I had a question about the following theorem.(adsbygoogle = window.adsbygoogle || []).push({});

Basis Representation Theorem:Let [tex] k [/tex] be any integer larger than 1. Then, for each positive integer [tex] n [/tex], there exists a representation

[tex] n = a_{0}k^{s} + a_{1}k^{s-1} + ... + a_{s} [/tex]

where [tex] a_{0} \neq 0 [/tex], and where each [tex] a_{i} [/tex] is nonnegative and less than [tex] k [/tex]. Furthermore, this representation of [tex] n [/tex] is unique; it is called the representation of [tex] n [/tex] to base [tex] k [/tex].

Proof:Let [tex] b_{k}\left(n\right) [/tex] denote the number of representations of [tex] n [/tex] to the base [tex] k [/tex]. We must show that [tex] b_{k}\left(n\right) [/tex] always equals 1.

Suppose that

[tex] n = a_{0}k^{s} + a_{1}k^{s-1} + ... + a_{s-t}k^{t} [/tex]

where neither [tex] a_{0} [/tex] nor [tex] a_{s-t} [/tex] equals zero. Then

[tex] n - 1 = a_{0}k^{s} + a_{1}k^{s-1} + ... + a_{s-t}k^{t} - 1 = a_{0}k^{s} + a_{1}k^{s-1} + ... + \left(a_{s-t} - 1\right)k^{t} + k^{t} - 1 = a_{0}k^{s} + a_{1}k^{s-1} + ... + \left(a_{s-t} - 1\right)k^{t} + \sum_{j=0}^{t-1} \left(k - 1\right)k^{t}[/tex]

Thus we see that for each representation of [tex] n [/tex] to the base [tex] k [/tex], we can find a representation of [tex] n-1 [/tex]. Consequently,

[tex] b_{k}\left(n\right) \leq b_{k}\left(n-1\right) [/tex]

Question:In the previous line, why is there a "less than or equal to" sign rather than an "equal" or "greater than or equal to" sign?

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# Proof: Basis Representation Theorem

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