# Real Analysis/Section 1 Exercises/Answers

Answers in this wikibook work on the didactic that you should still be able to logically piece together a proof that can sufficiently answer a question. Thus, no copy-paste answer is available on this page. Instead, the answer section simply lays out all the tools to prove the problem sufficiently. Many answers will provide hidden questions since some assumptions may not be known to you. Luckily, some of these assertions will have answers elsewhere in the wikibook or can be solved independently.
Note that all solutions are suggestions. There can be multiple ways to solve a question, and as this is a collaborative wikibook, you are free to put down your answer (as long as it follows out guidelines).

## Unsorted
### Problem 4: Prove that $0=-0$

* One algebraic property of zero is $\forall x\in \mathbb {R} :x-x=0$
* And we can use either valid theorem
** Subtracting both sides of the equation by the same variable is valid.
** Multiplying both sides by negative 1 is valid.
* And finally, substitution of whole terms is valid.

## Algebra

* The following are answers.
** As a side note, this is why you do the "inequality sign flip" when you "multiply by -1" ${\begin{aligned}a&<b\\a-a&<b-a\\0&<b-a\\-b&<b-a-b\\-b&<(b-b)-a\\-b&<0-a\\-b&<-a\\\square \end{aligned}}$
** Intuition solved! ${\begin{aligned}a&<b\\a+c&<b+c\\a+c&<b+c<b+d\\a+c&<b+d\\\square \end{aligned}}$
** Part 1: Let's get a new property from the given statement by applying (II) ${\begin{aligned}c&>d\\-c&<-d\end{aligned}}$ Part 2: Let's solve it now ${\begin{aligned}a&<b\\a+(-c)&<b+(-c)\\a-c&<b+(-d)\\a-c&<b-d\\\square \end{aligned}}$
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* The following answers only outline a method to answer the question
* The following answers only outline a method to answer the question
** The largest hurdle is thinking of ways to prove this since distribution of the power term has not been established yet. However, there is a method:
** Start off with $1=1$
** Apply the existence of an inverse as such: $(a\cdot 1/a)\cdot (b\cdot 1/b)=1$
** Redistribute and swap over variables to the other side until the equivalency shows up. As a last resort, consult the following whited out hint in text format: <pre style="text-color: white;">(ab) × 1/a 1/b = 1 → 1/a × 1/b = 1/ab</pre>
** Assuming problem 4I has been solved, this question is relatively easy.
** Assuming problem 4I has been solved, this question is relatively easy.
** Answer me

## Absolutes

* 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. ${\begin{aligned}|a+b|&\leq |a|+|b|\\|a+b|^{2}&\leq (|a|+|b|)^{2}\\a^{2}+2ab+b^{2}&\leq |a|^{2}+2|a||b|+|b|^{2}\\a^{2}+2ab+b^{2}&\leq a^{2}+2|ab|+b^{2}\\2ab&\leq 2|ab|\\ab&\leq |ab|\\\square \end{aligned}}$

## Number Theory
### Problem 1: Prove the following properties on even and odd numbers

* These definitions can be substituted in for each of the following problems.
** The definition of an even number e is $\exists k\in \mathbb {Z} :e=2k$
** The definition of an odd number o is $\exists j\in \mathbb {Z} :o=2j+1$
* Multiplication and addition are closed under the integers.
* The distributive law extends to integers, and this axiom is probably the most difficult theorem that is used to solve these problems.
* Any necessary factor can be expressed as a variable to clean up interpretation.

### Problem 2: Prove that no consecutive number of a perfect square is also a perfect square

* In math notation, the question asks $\forall x\in \{x^{2}:x\in \mathbb {N} \}:[(x+1)^{2}\neq x^{2}+1]\land [(x-1)^{2}\neq x^{2}-1]$
* Some valid theorems are the distributive law for natural numbers and adding the same number over both sides of an equation.

### Problem 6: Prove that any square root of a prime number is irrational

* A prime number does not include 1 and should only have itself as a factor.
* If ${\sqrt {p}}$ is rational, then ${\sqrt {p}}=r/s$ for some coprime integers $s$ and $r$; $s$ and $r$ are in its lowest terms.
* Types of ways to reach a contradiction:
** Both $s$ and $r$ can be shown to be divisible by the prime number $p$ after exposing the prime number using algebraic manipulation, thereby breaking the coprime definition.
*** "Any natural number has a prime number as a factor" is valid.
** The prime number $p$, after exposing the prime number using algebraic manipulation, can be shown to have a factor of two of the same natural number squared, which breaks the prime number definition.
*** "Any natural number cannot be expressed as a fraction in its lowest common factor form and have its denominator anything other than 1" is valid.