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Proposition: Let n $\in$ N. If $n = \sum_{i=0}^{p} x_i10^{i} = \sum_{i=0}^{q}y_i10^{i} $ where $p,q, \in Z_{\geq0}$, each $x_i$ and each $y_i$ is a digit, $x_p\neq0$ and $y_q\neq0$,

then $p=q$, and $x_i = y_i$ for all i

So far I have:

$n = \sum_{i=0}^{p} x_i10^{i} = \sum_{i=0}^{q}y_i10^{i} $

$\sum_{i=0}^{p} x_i = \sum_{i=0}^{q}y_i $

$\sum_{i=0}^{p} x_i - \sum_{i=0}^{q}y_i = 0$

I feel like I can perform some summation algebra to write this as one summation, but I'm not sure how.

How do I prove that $p=q$ and the sequence $x_i = y_i$ ?

beepbeepboop123123
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    Proof by contradiction, find the first difference, show that its impossible. – Rushabh Mehta Mar 29 '19 at 22:30
  • Let $x = \sum\limits_{i=0}^{p}x_i10^i$ and $y=\sum\limits_{i=0}^q y_i 10^i$. To show $p=q$: supposing that $q > p$, can you work out the minimum possible value of $y$ and the maximum possible value of $x$ and hence show that $y > x$ (so $y\ne x$)? (It might help if you think in terms of specific numbers a bit: "clearly" a $4$ digit number is greater than any $3$ digit number for example, since a $4$ digit number is at least $1000$ and a $3$ digit number is at most $999$.) – Minus One-Twelfth Mar 29 '19 at 22:33
  • This is a little confusing to me; I tried going along this path and I proved that the last term of each of the different sums, are unequal to each other. How does this help proving p = q and $x_i = y_i$? – beepbeepboop123123 Mar 30 '19 at 00:54
  • As you well know, the proof won’t be too much good for infinite expansions. How were you planning to use it to get to the claim in your title ? – Lubin Mar 30 '19 at 01:49
  • Possible duplicate: https://math.stackexchange.com/questions/3017323/is-there-an-easy-way-to-see-that-binary-expansion-is-unique/3017368#3017368 – Ethan Bolker Mar 30 '19 at 02:54

2 Answers2

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The key issue with the base $10$ representation being unique is that the digits are restricted to being between $0$ and $9$ inclusive (note that without this limitation, the representations may not necessarily be unique). The main reason this is important is that the difference in digits is limited to being between $-9$ and $9$, inclusive. To see why this is important, consider what you've stated, i.e.,

$$n = \sum_{i=0}^{p} x_i 10^{i} = \sum_{i=0}^{q} y_i 10^{i} \tag{1}\label{eq1}$$

where $p,q, \in Z_{\geq 0}$, each $x_i$ & $y_i$ is a digit of $0$ to $9$, $x_p \neq 0$ and $y_q \neq 0$. To make it a bit easier for processing in case $p \neq q$, let $r = \max(p,q)$ and extend $x_i$ to be $0$ for any $i \gt p$ and $y_i$ to be $0$ for any $i \gt q$, so \eqref{eq1} becomes

$$n = \sum_{i=0}^{r} x_i 10^{i} = \sum_{i=0}^{r} y_i 10^{i} \tag{2}\label{eq2}$$

Subtracting the middle & LHS parts of \eqref{eq2} gives

$$0 = \sum_{i = 0}^{r} \left(x_i - y_i\right) 10^i \tag{3}\label{eq3}$$

If $x_i \neq y_i$ for one or more $i$, let $k$ be the smallest one. Thus, for all $i \lt k$, $x_i = y_i$ so $x_i - y_i = 0$, causing those terms to all become $0$. They can therefore be ignored, simplifying \eqref{eq3} to

$$0 = \sum_{i = k}^{r} \left(x_i - y_i\right) 10^i \tag{4}\label{eq4}$$

Next, dividing both sides by $10^k$ gives

$$0 = \sum_{i = k}^{r} \left(x_i - y_i\right) 10^{i - k} \tag{5}\label{eq5}$$

Note that all of the terms on the RHS of \eqref{eq5} have at least one factor of $10$ except for the first one. Since $10$ divides $0$, plus all of the terms on the RHS of \eqref{eq5} after the first term, this means the first term of $\left(x_k - y_k\right)10^0 = x_k - y_k$ must be a multiple of $10$. Since $-9 \le x_k - y_k \le 9$, this means that $x_k - y_k = 0$, i.e., $x_k = y_k$, so they can't be different. This contradicts the assumption that there is at least one digit different, thus showing that all of the digits up to $r$ are the same. This also means the highest non-zero ones must be the same, i.e., $p = q$.

This same basic argument can be used to show that an integer can be uniquely represented in any base $b \ge 2$ as long as the digits used are in the range of $0$ to $b - 1$, inclusive (e.g., binary using $0$ and $1$, or hexadecimal using $0$ to $F$).

John Omielan
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  • If $xi≠yi$ for one or more i, let k be the smallest one. Thus, all lower differences will be 0... What do you mean by all the lower differences will be 0? – beepbeepboop123123 Mar 31 '19 at 20:45
  • @beepbeepboop123123 Since $k$ is the smallest value of $i$ where $x_i \neq y_i$, then for all $i$ (if any) less than $k$, we have that $x_i = y_i$, so $x_i - y_i = 0$, i.e., the differences are $0$. – John Omielan Mar 31 '19 at 20:49
  • @beepbeepboop123123 I used the term "lower" to mean for all smaller indices, but that wasn't clear to you, so I've updated my answer to try to explain that section more clearly. – John Omielan Mar 31 '19 at 21:08
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Rather the proving any two representations on $N$ must be equal, try proving $N$ has exactly one unique representation.

But the division theorem, for any integer $N$ the are are unique integers $M_1$ and $x_0$ so that $N = 10M_1 + x_0; 0 \le x_0 < 10$. This proves that $x_0$ must be uniquely determined.

Inductively for each $M_k$ there are unique $M_{k+1}$ and $x_{k}$ so that $M_k = 10M_{k+1} + x_k; 0\le x_k < 10$.

And that's it really. As $N$ is bounded above by some $10^{p+1}$ this will have a finite number of steps.

fleablood
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