fraction of two integers.
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Assume a rational number \( r \) and an irrational number \( x \):
- Let \( r = \frac{a}{b} \) (where \( a \) and \( b \) are integers, and \( b \neq 0 \)) and \( x \) be an irrational number.
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Consider the product \( r \cdot x \):
- The product \( r \cdot x = \frac{a}{b} \cdot x = \frac{a \cdot x}{b} \).
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Assume for contradiction that the product \( r \cdot x \) is rational:
- If \( r \cdot x \) is rational, then \( \frac{a \cdot x}{b} \) must also be rational. This implies that \( a \cdot x = q \cdot b \) for some integer \( q \).
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From the above, isolate \( x \):
- Rearranging gives \( x = \frac{q \cdot b}{a} \).
- Since \( q \), \( b \), and \( a \) are integers (and \( a \neq 0 \)), \( \frac{q \cdot b}{a} \) is a rational number.
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Contradiction:
- However, this contradicts the assumption that \( x \) is irrational.
Thus, we have shown that if \( r \cdot x \) were rational, it would force \( x \) to also be rational, which is a contradiction.
Therefore, it concludes that the product of a rational number and an irrational number must be irrational.
This method of direct proof demonstrates the truth of the claim and effectively disproves Ryiaz's assertion.
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