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RolyPoly

  • 11 months ago

"The numbers 1, 1+1=2, 2+1=3, etc. are said to be natural; it is assumed that none of these numbers is zero". Why do we have/need such an assumption?

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  1. kc_kennylau
    • 11 months ago
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    Because it makes that \(\forall A,B\in\mathbb N, \frac AB\in\mathbb N\). @ikram002p am I right?

  2. RolyPoly
    • 11 months ago
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    We can also have A = 1, B = 3, but 1/3 \(\notin \mathbb N\)?

  3. kc_kennylau
    • 11 months ago
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    then i'm not right lol

  4. kc_kennylau
    • 11 months ago
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    oh, i see why 0's not natural, coz you can't have 0 things.

  5. kc_kennylau
    • 11 months ago
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    NATURal numbers are numbers that occur in the NATURE.

  6. RolyPoly
    • 11 months ago
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    -_- You can have nothing, right?

  7. RolyPoly
    • 11 months ago
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    There's something you CAN'T find in the nature though!

  8. kc_kennylau
    • 11 months ago
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    How do you count?

  9. RolyPoly
    • 11 months ago
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    From 0

  10. kc_kennylau
    • 11 months ago
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    cool

  11. kc_kennylau
    • 11 months ago
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    "There is no universal agreement about whether to include zero in the set of natural numbers: some define the natural numbers to be the positive integers {1, 2, 3, ...}, while for others the term designates the non-negative integers {0, 1, 2, 3, ...}. The former definition is the traditional one, with the latter definition having first appeared in the 19th century." https://en.wikipedia.org/wiki/Natural_number

  12. Spacelimbus
    • 11 months ago
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    Not sure if I understand the concept of this question. But I remember the following words from my Professor "First thing, you always do when you buy a boot about Mathematics, check on their definition of \(\mathbb{N}\)" What he meant by that, was that whether or not \(0 \in \mathbb{N}\) or not is still a big discussion. For references check for example Zorich Analysis 1, Blatter Analysis and so on. Also consider the notation \(\mathbb{N}_0\)

  13. ikram002p
    • 11 months ago
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    cuz 0 is the identity of any groupe on the binary operation of addition , so it must be UNIQE its thm from (elementary properties of groups)

  14. RolyPoly
    • 11 months ago
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    The footnote of this quotation was as follows: "Given two elements N and E, say, we can construct a field by the rules N+N=N, N+E=E, E+E=N, N\(\cdot\)N=N, N\(\cdot\)E=N, E\(\cdot\)E=E. Then, in keeping with our notation, we should write N=0, E=1 and hence 2=1+1=0. To exclude such number systems, we require that all natural field elements be nonzero" Basically, I don't know what's going on here.

  15. kc_kennylau
    • 11 months ago
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    The problem is that the question itself is already wrong. It isn't universal to exclude 0 from the list of natural numbers.

  16. RolyPoly
    • 11 months ago
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    But it also isn't universal to include 0 as well?

  17. RolyPoly
    • 11 months ago
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    Warning: Please ignore the following. ---------------------------------------------------------------- Should 0 be included in the set of natural numbers? ---------------------------------------------------------------- <Saying I> No, \(0\notin \mathbb N\) The numbers 1, 1+1=2, 2+1=3, ... are said to be natural. Given two elements N and E, we can struct a field by the rules (i) N+N=N (ii) N+E=E (iii) E+E=N (iv) N\(\cdot\)N = N (v) N\(\cdot\)E = N (vi) E\(\cdot\)E = E Then in keeping with our notation, we should be N=0, E=1. From (iii), E+E = N, so, we have 1+1 = 0 However from the first statement, we have 1+1=2 \(\ne\) 0, so we need to exclude 0. Problems: 1) Why would we construct such a field with rules (i) to (vi), particularly with rule (iii)? 2) What would happen if we picked some other values for N and E?

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spraguer (Moderator)
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