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Sorry in advance for my lack of mathematical knowledge, I am very new to it.


Yesterday, I posed this question to myself:

"In a world without addition or subtraction, how could we derive the next value in the sequence of natural numbers from $1\to\infty$ with a step size of $1$?"

This lead me to the idea of multiplication to find the next value in a sequence. After analyzing the multipliers between each natural value using: $$ \frac{(n+1)}{n} $$

I noticed the pattern of this sequence starts at the high values of $2$ and $1.5$, then converges to a value of $1$.


My two questions:

  • Is it right to assume that the sequence of multipliers should have a more predictable sequence?
  • Are there more elegant ways of producing the next natural number without addition or subtraction?
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    $\begingroup$ Hi Vita. First of all let me congratulate you with your mathematical curiosity! Keep asking yourself questions like these, and never worry about whether or not they are stupid questions! $\endgroup$ – Joachim Jul 19 '15 at 19:40
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    $\begingroup$ @john why not ? $p=1\cdot p$. I don't see why we need two "non-trivial numbers". $\endgroup$ – Dietrich Burde Jul 19 '15 at 19:54
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    $\begingroup$ How do you define "next natural number" in a world without +1? $\endgroup$ – Siméon Jul 19 '15 at 19:59
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    $\begingroup$ @siméon: hadn't just this been the point/the nerve of the couriosity of the OP? $\endgroup$ – Gottfried Helms Jul 19 '15 at 20:10
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    $\begingroup$ Something like $n+1=\log_2(2\cdot2^n)$? $\endgroup$ – g.kov Jul 19 '15 at 20:56
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With the function $2^n$ and it's inverse, $\log_2$ available, $n+1=\log_2(2\cdot2^n)$.

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If we were allowed to use floor and ceiling functions in this strange world without $+$ and $-$, then perhaps we could use the following function which would generate the next integer after $n$:

$$f(n)=\lceil{(n\times\frac{\lceil n\sqrt2\rceil}{\lfloor n\sqrt2\rfloor})\rceil}$$

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  • $\begingroup$ This is very nice! Thank you for this answer! $\endgroup$ – Vita Pluvia Jul 19 '15 at 21:52

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