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

$${\displaystyle n!=1\cdots k\cdots n}$$ therefore $${\displaystyle \frac{n!}{k}=1\cdots \left(\frac{k}{k}\right)\cdots n\in \mathbb{Z}}$$ so therefore $k\mid n!$

In the definition of divides take $k=j$, then we can see that $j=j·1$ thus we can say that $1|j$

We have that $1=kd$ but the only numbers in $\mathbb{Z}$ that have a multiplicative inverse are $1,-1$ so $d$ must be one of those or else we have a contradiction.

By definition we have $d=k·0=0$ so $d=0$

In the definition of divides take $k=0$, then we can see that $0=0·d$ thus we can say that $d|0$

Clearly $1·a=a$ therefore $a\mid a$

We have that $b=k_{a}a$ and $c=k_{b}b$ therefore $c=k_b\left(k_{a}a\right)$ therefore by taking $k_c=\left(k_a{k}_b\right)$ we see that $a\mid c$

Suppose that $a\mid b$ therefore there is $k\in \mathbb{Z}$ such that $b=ka$, therefore multiply both sides to get $bc=\left(kc\right)a$ and therefore $a\mid bc$ as needed.

We see that $n=kd$ for some $k\in \mathbb{Z}$, since $n\ne 0$ then also we must have $k,d\ne 0$ specifically note that $\left|k\right|\ge 1$ so that we have $${\displaystyle \left|n\right|=\left|kd\right|=\left|k\right|·\left|d\right|\ge 1\left|d\right|=\left|d\right|}$$ so that $\left|n\right|\ge \left|d\right|$

If $a=0$ then because $a\mid b$ then this forces $b=0$ so that $a=b=0$, also if we assumed $b=0$ the same thing occurs.

In the case that neither of $a,b=0$ then recall that if $a\mid b$ we have $\left|a\right|\le \left|b\right|$ also since $b\mid a$ then also $\left|a\right|\le \left|b\right|$ so that $\left|a\right|=\left|b\right|$.

It is reflexive because we know that everything divides itself, we know it is transitive, and we also know that it is anti-symmetric, because for any $n\in {\mathbb{N}}_0,\left|n\right|=n$ and using this proposition.

For the base case of $n=2$ if $a\mid b_1$ and $a\mid b_2$ then we know that there are $k_1,k_2\in \mathbb{Z}$ such that $b_1=k_{1}a$ and $b_2=k_{2}a$ so that $c_1{b}_1+c_2{b}_2=c_1\left(k_{1}a\right)+c_2\left(k_{2}a\right)=a(c_1{k}_1+c_2{k}_2)$.

For the induction step, assume the statement holds true for $k\in {\mathbb{N}}_1$ and we'll show that it holds true for $k+1$ so suppose that $a\mid b_1,\dots b_{k+1}$ by the induction hypothesis we know that $a\mid c_1{b}_1+\cdots +c_k{b}_k$ so we get some $j^{\prime }\in \mathbb{Z}$ such that $c_1{b}_1+\cdots +c_k{b}_k=j^{\prime }a$ and also some $j^{\prime \prime }\in \mathbb{Z}$ such that $b_{k+1}=j^{\prime \prime }a$ and therefore $${\displaystyle c_1{b}_1+\cdots +c_k{b}_k+c_{k+1}{b}_{k+1}=j^{\prime }a+j^{\prime \prime }a=a(j^{\prime }+j^{\prime \prime })}$$ so therefore $a\mid c_1{b}_1+\cdots +c_{k+1}{b}_{k+1}$

If $n$ is even then there is some $k\in \mathbb{Z}$ such that $n=2k$ and therefore $n^m={\left(2k\right)}^m=2^m{k}^m=2\left(2^{m-1}{k}^m\right)$ so that $n^m$ is also even.

Base case, set $m=0$ then $n^m=1$ which is odd. For the induction step assume it holds true for $k\in {\mathbb{N}}_0$ so that $n^k$ is odd, which means that there is some $j\in \mathbb{Z}$ such that $n^k=2j+1$ therefore $n^{k+1}=(2j+1)n$ but $n$ is odd so there is some $j^{\prime }\in \mathbb{Z}$ such that $n=2j^{\prime }+1$ and thus $${\displaystyle n^{k+1}=(2j+1)(2j^{\prime }+1)=4j{j}^{\prime }+2j+2j^{\prime }+1=2(2j{j}^{\prime }+j+j^{\prime })+1}$$ hence $n^{k+1}$ is prime so the statement follows from the principle of induction.

Suppose for the sake of contradiction that it was not true, so that we knew that $${\displaystyle (b\le \sqrt{a}\vee c\ge \sqrt{a})\wedge (b\ge \sqrt{a}\vee c\le \sqrt{a})}$$ Suppose that $b\le \sqrt{a}$ and $b\ge \sqrt{a}$ holds true, then $b=\sqrt{a}$ since $c\ne b$ then if $c<b$ we see that $bc<\sqrt{a}\sqrt{a}=a$ which is a contradiction since we know that $bc=a$, if it turns out that $c\le \sqrt{a}$ and $c\ge \sqrt{a}$ it is handeled symetrically.

On the other hand if we know that $b\le \sqrt{a}$ and that $c\le \sqrt{a}$, since $b\ne a$ then we know that $b\ne \sqrt{a}$, therefore more specifically we see that $b<\sqrt{a}$ this implies that $a=bc<\sqrt{a}\sqrt{a}=a$ which is a contradiction. In the case of $c\ge \sqrt{a}$ and $b\ge \sqrt{a}$ it follows in a symmetrical manner.

We have that $n=mk$ for some $k\in \mathbb{Z}$ and we want to prove that $a^{mk}-1=(a^m-1)j$ for some $j\in \mathbb{Z}$, a good guess would be that $j=a^{(m-1)k}$ which produces $(a^m-1)\left(a^{(m-1)k}\right)=a^{mk}-a^{(m-1)k}$ and so we have to get rid of the $a^{(m-1)k}$ term, which inspires us to try $j=a^{(m-1)k}+a^{(m-2)k}$ which yields $${\displaystyle (a^m-1)(a^{(m-1)k}+a^{(m-2)k})=a^{km}+a^{(m-1)k}-(a^{(m-1)k}+a^{(m-2)k})=a^{km}-a^{(m-2)k}}$$

As we've just run into the same problem as last time we see that adding $a^{(m-3)k}$ would remove the most recent residue term, but then leave the new residue term $a^{(m-3)k}$ . Therefore formally using induction we can show that $${\displaystyle (a^m-1)\left({\sum }_{i=0}^{m-1}{a}^{ik}\right)=a^{mk}-1=a^n-1}$$ where $1$ is really the residue term $a^{\left(0\right)k}$