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Spivak calculus, page 22( 3rd edition).

  1. Oct 20, 2015 #1
    Iam using Spivak these days for learning calculus. In page 22, I have difficulty understanding. He speaks about natural numbers. Do natural numbers always start with 1?
    He talks about the definition of a set of natural numbers as having
    1. Always 1 in set.
    2. If k is present, k+1is also present.
    Then cant I say, since 1 were present, 0 must also be present. Else, the second rule is not satisfied. Am I missing something?
  2. jcsd
  3. Oct 20, 2015 #2


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    I don't see a contradiction with the second rule. That rule doesn't say: "If ##k## is present, then ##k - 1## is also present". Whether ##\mathbb{N}## includes ##0## seems largely a matter of convention. In almost all literature that I read the convention is that ##0 \not\in \mathbb{N}##.
  4. Oct 20, 2015 #3
    Let us consider a set -2,-1,0,1,2,3,...
    This set also satisfies the two properties. Shouldnt we call them as natural numbers?
  5. Oct 20, 2015 #4


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    You can include or not include the ##0## in the set of natural number. It is a convention, sometimes they denote with ##\mathbb{N}^{*}## the set of natural numbers without the zero.
  6. Oct 20, 2015 #5


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    No, I'd rather not. Still, I don't see how the absence of ##0## from ##\mathbb{N}## would contradict the two rules in your OP or, put differently, I don't see how the second rule would necessarily imply the presence of ##0## in ##\mathbb{N}##.

    I don't have Spivak's book but you are right: If these are the only rules he states in his definition of ##\mathbb{N}##, then something must be missing. In fact, I have a rather practical stance towards these things and like Kronecker's often-quoted comment: "God made the natural numbers; all else is the work of man."
  7. Oct 20, 2015 #6


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    I don't have Spivak here with me but what you give as a "definition" of the natural numbers is not a definition at all because it does not define "1" before it uses it. "Peano's axioms", that these are presumably derived from, are:
    There exist a set of objects, N, called "numbers", together with a function, s, called the "successor function", from N to itself, such that
    1) There exist a unique object in N, 1, such that s maps N one to one and onto N- {1}.
    2) If X is a subset of N such that if [itex]1\in X[/itex] and whenever [itex]x\in X[/itex], [itex]s(x)\in X[/itex] then X= N.

    It follows immediately that if x is any natural number other than 1, there exist y such that x= s(y).

    We define "addition", a+ b, by
    "If b= 1 then a+ b= a+ 1= s(a). If b is not 1 then there exist c such that b= s(c) and we define a+ b= a+ s(c)= s(a+ c)."
  8. Oct 20, 2015 #7


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    Spivak is not defining the natural numbers there but stating the principle of mathematical induction. this principle assumes you know what the natural numbers are, and want a criterion that guarantees when a given subset of the natural numbers is actually all of them. So he states that any subset of natural numbers which has the two properties you gave above actually equals all natural numbers. and yes, for Spivak natural numbers begin with 1.
  9. Oct 20, 2015 #8
    Thank you...cleared my doubts
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