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I dont want the proof of this statement unless it is necessary for my questions, I just want some clarification.

  1. If cr = 1 would cr-1 = 0?

  2. How is cj 1 or 0? I understand cj is an arbitrarily picked number within the theorem 'line' but why is it limited to a rather boolean value?

Sad CRUD Developer
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2 Answers2

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The theorem is precisely asserting the existence and uniqueness of binary expansions of any positive integer $n$. That's what the $c_j$'s are: the digits of the binary expansion.

I hope the previous paragraph answers your second question. As for the first: no, the $c_r$'s can be arbitrary elements of $\{0,1\}$, with the one constraint that the last one of them is $1$ (if they were all $0$ we would get $0$, not a positive integer; and since the expansion is finite there must be a last nonzero term). So for instance $(c_0,c_1,c_2,c_3,c_4) = (1,1,1,1,1)$ is the binary expansion of $1+2+4+8+16= 31$.

Pete L. Clark
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  • Just a quick comment for [ 1 + 2 + 4 + 8 + 16] - for cr * 2^r (in the expression) it would only make sense if r is 4 so its 1 * 2^4 = 16.. and continue for r being 3, then 2 all the way to 1. Is this right? I get the feeling r should be 5 since we are dealing with 5 numbers. – Sad CRUD Developer Jan 30 '14 at 00:17
  • Yes, in my example $r = 4$ since the highest power of $2$ that appears is $2^4 = 16$. (Yes there are five terms...but we start at $0$.) – Pete L. Clark Jan 30 '14 at 00:33
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If you mean $c_r-1$ would be zero, yes. If you mean must $c_{r-1}$ must be zero, no.

And since we're proving the existence of a binary representation of an integer, then it should not be all that surprising that we have only 0 and 1 available.

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