To prove that the sum \( t = r + s \) is irrational when \( r \) is rational and \( s \) is irrational, you can follow these steps in the correct sequence:
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Assume \( r \) is rational: By definition, this means that \( r \) can be expressed as \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
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Assume \( s \) is irrational: This means that \( s \) cannot be expressed as a fraction of integers; it does not equal \( \frac{c}{d} \) for any integers \( c \) and \( d \).
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Assume, for the sake of contradiction, that \( t = r + s \) is rational: This means \( t \) can also be expressed as \( \frac{m}{n} \), where \( m \) and \( n \) are integers and \( n \neq 0 \).
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Rearrange the equation: From the assumption, we can write: \[ s = t - r \]
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Substitute the expressions: Substitute \( t \) and \( r \) with their respective forms: \[ s = \frac{m}{n} - \frac{a}{b} \] To combine these fractions, find a common denominator: \[ s = \frac{mb - a n}{nb} \]
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Analyze the expression: Since \( mb \), \( a \), \( n \), and \( b \) are integers, the result of \(\frac{mb - an}{nb}\) is a fraction of two integers, which makes \( s \) rational.
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Reach a contradiction: This contradicts our earlier assumption that \( s \) is irrational.
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Conclude that \( t = r + s \) must be irrational: Since assuming \( t \) is rational leads to a contradiction, we conclude that \( t \) must be irrational.
By following these steps in order, you've successfully demonstrated that the sum of a rational number and an irrational number is irrational.