To prove that the sum \( t = r + s \) is irrational when \( r \) is irrational and \( s \) is rational, we can follow these structured steps for the proof:
Step 1: Express \( s \) as a fraction
Since \( s \) is rational, we can express it as a fraction: \[ s = \frac{a}{b} \] where \( a \) and \( b \) are integers and \( b \neq 0 \).
Step 2: Assume for contradiction that \( t \) is rational
Assume, for the sake of contradiction, that \( t \) is rational. If \( t \) is rational, it can also be expressed as a fraction: \[ t = \frac{c}{d} \] where \( c \) and \( d \) are integers and \( d \neq 0 \).
Step 3: Rearrange the expression
Since we have defined \( t \) as the sum of \( r \) and \( s \): \[ t = r + s \] Substituting the expression for \( s \) into this equation gives: \[ t = r + \frac{a}{b} \] Rearranging this expression leads to: \[ r = t - s \] Substituting for \( t \) leads us to: \[ r = \frac{c}{d} - \frac{a}{b} \]
Step 4: Find a common denominator
To combine the fractions, we find a common denominator: \[ r = \frac{cb - ad}{bd} \] Here, \( cb - ad \) is an integer since it is a linear combination of integers, and \( bd \) is also an integer (and non-zero). This means that \( r \) can be expressed as a fraction where both the numerator and the denominator are integers.
Step 5: Realize the contradiction
Given that \( r \) is expressed as a fraction, we conclude that \( r \) is rational. However, this directly contradicts our original assumption that \( r \) is irrational.
Conclusion
Since our assumption that \( t \) is rational leads to a contradiction, we must conclude that \( t = r + s \) is irrational. Therefore, the statement holds: \[ \text{If } r \text{ is irrational and } s \text{ is rational, then } t = r + s \text{ is irrational.} \]
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