Claim:The base b representation is unique

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Revision as of 16:14, 8 October 2018 by {{GENDER:Mdg39|[math]'"2}} [/math]'"7
(<math>1) </math>2 | <math>3 (</math>4) | <math>5 (</math>6)

As usual, in order to justify calling the base b representation of [math]a [/math] "the" base b representation, we need to prove that it is unique. In fact, it is not necessarily unique if we allow leading 0's (for example, [math](0123)_{10} = (123)_{10} [/math]), but if we rule them out, then we have the following fact:

For all [math]k [/math], if [math]\href{/cs2800/wiki/index.php/Base}{(d_kd_{k-1}\dots{}d_1d_0)_b} = \href{/cs2800/wiki/index.php/Base}{(d_k'd_{k-1}'\dots{}d_1'd_0')_b} [/math] and if [math]d_k \neq 0 [/math] and if [math]d_k' \neq 0 [/math] then [math]\href{/cs2800/wiki/index.php/Sequence_notation}{(d_i)} = \href{/cs2800/wiki/index.php/Sequence_notation}{(d_i')} [/math].