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Spivak bad first impression

  1. Jun 7, 2014 #1
    Hi everybody!

    I have started working through Spivak's "Calculus". I was slightly upset with the first chapter. The book made a few assumptions. (The concept of a number and that if a=b then a+c=b+c, stuff like that) I am fine with it for now because I know where the gap in knowledge is but when I get further in the book, I won't be able to recognize things like that.

    I was wondering if the book kept assuming things like that later in the book?

    Thank you for your time!
     
  2. jcsd
  3. Jun 7, 2014 #2

    adjacent

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    It's not difficult.It's elementary algebra.
    if a=b then you can replace one of those variables:
    a+c=b+c - replace a with b because a=b.
    Then you will get b+c=b+c
    Two b's cancel out so you have c=c. :wink:
     
  4. Jun 7, 2014 #3
    I apologize if this is rude, but I really can't tell, and I am sorry if it is. Are you making a joke?
     
  5. Jun 7, 2014 #4

    adjacent

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    Why do you think I am making a joke? I am sorry if I offended you by saying it's elementary algebra.

    I said that because it's simple.
    You were given that if ##a=b## then ##a+c=b+c##
    Here ##c## can be anything and if ##a=b## we can interchange ##a## and ##b##.
    So we can either write ##b+c=b+c## or ##a+c=a+c##.
    In both cases, the other variables cancel out and what's left is ##c=c## so it is correct to say that if ##a=b## then ##a+c=b+c##.

    Why were you uncomfortable with assumptions like that?
     
  6. Jun 7, 2014 #5
    Ahhh ok.

    Truly sorry about that. Really I am.

    I misunderstood your proof.

    I would actually rewrite it like this.

    ##a = b## - Given
    ##a + c = a + c## (I am not not quite sure what this is called)
    ##a + c = b + c## because a = b.

    I was a little confused when you talked about cancelling things out but this Theorem allows for one to cancel, and the conclusion wasn't ##a + c = b + c##.

    But anyways, I was able to catch that assumption when reading the book. I might not be able to catch a similar assumption later in the book. I am wondering if the rest of the book has these little assumptions.

    EDIT : Also it assumes the definition of even or odd.
     
  7. Jun 7, 2014 #6

    462chevelle

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    There is no need to rewrite anything. its just a fact. if c is equal to say.. 2.
    that +2 on both sides doesn't even matter.since you could just subtract it from both sides and it states that a=b because c=c
     
  8. Jun 7, 2014 #7
    The problem is you can't subtract yet. We are trying to prove the theorem which allows us to do that.
     
  9. Jun 7, 2014 #8
    We can't add a number to both sides yet, that is the theorem we are trying to prove. Subtraction is just a consequence of being able to add, after we have the axiom of ## a + (-a) = 0##
     
  10. Jun 7, 2014 #9

    462chevelle

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    I wasn't saying to subtract it, just illustrating why the c is redundant.
     
  11. Jun 7, 2014 #10
    I think there is some disconnect here. Our tangent was that given a=b, we had to prove a+c=b+c.

    But anyways, does Spivak keep doing this throughout the book?
     
  12. Jun 7, 2014 #11

    micromass

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    Ok, I see what you mean. It is true that Spivak doesn't prove it, and it cannot be proven. But you can take it as an axiom. Spivak should have stated it, but very few books do. If you ever study mathematical logic in depth then you might see a justification of this.
     
  13. Jun 7, 2014 #12

    micromass

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    No, he doesn't. This is just one situation where he does it. After the first two chapters, such a thing shouldn't happen anymore.
     
  14. Jun 7, 2014 #13
    I worked through the historical section of Apostol I think I saw it in there. I had to return Apostol back to the library so I can't check now.

    Where is the fallacy in my revision of adjacent's proof?

    Does
    ##a+c = a+c## assume the conclusion?

    And to the original point, thank you Micromass.
     
    Last edited: Jun 7, 2014
  15. Jun 7, 2014 #14

    micromass

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    OK. (the second line is an axiom of equality by the way.

    I don't see how this follows from the previous two lines. You need an axiom or inference rule that does this.

    It is very rare that math books start by stating all the axioms. The only place you'll see this happening is in mathematical logic books. There, everything is built from the axioms rigorously. A site that does this is metamath: http://us.metamath.org/
    But usually books assume some axioms. In this case, Spivak assumes the axiomatics of sets and of equality. This means specifically, that he assumes that if ##A=B## and that if ##f## is a function, then ##f(A) = f(B)##. You can take this as an axiom. Furthermore, he assumes that addition is a function, thus ##f(B) = B+C## is a function. So it follows that if ##A=B##, then ##f(B) = f(A)##, and thus ##B+C = A+C##.
     
  16. Jun 7, 2014 #15
    Because the a=b, I replaced the second a with b.

    I will certainly check the website out!
     
  17. Jun 7, 2014 #16

    adjacent

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    So you can take anything as a axiom? Like ##1=3##?
    You said that he assumes addition as a function. How does he assume a function then? Is everything ultimately reduced to mere axioms?
     
  18. Jun 7, 2014 #17

    micromass

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    Sure, I see that. But who says you can do that?
     
  19. Jun 7, 2014 #18

    micromass

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    Yes, that would be a valid axiom, but one that does not reflect our world.

    By defintion, he considers a set ##F##, equipped with a function ##+:F\times F\rightarrow F## that satisfies some axioms. This is the mathematical way of saying the things that Spivak means. There is also a function ##\cdot: F\times F\rightarrow F## and a relation ##<##. All of these satsify some axioms.

    It should be, yes.
     
  20. Jun 7, 2014 #19
    Hmmm, I see. I am new to this so this is probably very naive. If a=b, isn't a and b interchangeable?

    EDIT : Nevermind, I looked at the page.
     
    Last edited: Jun 7, 2014
  21. Jun 7, 2014 #20

    micromass

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    Sure, it should be. But that's why you need axioms.

    Here is a simple example, consider the operation ##*## on ##\mathbb{Q}## defined as follows:

    [tex]\frac{a}{b}*\frac{c}{d} = \frac{a+c}{b+d}[/tex]

    This satisfies all of Spivak's axioms for ##+## (associative, zero element, inverses, commutative).

    But then for example, we have ##A = \frac{0}{1}##, ##B=\frac{0}{2}## and ##C=\frac{1}{1}##. Certainly, ##A=B##, but

    [tex]A*C = \frac{0}{1}*\frac{1}{1} = \frac{0+1}{1+1} = \frac{1}{2}[/tex]

    and

    [tex]B*C = \frac{0}{2}*\frac{1}{1} = \frac{0+1}{2+1} = \frac{1}{3}[/tex]

    So ##A=B##, but ##A*C\neq B*C##. So there are funny examples of "operations" that do not satisfy this. What goes wrong of course is that the function ##f(A) = A*C## is not a function, since ##A=B## does not imply here that ##f(A) = f(B)##.

    So Spivak should have added an extra axiom that ##+## is a function. Together with the axiom of equality that if ##A=B##, then ##f(A) = f(B)## for all functions ##f##, this proves what you want.
     
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