To prove that the sum \( t = r + s \) is irrational if \( r \) is rational and \( s \) is irrational, you can follow these steps:
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Assume \( r \) is rational: Start by stating that since \( r \) is rational, it can be expressed as a fraction, \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
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Assume \( s \) is irrational: State that \( s \) is an irrational number, meaning it cannot be expressed as a fraction of two integers.
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Assume, for contradiction, that \( t \) is rational: Suppose that the sum \( t = r + s \) is rational. By the definition of rational numbers, \( t \) can also be expressed as a fraction, \( t = \frac{c}{d} \), where \( c \) and \( d \) are integers and \( d \neq 0 \).
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Rearrange the equation: From the assumption that \( t \) is rational, rewrite it as: \[ s = t - r \] Since \( r \) is rational (and can be written as \( \frac{a}{b} \)), we have: \[ s = \frac{c}{d} - \frac{a}{b} \]
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Combine the right-hand side: Find a common denominator to combine the two fractions: \[ s = \frac{cb - ad}{bd} \] Here, \( cb - ad \) is an integer (since both \( cb \) and \( ad \) are products of integers), and \( bd \neq 0 \) since both \( b \) and \( d \) are non-zero.
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Conclude that \( s \) is rational: The expression \( s = \frac{cb - ad}{bd} \) shows that \( s \) can be expressed as a fraction of two integers, which contradicts the assumption that \( s \) is irrational.
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Conclude the proof: Since the assumption that \( t \) is rational leads to a contradiction, we conclude that the original assumption must be false. Therefore, the sum \( t = r + s \) is irrational.
By following these steps, you have logically arrived at the conclusion that the sum \( t = r + s \) is indeed irrational when \( r \) is rational and \( s \) is irrational.
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