## General

6. With the view of getting a contradiction, assume $\sqrt{p}$ is rational. Then $\sqrt{p}=r/s$ for some integers $s$ and $r$ such that they have no common factor other than one, (that is, $s$ and $r$ are in lowest terms). Squaring both sides and rearranging terms gives $s^2p=r^2.$ Since $p$ is prime, $r$ must be divisible by $p$, say $r=pt$ for some integer $t.$ By substitution, $s^2p=p^2t^2$, so that $s^2=pt^2$, and thus $s$ must also be divisible by $p$, a contradiction of $r$ and $s$ being coprime.

## Inequalities

1. As a side note, this is why you do the "inequality sign flip" when you "multiply by -1"

\begin{align}a &< b \\ a - a &< b - a \\ 0 &< b - a \\ -b &< b - a - b \\ -b &< (b - b) - a \\ -b &< 0 - a \\ -b &< -a \\ \square \end{align}

2. Intuition solved!

\begin{align}a &< b \\ a + c &< b + c \\ a + c &< b + c < b + d \\ a + c &< b + d \\ \square\end{align}

3. Part 1: Let's get a new property from the given statement by applying (II)

\begin{align}c &> d \\ -c &< -d\end{align}

Part 2: Let's solve it now

\begin{align}a &< b \\ a + (-c) &< b + (-c) \\ a - c &< b + (-d) \\ a - c &< b - d \\ \square\end{align}

\begin{align}|a + b| &\le |a| + |b| \\ |a + b|^2 &\le (|a| + |b|)^2 \\ a^2 + 2ab + b^2 &\le |a|^2 + 2|a||b| + |b|^2 \\ a^2 + 2ab + b^2 &\le a^2 + 2|ab| + b^2 \\ 2ab &\le 2|ab| \\ ab &\le |ab| \\ \square \end{align}