To prove that the sum \( t = r + s \) is irrational under the assumption that both \( r \) and \( s \) are rational, it seems there may be confusion. In fact, the correct statement is:
If \( r \) and \( s \) are both rational numbers, then their sum \( t = r + s \) is also rational.
However, if you meant to prove that the sum \( t = r + s \) is irrational under the assumption that at least one of them is irrational, here are the series of steps for that proof:
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Define Rational and Irrational Numbers: Recall that 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 in this form.
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Assume Rational and Irrational: Let \( r \) be a rational number, which can be written as \( \frac{a}{b} \), and let \( s \) be an irrational number.
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Express the Sum: Consider the sum \( t = r + s = \frac{a}{b} + s \).
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Rearranging: Rearranging gives us \( s = t - \frac{a}{b} \).
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Assume \( t \) is Rational: If \( t \) were rational, then both \( t \) and \( \frac{a}{b} \) would be rational.
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Conclude that \( s \) is Rational: If \( t \) is rational and \( \frac{a}{b} \) is rational, then \( s = t - \frac{a}{b} \) would also have to be rational because the set of rational numbers is closed under subtraction.
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Contradiction: This contradicts the assumption that \( s \) is irrational.
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Final Conclusion: Therefore, the original assumption that \( t = r + s \) is rational must be false, and we conclude that \( t = r + s \) is irrational if \( r \) is rational and \( s \) is irrational.
Here's the sequence summarized:
- Define rational and irrational numbers.
- Assume \( r \) is rational, \( s \) is irrational.
- Consider \( t = r + s = \frac{a}{b} + s \).
- Rearrange to get \( s = t - \frac{a}{b} \).
- Assume \( t \) is rational.
- Conclude \( s \) must be rational.
- Find a contradiction.
- Conclude \( t = r + s \) is irrational.
If your question intended something different, please clarify!
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