# Real Analysis/Section 1 Exercises/Answers

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## General[edit]

*Answer me**Answer me**Answer me**Answer me**Answer me*- With the view of getting a contradiction, assume is rational. Then for some integers and such that they have no common factor other than one, (that is, and are in lowest terms). Squaring both sides and rearranging terms gives Since is prime, must be divisible by , say for some integer By substitution, , so that , and thus must also be divisible by , a contradiction of and being coprime.
*Answer me**Answer me**Answer me**Answer me*

## Inequalities[edit]

- The following are answers.
- As a side note, this is why you do the "inequality sign flip" when you "multiply by -1"
- Intuition solved!
- Part 1: Let's get a new property from the given statement by applying (II)
Part 2: Let's solve it now

*Answer me**Answer me**Answer me**Answer me**Answer me**Answer me**Answer me**Answer me*

## Absolutes[edit]

- The following are answers.
- As a side note, you should
*generally*not use squares in your proof! However, it works here just because we're dealing with absolute values, the resulting number when you reverse the squaring operation.

- As a side note, you should