Question

Put the steps to the proof that the sum t=r+s is irrational if r is a rational and s is irrational in the correct sequence

Answer 1

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:

1. **Assume \( r \) is rational**: By definition, this means that \( r \) can be expressed as \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).

2. **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 \( d \).

3. **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 \( n \neq 0 \).

4. **Rearrange the equation**: From the assumption, we can write:
\[
s = t - r
\]

5. **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:
\[
s = \frac{mb - a n}{nb}
\]

6. **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.

7. **Reach a contradiction**: This contradicts our earlier assumption that \( s \) is irrational.

8. **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.