"Arithmetic?" proof

  • #1
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This is a self-assigned question. Not homework.

I may have the right answer, but would like some reviewing. It came to me while reading on a CS topic, this did not come from a math textbook, otherwise it would be slightly more in context.

I did not use any formula (so 2 would be empty) and 1 and 3 are in a PDF here.

1. Is my work correct?
2. Is there an easier way to prove it?
3. What exactly is this? I called it an "'Arithmetic?' proof", but I don't think that this would be its name.
 

Attachments

  • Proof.pdf
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Answers and Replies

  • #2
haruspex
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Looks right, and fairly straightforward. I doubt there's an easier way.
How about generalising it?
 
  • #3
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Dream case scenario I would get that
[tex]\text{Given that } S \text{ is an array of } m \text{ small nonnegative integers and } \alpha \text{ is a prime larger than the largest integer in } S[/tex]
[tex]\sum_{i=0}^{m-1}\left(\alpha^{m-(i+1)}\times S_{1_i} \right) \ne \sum_{i=0}^{m-1}\left(\alpha^{m-(i+1)}\times S_{2_i}\right) \forall S_1 \ne S_2[/tex]
Obs.: for convenience, S is zero-based (lame math).

I think that this is going to the bottom of my TODO list.
 
  • #4
RUber
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Well, that is clearly true.
Alpha doesn't even need to be prime for that to be the case. It just has to be larger than the largest integer in either S1 or S2.
Think about alpha = 10 and our number system where S is the index set of digits from 0 to 9. No two numbers are the same unless they have the same digits becuase 10 is bigger than the largest member of the index set.
 

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