Question

Put the steps to the proof that the sum t equals r + s is rational if r is irrational and s is rational and correct sequence step 1 subtract writing s as a fraction step 2 for a contradiction assume that he is rational and right over and t is fraction step 3 realize a contraction step four find a

Answer 1

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:
\[
\text{If } r \text{ is irrational and } s \text{ is rational, then } t = r + s \text{ is irrational.}
\]