Base Representation
Suppose that $$m , k \in \mathbb{N}_{1}$$ and $$\left ( a_{k} , a_{k - 1} , \ldots , a_{1} \right ) \in \left \lbrace 0 , 1 , 2 , \ldots , m - 1 \right \rbrace^{k}$$. Then we define $$n := \sum_{i = 0}^{k - 1} a_{i} m^{i}$$ and we say that the string $$a_{k } a_{k - 1} \ldots a_{1}$$ is the representation of $$n$$ in the base $$m$$ and define $\operatorname{ base\_repr } \left( n, m \right) := a _ { k } a _ { k - 1 } \ldots a _ 1$
Every Number has a Base $$b$$ Representation
Let $$b \in \mathbb{ N } _ 2$$ and $$n \in \mathbb{ N } _ 0$$ then there exists a $$k \in \mathbb{ N } _ 1$$ and $$a _ 1, \ldots , a _ k \in \left[ 0, \ldots , b - 1 \right]$$ such that $n = \left( a _ k \ldots a _ 1 \right) _ b$

Let $$b \in \mathbb{ N } _ 2$$ and we continue with induction. In the base case when $$n = 0$$ then the representation $$\left( 0 \right) _ b$$ works. Now let $$j \in \mathbb{ N } _ 0$$ and and let $$i \in \left[ 0, \ldots, j \right]$$ and assume that it holds for $$i$$, now we'll show it holds true for $$j + 1$$.

What we will try to do is extract the largest $$b ^ m$$ we can from $$j + 1$$ and then use the induction hypothesis. So let $$J := \left\{ m \in \mathbb{ N } _ 1: b ^ m \le j + 1 \right\}$$, this set has a maximum element because $$j + 1$$ is an upper bound, it is because we know that $$1 \le j + 1 \le 2 ^ { j + 1 } \le b ^ { j + 1 }$$ therefore $$j + 1$$ is indeed an upper bound to $$J$$, let $$l$$ be the max element of $$J$$ and then by the quotient remainder theorem we have some $$q, r \in \mathbb{ Z }$$ such that $j + 1 = q b ^ l + r \text{ with } 0 \le r \lt b ^ l$ since $$r \lt b ^ l \le j + 1$$ then the induction hypothesis holds on $$r$$ so that $$r$$ has a base $$b$$ representation so that $r = \sum _ { i = 0 } ^ { p } c _ i b ^ i$ where $$p \lt l$$, because if $$p \ge l$$ then that makes $$r = \sum _ { i = 0 } ^ { p } x _ i b ^ i \ge b ^ l$$ which is a contradiction since we know that $$r \lt b ^ l$$ this means that the representation is exactly $\left( q c _ p \ldots c _ 0 \right) _ b$

Every Base $$b$$ Representation is Unique
For any $$b \in \mathbb{ N } _ 2$$ and $$n \in \mathbb{ N } _ 0$$ there is exactly one base $$b$$ representation of $$n$$

Let $$b \in \mathbb{ N } _ 2$$ and perform induction on $$n$$. So for $$n = 0$$ we know that the representation $$\left( 0 \right) _ b$$ and for any other representation where $$a \neq 0$$ then $$\left( a \right) _ b \neq 0$$ there is exactly one representation as needed.

Now suppose that $$n \in \mathbb{ N } _ 0$$ and that for any $$k \in \left[ 0, \ldots n \right]$$ has a unique base $$b$$ representation, let's prove that $$n + 1$$ has a unique base $$b$$ representation, for the sake of contradiction suppose that it has two different representations, which implies that $\sum _ { i = 0 } ^ { m } a _ i b ^ i = n + 1 = \sum _ { i = 0 } ^ { k } c _ i b ^ i$ firstly we claim that $$m = k$$ for if that were not the case then without loss of generality $$m \lt k$$ and note the following \begin{align} \sum _ { i = 0 } ^ { m } a _ i b ^ i &\le \sum _ { i = 0 } ^ { m } \left( b - 1 \right) b _ i \\ &\le \left( b - 1 \right) \sum _ { i = 0 } ^ { m } b _ i \\ &\le \left( b - 1 \right) \frac{b ^ { m + 1 } - 1 }{b - 1} \\ &= b ^ { m + 1 } - 1 \lt b ^ { m + 1 } \end{align} Note that since $$m \lt k$$ we know that $$m + 1 \le k$$ then we know that $b ^ { m + 1 } \le b ^ k \le \sum _ { i = 0 } ^ { k } c _ i b ^ i$ which is a contradiction because chaining the last few inequalities we obtain $n + 1 = \sum _ { i = 0 } ^ { m } a _ i b ^ i \lt \sum _ { i = 0 } ^ { k } c _ i b _ i = n + 1$ thus we deduce that $$m = k$$.

Now we additionally claim that $$a _ m = c _ k$$ for if that were not the case then without loss of generality we assume that $$a _ m \lt c _ k$$ and therefore $\left( n + 1 \right) - a _ m b ^ m = \sum _ { i = 0 } ^ { m } a _ i b ^ i - a _ m b ^ m = \sum _ { i = 0 } ^ { m - 1 } a _ i b ^ i$ but also \begin{align} \left( n + 1 \right) - a _ m b ^ m &= \sum _ { i = 0 } ^ { k } c _ i b ^ i - a _ m b ^ m \\ &= c _ k b ^ k + \cdots + \left( c _ m - a _ m \right) b ^ m + \cdots + c _ 0 b ^ 0 \end{align} and since $$c _ m - a _ m \gt 0$$ this is valid base $$b$$ representation of $$\left( n + 1 \right) - a _ m b ^ m$$, but clearly they differ as $$a _ k \neq 0$$ and so the length of the two representations are different. This a contradiction because by our induction hypothesis they must have the same base $$b$$ representation, therefore our assumption that $$n + 1$$ has two different base $$b$$ representations is false and therefore by strong induction the original statement is true.

Decimal Representation
The decimal representation of a number $$n$$ is the representation of $$n$$ in the base $$10$$
The decimal representation of a number is the usual way you've most likely dealt with numbers so far, for example when you write $$123$$ you are referring to the number $$1 \cdot 10^{2} + 2 \cdot 10^{1} + 3 \cdot 10^{0}$$. Note
binary representation
The binary representation of a number $$n$$ is the represetnation of $$n$$ in the base $$2$$
Binary Natural Numbers
We define the following: $\mathbb{ B } := \operatorname{ base\_repr } \left( \mathbb{ N } _ 0, 2 \right)$
Number of Digits Required in a Base 2 Representation
Given any $$n \in \mathbb{ N } _ 0$$ prove that the number of digits in it's base 2 representation is given by $\left\lfloor \log _ 2 \left( n \right) \right\rfloor + 1$

Let $$n \in \mathbb{N}$$, suppose $$n$$ requires $$d$$ digits in it's base 2 representation, we'd like to show that $$d = \lfloor \log_2(n) \rfloor+1$$.

At most we have $$n = 1 \ldots 1$$, that is $$d$$ 1's in a row. But we know that is

$\sum_{i=0}^{d-1} 2^i = 2^d - 1$

At a minimum the first digit is a 1 and the rest are zeros (since the powers start at 0 right to left, this is $$2^{d - 1}$$).

We now have the following bound on $$n$$.

$2^{d-1 } \le n \le 2^d - 1$

Taking log base 2 on on the above inequality yields: (Call this inequality $\beta$)

$d - 1 \le \log_2 (n) \le \log_2(2^d -1 )$

*Note*: Taking the log respects the inequalities because $\log_2(\cdot)$ is a strictly increasing function (check it's derivative)

We will attempt to take the floor of $$\beta$$. Since $$d-1$$ is an integer $$\lfloor d-1 \rfloor = d-1$$, as for $$\log_2(2^d-1)$$, we must look a little closer.

Since $$2^{d-1} \le 2^d - 1 < 2^d$$ and , we know that

$d-1 = \log_2(2^{d-1}) \le \log_2(2^d - 1) < log_2(2^d) = d$

Therefore $$\lfloor \log_2(2^d - 1) \rfloor = d-1$$ and so the result of taking the floor of $$\beta$$ yields

$d - 1 \le \lfloor \log_2 (n) \rfloor \le \lfloor \log_2(2^d -1 ) \rfloor = d-1$

In other words

$d - 1 \le \lfloor \log_2 (n) \rfloor \le d-1$

So

$d - 1 = \lfloor \log_2 (n) \rfloor \Leftrightarrow d = \lfloor \log_2 (n) \rfloor + 1$
Base Plus Minus One To a Power is Congruent to Plus Minus One
For any $$b \in \mathbb{ N } _ 2$$ and for any $$n \in \mathbb{ N } _ 1$$ and let $$x$$ be the character we use to represent $$b - 1$$ then $\left( 10 ^ n \right) _ b = \left( x \ldots x \right) + 1$
Division by 9 Rule
Suppose that $$n = \left( a _ k \ldots a _ 0 \right)$$ then $$9 | n$$ if and only if $9 | \sum_{i = 0}^{k - 1} a_{i}$
We know that $$10 ^ n \equiv 1 ^ n \left( \operatorname{ mod } 10 \right)$$ and therefore \begin{align} n &= \sum _ { i = 0 } ^ { k } a _ i 10 ^ i \\ &\equiv \sum _ { i = 0 } ^ {k } a _ i \left( \operatorname{ mod } 9 \right) \end{align}