Different Representations to Number Base represent Different Integers
Theorem
Let $k \in \Z$ such that $k \ge 2$.
Let $a$ and $b$ be representations of integers in base $k$ notation:
- $a = \ds \sum_{j \mathop = 0}^r a_j k^j$
- $b = \ds \sum_{j \mathop = 0}^s b_j k^j$
such that either:
- $r \ne s$
or:
- $\exists j \in \set {0, 1, \ldots, r}: a_j \ne b_j$
Then $a$ and $b$ represent different integers.
Proof
First suppose that $r \ne s$.
Without loss of generality, suppose $r > s$.
Then from Bounds for Integer Expressed in Base k:
\(\ds a_r k^r\) | \(>\) | \(\ds b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(>\) | \(\ds b\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(\ne\) | \(\ds b\) |
Otherwise $r = s$.
Let $l$ be the largest such that $a_l \ne b_l$.
Without loss of generality, suppose $a_l > b_l$.
For all $j > l$, let $a_1 = a - a_j k_j$ and $b_1 = b - b_j k_j$.
As $a_j = b_j$ in this range, $a_j k_j = b_j k_j$ and so the same amount is being subtracted from both.
So consider $\paren {a_l - b_l} k_l$.
From Bounds for Integer Expressed in Base k:
- $\paren {a_l - b_l} k_l > \ds \sum_{j \mathop = 0}^{l - 1} a_j k^j$
and:
- $\paren {a_l - b_l} k_l > \ds \sum_{j \mathop = 0}^{l - 1} b_j k^j$
and so $a_1 > b_1$.
Hence:
- $a_1 + a_j k_j > b_1 + b_j k_j$
Hence:
- $a \ne b$
and hence the result.
$\blacksquare$
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Sources
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {1-2}$ The Basis Representation Theorem: Exercise $7$