Is the proof given in the next section valid?
I've seen the proof of the AM–GM inequality using forward-backward-induction and I'm trusting that the inductive reasoning of my proof can hold up to scrutiny.
Here we insist that all representations are expressed using the familiar capital sigma notation with the lower bound of summation starting at 0. So if we have a representation for a natural number $m$ it is converted and we can write
$$\tag 1 m = \sum_{i=0}^k x_i b^i$$
So for example,
$$ b^{n+1} = \sum_{i=0}^{n+1} x_i b^i \text{ where } x_i = 0 \text{ for } i \le n \text{ and } x_{n+1} = 1$$
When we have a $\text{Base-b}$ representation, $m = \sum_{i=0}^k x_i b^i$ and $x_k \ne 0$, then we say that the representation has a leading coefficient of $x_k$ and if $x_k = 0$ that the representation has padding.
Given two representations, by allowing padding, we can insist that they both have the same upper bound of summation. By uniqueness we mean that any two representations of a number $m$ with the same upper bound of summation have the same coefficients.
For $n \ge 0$ construct the representations
$$ p_{(n,n+1)} = \sum_{i=0}^n (b - 1) b^i$$
Lemma 1: For $n \ge 0$, $\;p_{(n,n+1)} + 1 = b^{n+1}$.
Proof
We use induction.
For $k = 0$, $\;p_{(0,1)} + 1 = (b-1) + 1 = b^{0+1}$ and the base case is true.
Assume $\;p_{(k,k+1)} + 1 = b^{k+1}$. Then
$$ p_{(k+1,k+2)} + 1 = p_{(k,k+1)} + (b-1)b^{k+1} + 1 = b^{k+1} + (b-1)b^{k+1} = b^{k+2} $$
and the proof is complete. $\quad \blacksquare$
If all the coefficients of a representation for $m$ are equal to $b -1$ use padding to get another representation where the coefficient corresponding to the index of the upper bound of summation is $0$. Then by working with the 'carry over' logic found in lemma 1, we can get a representation for $m +1$. If $m$ is represented by $\text{(1)}$, then with $\gamma$ denoting the first index where $x_{\gamma} \lt b - 1$:
$\bullet$ Set $y_\gamma = x_{\gamma} + 1$
$\bullet$ For $i \lt \gamma$ set $y_i = 0$
$\bullet$ For $i \gt \gamma$ set $y_i = x_i$
to form the next representation
$$\tag 2 m + 1 = \sum_{i=0}^k y_i b^i$$
Moreover, it can be shown that if we have two different representation for $m$, these constructions create two different representations for $m + 1$.
Proposition 2: Every natural number has a $\text{Base-b}$ representation.
Proof
We prove this using induction.
For $m = 0$ define $x_0 = 0$ giving $m = \sum_{i=0}^0 x_i b^i$.
If we have a representation for $m$ then construct $m + 1$ using $\text{(2)}$. $\quad \blacksquare$
Lemma 3: The number $m = b^n$ has a unique representation.
Proof
We claim that any other representation has the same coefficients.
Let
$$\tag 3 b^n = \sum_{i=0}^k x_i b^i = \sum_{i=0}^{n-1} x_i b^i + x_n b^n +\sum_{i=n+1}^{k} x_i b^i$$
The rhs can't be greater than $b^n$ and using simple properties of the numbers we can drop the summation (if it is present) with a lower bound of $n + 1$, so we have
$$\tag 3 b^n = \sum_{i=0}^{n-1} x_i b^i + x_n b^n $$
Again, since the rhs can't be greater that $b^n$, $x_n = 0$ or $x_n = 1$
Case 1: $x_n = 0$
Since
$\quad \sum_{i=0}^{n-1} x_i b^i \le p_{(n-1,n)} \lt b^n$
this is not possible.
Case 2: $x_n = 1$
Then
$$\tag 4 b^n = \sum_{i=0}^{n-1} x_i b^i + b^n $$
and again, since the rhs can't be greater than $b_n$, $x_i = 0$ for $i \lt n$.
So the any other representation has $x_n = 1$ and all other coefficients must be equal to $0$. $\quad \blacksquare$
Theorem 4: Every natural number has one and only one representation $\text{Base-b}$ representation.
Proof
The existence is given by proposition 2.
To get a contradiction, let $m$ a natural number with two different representations. The exposition above shows that we can create two different representations for $m + 1$. We can continue doing this until we get two different representation for $b^n$. Applying lemma 3 we get a contradiction.
$\blacksquare$
