Spivak calculus, page 22( 3rd edition).

  • Context: Undergrad 
  • Thread starter Thread starter Alpharup
  • Start date Start date
  • Tags Tags
    Calculus Spivak
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 2K views
Alpharup
Messages
226
Reaction score
17
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 can't I say, since 1 were present, 0 must also be present. Else, the second rule is not satisfied. Am I missing something?
 
Physics news on Phys.org
sharan swarup said:
Else, the second rule is not satisfied.
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}##.
 
  • Like
Likes   Reactions: Alpharup
Krylov said:
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}##.
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?
 
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.
 
  • Like
Likes   Reactions: Alpharup
sharan swarup said:
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?
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."
 
  • Like
Likes   Reactions: Alpharup
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)."
 
  • Like
Likes   Reactions: jim mcnamara and Alpharup
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.
 
  • Like
Likes   Reactions: Alpharup
Thank you...cleared my doubts