🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

Observe that $x\ge 0$ iff $x+(-x)\ge 0+(-x)$ iff $0\ge -x$ as needed.

Since $0$ is an identity element with respect to $+$ then $x+0=x$, specifically for $x=0$ by definition this means that $-0=0$ as needed.

Observe that $x\ge y$ is the same as $x-y\ge 0$ is the same as $0\ge y-x$ is the same as $-x\le -y$ as needed.

Let $x\in ℝ$ if $x>;0$, then $x=\left|x\right|$ by definition, if $x<;0$, then $\left|x\right|=-x$, or rather $x=-\left|x\right|$

Recall the definition of $max(x,-x)$ which is that it's value is $x$ if $x\ge -x$ and $-x$ if $x\le -x$. Suppose that $x\ge 0$ then $\left|x\right|=x$ and since $x\ge -x$ then we know that $max(x,-x)=x$ and so in this case we see that $\left|x\right|=max\left(x\right)$.

Now in the case that $x<0$ then $-x>0$ and therefore $-x>x$ therefore $max(x,-x)=-x$ and since $x<0$ then $\left|x\right|=-x$.

Therefore it's been verified in any case so that $\left|x\right|=max(x,-x)$.

$⟹$ we want to prove that $\left|a\right|\le x$. Case one, if $a\ge 0$ then $\left|a\right|=a$ and since we already know that $a\le x$ then we obtain that $\left|a\right|\le x$ as needed. In the second case if $a<0$ then $\left|a\right|=-a$ so we want to show that $-a\le x$ which is the same as $a\ge -x$ which we know by assumption as well.

$⟸$ Suppose that $\left|a\right|\le x$ if $a\ge 0$ then $\left|a\right|=a$ and so we're really assuming that $a\le x$ therefore $-a\ge -x$ since $a\ge -a$ then by chaining inequalities we get $x\ge a\ge -a\ge -x$ which implies $x\ge a\ge -x$ as needed. Now for the case of $a\le 0$ then we've assumed that $-a\le x$ now $-a$ is positive and by symmetrically following our previous case with $-a$ in place of $a$ we see that $x\ge -a\ge a\ge -x$ and so $x\ge a\ge -x$ again, as needed.

Let's recall that $c\le \left|c\right|\le max\left(\right|c|,|a\left|\right)$, and on the other hand that $a\ge -\left|a\right|\ge -max\left(\right|a|,|c\left|\right)$ combining inequalities, that says $${\displaystyle -max\left(\right|a|,|c\left|\right)\le b\le max\left(\right|a|,|c\left|\right)}$$ therefore $${\displaystyle \left|b\right|\le |max\left(\right|a|,|c\left|\right)|=max\left(\right|a|,|c\left|\right)}$$ as needed.

Since $1<\left|c\right|$ then $1·\left|x\right|\le \left|c\right|·\left|x\right|=\left|cx\right|$

TODO

Suppose that $a=b$ for contradiction, then $-\left|a\right|<a<\left|a\right|$ , but either $a=\left|a\right|$ or $a=-\left|a\right|$, so this yields a contradiction.

We prove it true by induction based on the size of $X$. Base case $\left|X\right|=2$ is covered by the regular triangle inequality. Now suppose that it holds true for $\left|Y\right|=k$ we'll show it holds true for $\left|Z\right|=k+1$, let $z\in Z$, then we have $|\sum \left(Z⧵\left\{z\right\}\right)|\le {\sum }_{a\in Z⧵\left\{z\right\}}\left|a\right|$ since $\left|Z⧵\left\{z\right\}\right|=k$. Now by the regular triangle inequality we obtain that $$\begin{gathered} {\displaystyle |\sum Z|\&=|\sum \left(Z⧵\left\{z\right\}\right)+z| \\ \&\le |\sum \left(Z⧵\left\{z\right\}\right)|+\left|z\right| \\ \&\le {\sum }_{a\in Z⧵\left\{z\right\}}\left|a\right|+\left|z\right| \\ \&={\sum }_{z\in Z}\left|z\right|} \end{gathered}$$

We have the following: $$\begin{gathered} {\displaystyle |a-b|\le c \\ \Updownarrow \\ |a-b|+\left|b\right|\le c+\left|b\right| \\ \Updownarrow \\ |a-b+b|\le |a-b|+\left|b\right|\le c+\left|b\right| \\ \Updownarrow \\ \left|a\right|\le c+\left|b\right|} \end{gathered}$$ Where in the second last line we used the triangle inequality