Prove that there is only one unique base b representation of any natural number.

Question

I have been asked to prove that for any integer base $b \geqslant 2$, every natural number has a unique base $b$ representation. I am not sure if this has been answered somewhere already, but I could not find a general answer that can be applied to any base.

Would the division algorithm apply here? I know that it would serve to give every number a unique quotient and remainder, but I'm not sure how exactly to craft the proof. I appreciate any input, thanks in advance!!

Answer 1

Let $N$ be a natural number. You want to write it as $$ N = a_{0} + a_{1} b + a_{2} b^{2} + \dots + a_{k} b^{k} $$ for a suitable $k$, and $0 \le a_{i} < b$.

Now rewrite the above as $$ \begin{cases} N = a_{0} + q b,\\ 0 \le a_{0} < b \end{cases}$$ to see that $a_{0}$ is uniquely determined as the remainder of the division of $N$ by $b$.

Now consider $$ q = a_{1} + a_{2} b + \dots + a_{k} b^{k-1} $$ and repeat, that is, use induction.

Answer 2

Below we show uniqueness of radix rep is a special (integer root) case of the Rational Root Test.

If $\,g(x) = \sum g_i \:\! x^i$ is a polynomial with integer coefficients $\,g_i\,$ with $\,0\le g_i < b\,$ and $\,g(b) = n\,$ then we call $\,(g,b)\,$ a radix $\,b\,$ representation of $\,n.\,$ It is unique: $ $ if $\,n\,$ has another rep $\,(h,b)\,$ with $\,g \ne h,\,$ then $\,f := g-h\ne 0\,$ has root $\,b\,$ and all coefficients $\,\color{#c00}{|f_i| < b},\,$ contra the below slight generalization of: $ $ integer roots of integer coef poly's $f(x)\in\Bbb Z[x]$ divide its constant term $\color{#0a0}{f(0)}\,$ [an obvious special case of the Rational Root Test].

Theorem $ $ If $\,f(x) = x^k(\color{#0a0}{f_0}\!+f_1 x +\cdots + f_n x^n)=x^k\bar f(x)\,$ is a polynomial with integer coefficients $\,f_i\,$ and with $\,\color{#0a0}{f_0\ne 0}\,$ then an integer root $\,b\ne 0\,$ satisfies $\,b\mid f_0,\,$ so $\,\color{#c00}{|b| \le |f_0|}$

Proof $\ \ 0 = f(b) = b^k \bar f(b)\overset{\large b\,\ne\, 0}\Longrightarrow\:\! 0 = \bar f(b),\,$ so subtracting $\,f_0\,$ from both sides yields $$\color{#0af}{-f_0 =\, b}\,(f_1\!+f_2 b+\,\cdots+f_n b^{n-1})\, \Rightarrow\,\color{#0af}{b\mid f_0} \underset{\large \color{#0a0}{f_0\,\ne\, 0}}\Longrightarrow\, |b| \le |f_0|\quad\small {\bf QED}$$

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Remark $ $ This is very closely related to the common proof using the uniqueness of the remainder (e.g. as in Andreas's answer) since the key result that the above proof depends on is this property: if $\,\color{#c00}{0\le g_i,h_i < b}\,$ then $\,h_i\neq g_i\Rightarrow b\nmid h_i-g_i ,\,$ i.e. remainders $\!\bmod b\,$ are unique (proof here).

In modular arithmetic language the key inference is

$$g_i\:\! b^i \equiv h_i\:\! b^i\!\!\!\!\pmod{\!b^{i+1}}\!\!\underset{{\rm cancel}\ \large b^{\large i}}\Longrightarrow g_i\equiv h_i\!\!\!\!\pmod{\!b}\ \ \color{#c00}{\rm so}\ \ g_i = h_i\qquad\qquad$$

Answer 3

Let two representations of a number be such that they differ in the $k^{th}$ digit ($k$ zero-based, from the right).

Then

$$(n\text{ div } b^k)\bmod b$$

has different values for these two representation, which is contradictory ($\text{div}$ denotes integer division).

Answer 4

The OP asked

Would the division algorithm apply here?

Here we provide, working in the natural numbers $\Bbb N =\{0,1,2,\dots\}$, a proof of uniqueness that doesn't use full blown Euclidean division; we only need the following weaker result.

Lemma 1: Let $b,k,h,r \in \Bbb N$ satisfy

$$\tag 1 kb = hb + r \; \text{ with } r \lt b$$

Then $k = h$ and $r = 0$.
Proof
If $k \lt h$ then $kb \lt hb \le hb + r$, so $\text{(1)}$ is false.
If $k \ge h$ then write $k = h + u$ with $u \ge 0$. Substituting into $\text{(1)}$,

$$ kb = (h + u)b = hb + ub = hb + r $$

so it must be true that $ub = r$. If $u \ge 1$, $\; ub \ge b$ and that can't happen since $r \lt b$. So $u = 0$ and $k = h$, and substituting into $\text{(1)}$,

$$\quad hb = hb + r $$

and so $r = 0$. $\quad \blacksquare$

Note: You can use an alternate proof for lemma 1: If both $k$ and $h$ are greater than $0$ they both have a predecessor, and lemma 1 is equivalent to the statement obtained by replacing both both $k$ and $h$ with these smaller numbers. So there is a 'truth chain' allowing us to assume that $k=0$ or $h=0$.
If $k = 0$ both $h$ and $r$ must be $0$.
If $h = 0$ then we have $r = kb$. If $k \ge 1$ then $r \ge b$ but that can't happen since $r$ is assumed to be less than $b$. So $k = h = r = 0$.

Let $b \ge 2$.

Theorem 2: Let a number $n \ge 0$ have two $\text{Base-b}$ representations $x$ and $y$. Then, using Capital sigma notation with $0\text{-padding}$ if necessary, we can express the equality of these two representations by writing

$$ \tag 2 \sum_{i=0}^k x_i b^i = \sum_{i=0}^k y_i b^i \; \text{ with the coefficients } x_i \text{ and } y_i \text{ all between } 0 \text{ and } b-1$$

Then if $j \in \{i \, | \, 0 \le i \le k\}$, $\;x_j = y_j$, i.e. the coefficients are all equal.
Proof
If the two families $(x_i)$ and $(y_i)$ are not identical, let $j$ be the first integer where $x_j \ne y_j$. Cancelling the initial (if any) equal terms from both sides of equation $\text{(2)}$, there would remain

$$ \tag 3 \sum_{i=j}^k x_i b^i = \sum_{i=j}^k y_i b^i $$

Assume that $x_j \gt y_j$ with $x_j = y_j + u$. Cancelling the $y_jb^j$ term from each side of $\text{(3)}$ allows us to write

$$ \tag 4 ub^j + \sum_{i={j+1}}^k x_i b^i = \sum_{i={j+1}}^k y_i b^i \text{ with } 0 \lt u \lt b$$

since it must be true that $k \ge j+1$. But then, by factoring out $b^j$ and using algebra, we can write

$$ \tag 5 u + kb = hb$$

for some numbers $k$ and $h$. Since $u \lt b$, applying lemma 1 we must conclude that $u = 0$. But that can't happen since we assumed that $x_j \ne x_j$.

The reductio ad absurdum argument is constructed in a like manner when $y_j \gt x_j$.

So assuming that $x_j \ne y_j$ leads to a contradiction.$\quad \blacksquare$

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BONUS MATERIAL

We can also prove the existence of $\text{Base-b}$ representations without using Euclidean division (see this). It is an easy exercise to show that the existence of $\text{Base-b}$ representations can be used to demonstrate the existence part of the Euclidean division theorem.

Exercise: Prove the uniqueness part of the Euclidean division theorem using lemma 1.
Hint: Use the same technique used in theorem 2, taking the difference $u$ between two numbers,

Answer 5

Here is an argument that works using generating functions. The coefficient of $x^m$ in the generating function needs to count the number of ways of writing $m$ as $$ m=a_0+a_1n+a_2n^2+\cdots+a_ln^l,\quad0\leq a_i\leq n-1$$ This generating function is $$(1+x+\cdots+x^{n-1})(1+x^n+\cdots+x^{(n-1)n})\cdots(1+x^{n^l}+\cdots+x^{(n-1)n^l})\cdots$$ Each factor is a geometric sum, therefore $$1+x^{n^l}+x^{2n^l}+\cdots+x^{(n-1)n^l}=\frac{1-(x^{n^l})^n}{1-x^{n^l}}=\frac{1-x^{n^{l+1}}}{1-x^{n^l}}$$ Therefore the generating function is $$\prod_{l=0}^\infty\frac{1-x^{n^{l+1}}}{1-x^{n^l}}=\frac{1-x^n}{1-x}\cdot\frac{1-x^{n^2}}{1-x^n}\cdot\frac{1-x^{n^3}}{1-x^{n^2}}\cdots=\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$ This shows that there is a unique representation of $m$ in base $n$.