part d
now examine the sum of a rational number, y, and an irrational number, x. the rational number y can be written as y = $\frac{a}{b}$, where a and b are integers and b ≠ 0. leave the irrational number x as x because it can’t be written as the ratio of two integers.
let’s look at a proof by contradiction. in other words, we’re trying to show that x + y is equal to a rational number instead of an irrational number. let the sum equal $\frac{m}{n}$, where m and n are integers and n ≠ 0. the process for rewriting the sum for x is shown.
| statement | reason |
| --- | --- |
| $x + \frac{a}{b} = \frac{m}{n}$ | substitution |
| $x + \frac{a}{b} - \frac{a}{b} = \frac{m}{n} - \frac{a}{b}$ | |
| $x = \frac{m}{n} - \frac{a}{b}$ | subtraction property of equality |
| $x = (\frac{b}{b})(\frac{m}{n}) - (\frac{n}{n})(\frac{a}{b})$ | |
| $x = \frac{bm}{bn} - \frac{na}{bn}$ | create common denominators. |
| $x = \frac{bm - na}{bn}$ | simplify. |
based on what we established about the classification of x and using the closure of integers, what does the equation tell you about the type of number x must be for the sum of x and y to be rational? what conclusion can you now make about the result of adding a rational and an irrational number?

Question

part d
now examine the sum of a rational number, y, and an irrational number, x. the rational number y can be written as y = $\frac{a}{b}$, where a and b are integers and b ≠ 0. leave the irrational number x as x because it can’t be written as the ratio of two integers.
let’s look at a proof by contradiction. in other words, we’re trying to show that x + y is equal to a rational number instead of an irrational number. let the sum equal $\frac{m}{n}$, where m and n are integers and n ≠ 0. the process for rewriting the sum for x is shown.
| statement | reason |
| --- | --- |
| $x + \frac{a}{b} = \frac{m}{n}$ | substitution |
| $x + \frac{a}{b} - \frac{a}{b} = \frac{m}{n} - \frac{a}{b}$ |  |
| $x = \frac{m}{n} - \frac{a}{b}$ | subtraction property of equality |
| $x = (\frac{b}{b})(\frac{m}{n}) - (\frac{n}{n})(\frac{a}{b})$ |  |
| $x = \frac{bm}{bn} - \frac{na}{bn}$ | create common denominators. |
| $x = \frac{bm - na}{bn}$ | simplify. |
based on what we established about the classification of x and using the closure of integers, what does the equation tell you about the type of number x must be for the sum of x and y to be rational? what conclusion can you now make about the result of adding a rational and an irrational number?

Accepted Answer

# Brief Explanations:
1. From the proof, we assumed $ x + y $ (where $ y=\frac{a}{b} $ is rational) is rational ($ \frac{m}{n} $) and derived $ x=\frac{bm - an}{bn} $. Since $ a,b,m,n $ are integers and $ b,n
eq0 $, $ bm - an $ and $ bn $ are integers (by closure of integers under multiplication and subtraction). Thus, $ x $ would be expressed as a ratio of two integers, meaning $ x $ is rational. But we initially defined $ x $ as irrational, so this is a contradiction.
2. The contradiction implies our assumption that $ x + y $ is rational is false. So, the sum of a rational number and an irrational number must be irrational.

# Answer:
- For the type of number $ x $ must be: If the sum $ x + y $ ( $ y $ rational) were rational, the proof shows $ x $ would be rational (as $ x=\frac{bm - an}{bn} $, with $ bm - an $ and $ bn $ integers, $ bn
eq0 $), contradicting $ x $ being irrational. Thus, our assumption is wrong.
- Conclusion about the sum: The sum of a rational number and an irrational number is irrational.

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QUESTION IMAGE

Question

Explanation:

Answer:

statement	reason
\\(x + \frac{a}{b} - \frac{a}{b} = \frac{m}{n} - \frac{a}{b}\\)
\\(x = \frac{m}{n} - \frac{a}{b}\\)	subtraction property of equality
\\(x = (\frac{b}{b})(\frac{m}{n}) - (\frac{n}{n})(\frac{a}{b})\\)
\\(x = \frac{bm}{bn} - \frac{na}{bn}\\)	create common denominators.
\\(x = \frac{bm - na}{bn}\\)	simplify.