Question
Arrange the steps to prove that the
sum t = r + s is irrational if r is rational and s is irrational.
sum t = r + s is irrational if r is rational and s is irrational.
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GPT-4o mini
To prove that the sum \( t = r + s \) is irrational if \( r \) is rational and \( s \) is irrational, you can follow these steps:
1. **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 \( b \neq 0 \).
2. **Assume \( s \) is irrational**: State that \( s \) is an irrational number, meaning it cannot be expressed as a fraction of two integers.
3. **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 \( d \neq 0 \).
4. **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:
\[
s = \frac{c}{d} - \frac{a}{b}
\]
5. **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.
6. **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.
7. **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.
1. **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 \( b \neq 0 \).
2. **Assume \( s \) is irrational**: State that \( s \) is an irrational number, meaning it cannot be expressed as a fraction of two integers.
3. **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 \( d \neq 0 \).
4. **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:
\[
s = \frac{c}{d} - \frac{a}{b}
\]
5. **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.
6. **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.
7. **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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