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I Is this a correct way to describe number sets?

  1. Sep 15, 2016 #1
    Hello,

    I am excited to be learning about number sets again, :P.

    Ok, so this is how I describe them:

    N={1,2,3,4,5,6,7,8,9,10….∞} Z={-∞….-3,-2,-1,0,1,2,3…∞}

    Q=a/b ,a∈Z,b∈{Z∖0}

    The first three are correct. However, how do we describe irrational numbers? Is it just the difference set of Real - rational - integer - natural? or something else?


     
  2. jcsd
  3. Sep 15, 2016 #2

    Math_QED

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    Irrational numbers don't have a symbol. For the irrational numbers, we write:

    ##\mathbb{R}\backslash\mathbb{Q}##

    Since we exclude ##\mathbb{Q}##, we also exclude ##\mathbb{N}## and ##\mathbb{Z}## as they are subsets of ##\mathbb{Q}##
     
  4. Sep 15, 2016 #3

    fresh_42

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    ##+ \infty## and ##-\infty## are not numbers and shouldn't appear in the definition. However, you may define sets which include them, but then they aren't number sets anymore and ##\frac{a}{b}## is not allowed in order to define ##\mathbb{Q}##, because ##a,b \in \{\pm \infty\}## don't allow ##\frac{a}{b}##.

    And as @Math_QED has said: irrational literally means not rational, i.e. not in ##\mathbb{Q}##.
     
  5. Sep 15, 2016 #4
    is it because the rational numbers nest integers which nest natural numbers? so bascially was i correct but i forgot about the "nesting" or this realtionship:

    N⊂Z⊂Q⊂R
     
  6. Sep 15, 2016 #5
    so how would one define rational numbers using proper set notation?
     
  7. Sep 15, 2016 #6

    Math_QED

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    I edited my first post, and as you can see you are right.
     
  8. Sep 15, 2016 #7
    but the number sets are infinite? isnt it important to indicate this?
     
  9. Sep 15, 2016 #8

    mfb

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    The dots do that.
    This should get brackets:

    Q={a/b ,a∈Z,b∈{Z∖0} }

    There is no similar way to define all irrational numbers, you have to introduce them as limit process (every irrational number can be seen as limit of a series of rational numbers) or via other things.
     
  10. Sep 15, 2016 #9

    Math_QED

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    Yes, but according your definition, for example ##\pm\infty \in \mathbb{Z}## and these are no numbers like ##1, -3 , 0, 325##. There are many reasons for this. For example, you may know that ##\forall a,b \in \mathbb{Z}: a + b \in \mathbb{Z}##. If ##\pm\infty \in \mathbb{Z}##, then this would be false since ##+\infty + (-\infty)## is undefined. @fresh_42 gave the example that ##\frac{\infty}{\infty}## does not make sense in ##\mathbb{Q}## (eg this is undefined).

    You are right though that these sets are infinite, but you indicate this in a wrong way.
     
  11. Sep 15, 2016 #10

    fresh_42

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    Just leave out ##\{\pm \infty\} ## and let the dots do their work.
    If you want to have a mathematical reasoning, then it goes like this:

    ##\mathbb{N} := \{1,2,3,4, \dots\}## is a half-group according to addition.

    ##\mathbb{Z} := \{\dots ,-3,-2,-1,0,1,2,3,4, \dots\}## is the group generated by the half-group ## \mathbb{N}##.

    It happens to be that ##\mathbb{Z} ## is also a ring and even an integral domain. Therefore it allows

    ##\mathbb{Q} := \{\frac{a}{b}\,\vert \, a,b \in \mathbb{Z}\,,\, b \neq 0\}## as its quotient field.

    ##\mathbb{R} := \overline{ \mathbb{Q} }## is the completion of ##\mathbb{Q}##, i.e. filling in the gaps.

    ##\mathbb{C} := \mathbb{R} [ i ] ## with ## i^2 + 1 = 0 ## is the algebraic closure of ##\mathbb{R}##, i.e. making all polynomial equations having a solution.
     
  12. Sep 15, 2016 #11

    Math_QED

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    I'm currently reading about the natural numbers and one can define ##\mathbb{N}## as the unique set that satisfies the Peano Postulates.

    Peano Postulates:

    There exists a set ##\mathbb{N}## with an element ##1 \in \mathbb{N}## and a function ##s: \mathbb{N} \rightarrow \mathbb{N}## satisfying 3 properties:

    1) There is no ##n## such that ##s(n) =1##
    2) ##s## is injective
    3) Let ##G \subset \mathbb{N}## be a set. Suppose that ##1 \in G##, and that if ##g \in G##, then ##s(g) \in G##. Then ##G = \mathbb{N}##
     
  13. Sep 15, 2016 #12

    fresh_42

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    It helps to call ##s## the successor function. Personally I can't memorize them in their exact wording. I like to think of ##\mathbb{N} ## as "there is a ##1## and all numbers have a successor (Peano)" which allows counting.
     
  14. Sep 15, 2016 #13

    Math_QED

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    Yes ##s## is the successor function, but it's hard to see (at least for me) from these axioms that ##s: n \mapsto n +1##. I'm actually currently reading the proof that this is possible, and it requires the recursion theorem.
     
  15. Sep 15, 2016 #14

    fresh_42

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    I guess, it would be for me, too. The good part is, that those abstract approaches force you not to take anything for granted. I once (started) to read a book about group theory that basically didn't contain any formulas. Almost everything has been said in plain text. Plus the groups have been defined differently than the Bourbaki version. And I remember when I learned the Dedekind cuts. I had trouble to see any real number at all.

    Have fun!
    (And as it is internet: this was not meant to be ironic.)
     
  16. Sep 15, 2016 #15

    micromass

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    There is no unique set satisfying Peano. There are many sets. For example the naturals satisfy it. But also ##\{\pi, 2\pi, 3\pi,...\}## or ##\{2,3,5,7,11,...\}## with ##s(p) = \text{the next prime}##.

    What is true is that the Peano axioms give rise to a set unique up to isomorphism. This is an important distinction! There is no way in math to define the natural numbers uniquely (although there are standard definitions, but these are not the only ones). We can only define them up to isomorphism.

    And this is an important theme in math. Mathematicians don't care about the exact nature of the objects, but rather how they behave. If an object behaves like the natural numbers should behave in all respects, then it is the natural numbers for us, no matter how it is defined.
     
  17. Sep 15, 2016 #16

    micromass

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    Yes. I personally don't think Dedekind cuts define real numbers. I think the real numbers are an object in our imagination, something we have a lot of intuition for. But not something that we can mathematically describe.
    The purpose of the Dedekind cuts is not to define real numbers. It is to define a structure inside our mathematical universe that has the same properties as a structure in our intuition. As such, we can easily reason with the real numbers and get results that would hold for the real numbers. But I would never say a Dedekind cut IS a real number.
     
  18. Sep 15, 2016 #17

    Math_QED

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    Those sets don't contain 1 as element, which the Peano Postulates require, but I assume that there are other sets that satisfy the postulates if you say so.
     
  19. Sep 15, 2016 #18

    micromass

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    The number ##1## is just a name for the first element in the Peano set. We can give this name to a lot of numbers. The Peano axioms assert the existence of some element that we call ##1## for convenience. We might as well call it ##\gamma##. The set ##\{\pi, 2\pi, 3\pi, ...\}## most definitely is a Peano structure with first element ##\gamma = 1##.

    In fact, without the Peano axioms, there is no number ##1##. We use the Peano axioms to define the number ##1##, and we can define it in a lot of ways.
     
  20. Sep 15, 2016 #19
    :eek:o_O I have understood all up to post #9..i do not think I can study enough in the time this debate lasts to understand the other posts..but feel free to discuss further.
     
  21. Sep 15, 2016 #20

    fresh_42

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    You would be surprised what might happen, if you asked something innocent like: "Did we invent or discovered natural numbers?" or even simpler "Is ##0## a natural number?" :biggrin:
     
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