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Spivak: Basic properties of numbers

  1. Aug 28, 2011 #1
    Hello all! :smile:

    I am (painfully) going through the first chapter of Spivak's Calculus. At one point he introduces the property: if a, b, and c are any numbers, then [itex]a\cdot(b+c) = a\cdot b +a\cdot c[/itex]. He then uses this property in an example in which he shows that the only time that [itex] a - b = b- a[/itex] is when a = b.


    [itex]a - b = b - a [/itex]
    [itex] \Rightarrow (a - b) + b = (b - a) + b = b + (b - a) [/itex]
    [itex]\Rightarrow a = b + b - a [/itex]
    [itex]\Rightarrow a + a = (b + b - a) + a = b + b [/itex]
    [itex]\mathbf{ \Rightarrow a \cdot (1 + 1) = b \cdot (1 + 1) } [/itex]
    [itex]\Rightarrow a = b [/itex]


    I am unsure of why the step in bold was necessary? Can someone elaborate as to why he included this step? In fact, he said it was necessary to include this step.
     
    Last edited: Aug 28, 2011
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  3. Aug 28, 2011 #2

    micromass

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    Well, how would you do it without the bold step??
     
  4. Aug 28, 2011 #3
    That step is basically justification of x + x = 2x.
     
  5. Aug 28, 2011 #4
    It would help greatly if you explained why you don't think it's necessary. How would you complete the proof?
     
  6. Aug 28, 2011 #5
    See edit.

    Hmmm. Good question :smile: It just seems weird to do that step (to me). I guess I see where this is going:

    I would have said that a + a = b + b simply means 2a = 2b but I suppose that is what he is driving at? That in terms of the properties he has just introduced a + a = a (1 + 1) which is 2a.


    EDIT: Wow: You guys pounced all over this one :smile: Better watch myself roaming into this forum.
     
  7. Aug 28, 2011 #6

    He's trying to be rigorous and follow from the propositions he proved before.
     
  8. Aug 28, 2011 #7

    micromass

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    Indeed, he just wants to do 2a=2b and eliminate the 2. But he didn't introduce 2 yet, so he calls it 1+1.

    The argument should be more rigorous though. He should also show that [itex]1+1\neq 0[/itex]. You should really prove everything from the axioms. Don't use anything else.
     
  9. Aug 28, 2011 #8
    I think that this is in an exercise somewhere. But seriously, why should this be proved? Of what benefit does this offer? And moreover, once one shows that [itex]1 + 1 \ne 0[/itex] then do I need to show that [itex] 1 + 1 + 1 \ne 0[/itex] and that [itex]1 +1 +1 +1 \ne 0 [/itex]? Where does it end?

    Also, I am being serious, not facetious. I have an engineer's background in math, so for me, there is really no need for all that rigor.
     
  10. Aug 28, 2011 #9
    It ends whenever you stop...

    Proof is the backbone of mathematics. It lets us know that we're correct and gives us a glimpse into why. Whether you need to prove that [itex]1 + 1 + 1 \ne 0[/itex] or not depends on how you go about your proofs. For instance, you can use mathematical induction to prove that no number of additions of the multiplicative identity of the real numbers will equal the additive identity. It's important to note that there are other number systems where this is not true: where you can add 1 a number of times to get 0, so it's nice to prove that this can't be the case for real numbers...

    I'm sure we all get that. I know I do. Do you understand that there is a place for this kind of rigor?

    Incidentally, why do you find going through Spivak "painful?"
     
  11. Aug 29, 2011 #10
    Just out of curiosity, how does one prove [itex]1 + 1 \neq 0[/itex]?
     
  12. Aug 29, 2011 #11
    You define what addition is...
     
  13. Aug 29, 2011 #12

    I like Serena

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    Is there an axiom that requires the additive inverse to be unique?
     
  14. Aug 29, 2011 #13
    So we basically define 1 + 1 = 2 and that 2 does not equal 0?
     
  15. Aug 29, 2011 #14
    No but you can prove this. For instance:

    Suppose we have a real number a and two additive inverses, b and c. By definition, a + b = 0 and a + c = 0. It follows that a + b = a + c. Therefore b = c.
    QED.
     
  16. Aug 29, 2011 #15
    Actually, I don't think you need to prove each specific example. By definition, [itex]1 + 1 = 2 \neq 0[/itex]. However, while you can do this for any number of additions of the multiplicative identity, this doesn't prove that it's true for any number of additions. This would require some kind of proof...

    Personally, with this, I would use induction and the ordering of the reals...

    I suspect that the reason Spivak didn't simply use a + a = 2a is because you'd derive that from the distributive property like so: a + a = (1 + 1) a = 2a. So for him, he's actually skipping a step by simply cancelling out (1 + 1)...
     
  17. Aug 29, 2011 #16
    Spivak is the route for many into formal maths.

    Saladsamurai, as an engineer you will have doubtless used the technique of practicing on a simple known and often trivial example before tackling more complicated stuuf or simply as a verification of a technique.

    I suggest taking Spivak's and JH examples in this vein.

    go well
     
  18. Aug 29, 2011 #17
    So is it actually defined that [itex]1 + 1 = 2 \neq 0[/itex] for most constructions of the reals? I only ask because in most constructions of the reals that I've seen this point seems largely neglected. In fact, I haven't bothered to think about it before now.
     
  19. Aug 29, 2011 #18
    Hi studiot :smile: Yes you are correct. I am taking a bit of a detour now that I am done with my formal engineering studies. It's just that Spivak has a lot of little side notes like the one that spawned this thread. I can certainly move forward without proving that [itex]1 + 1 \ne 0[/itex]. But as I move forward, I wish to know what motivates him to suggest that this is even worth worrying about (if only once).

    I am still not sure if I have gotten an answer to that question. Both Spivak and micromass found it non rigorous that the proof in post #1 did not include a step in which it was shown that [itex]1 + 1 \ne 0[/itex]. I suppose I am beginning to see it though.

    If 1+1=0 was true, then the step in bold could not be multiplied on both sides by (1+1)-1 and hence the proof would not hold.

    Just getting used to looking at things from a different perspective. :wink:
     
  20. Aug 29, 2011 #19
    It would be undefined.
     
  21. Aug 29, 2011 #20

    micromass

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    The thing is that mathematicians like Spivak try to prove statements without assuming anything but the axioms. Spivak has presented a number of axioms (about 15 or so), and we should prove everything from these 15 axioms. We cannot use anything else. That is what Spivak wants to do.

    Of course, you won't need to prove that [itex]1+1\neq 0[/itex] in order to proceed. You will still understand the theory. In fact, maybe you'll even find the book easier if you don't prove this. However, proving such a thing is benificial because it gives you acquaintance with easy proofs.

    I guess that it's a point-of-view thing. Spivak assumes nothing but the axioms. He assumes nothing else. That's his point-of-view. I hope that explains why Spivak does the things that he does...
     
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