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Proving that 1 is positive

  1. Jul 15, 2013 #1
    This is from Lang's "A First Couse in Calculus".

    In chapter 1 he makes two statements regarding positivity:

    POS 1. If ##a##, ##b## are positive, so is the product ##ab## and the sum ##a+b##.

    POS 2. If ##a## is a number, then either ##a## is positive, or ##a=0##, or ##-a## is positive, and these possibilities are mutually exclusive.

    ("POS" meaning positive or postulate I guess.)

    His proof that 1 is positive (using only the above statements) goes like this:

    "By POS 2, we know that either 1 or -1 is positive. If 1 is not positive, -1 is positive. By POS 1, it must then follow that (-1)(-1) is positive. But this product is equal to 1. Consequently, it must be 1 which is positive, and not -1."

    It seems like this proof hinges on that fact that (-1)(-1) is positive. However, since both (1)(1) and (-1)(-1) are positive, from this alone we really can't conclude that 1 or -1 are negative. Right? To me it'd make more sense to say that (-1)+(-1) is not positive, and by POS 1 -1 cannot be positive. By the same logic we can then claim that 1 is the positive one.

    I know this is probably an elementary mistake on my part, but I'm lost.
     
    Last edited: Jul 15, 2013
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  3. Jul 15, 2013 #2

    micromass

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    Do you understand this proof? Do you somehow disagree with it? It's not clear to me.

    Well, we know that ##(-1)(-1)=1##. So if ##(-1)(-1)## is positive, then ##1## is positive.

    How would you know that ##(-1)+(-1)## is not positive?
     
  4. Jul 15, 2013 #3
    I don't understand the reasoning in the proof; I'm not questioning it's validity. I apologize if I was unclear.

    (-1)+(-1) is not positive, since it's < 0. Or, geometrically one can view it as going further left on a number line, further from 0 (ie. becoming more negative).

    I think this is what I don't understand. Wouldn't (-1)(-1)=1 imply that -1 is positive, since it's product is positive? :confused:.
     
  5. Jul 15, 2013 #4

    micromass

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    Of course it's not positive. But my question is, how would you prove it using only POS 1 and POS 2?

    The axiom says: "If a and b are positive, then ab is positive". You somehow seem to think that if ab is positive then a and b are positive. That is not true at all.
    It's not because a product of two elements is positive, that the elements themselves are.
     
  6. Jul 15, 2013 #5
    I really don't see how you could.

    I understand that the product of two numbers may be positive if both are positive, or both are negative. But from the manner in which POS 1 is defined, it seems like he's saying that, because the product ab is positive, then a and b are positive. Obviously this isn't true, but I don't understand this logic.
     
  7. Jul 15, 2013 #6

    micromass

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    Exactly, this is why your proof doesn't work.

    POS 1 doesn't say that at all. I don't really understand why you think that it says that. POS 1 clearly says that if a and b are positive, then ab is positive. It says nothing about when ab is positive.
     
  8. Jul 15, 2013 #7
    I wasn't claiming to prove that in terms of only POS 1 and POS 2.

    I feel like I'm angering you. I'm just going to go think some more about it; I'm slow at these things.
     
  9. Jul 15, 2013 #8

    micromass

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    You're not angering me at all! My apologies if I gave you that impression.
    I simply don't understand what exactly your problem is, so I can't really help until I understand.
     
  10. Jul 15, 2013 #9

    lurflurf

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    The argument is
    suppose -1 is positive
    then by pos 1
    (-1)(-1)=1 is positive
    this violates pos 2
    thus -1 must not be positive

    another approach would be
    suppose either -1 or 1 is positive
    that is 1 is not 0 (pos 2)
    (-1)(-1)=1
    (1)(1)=1
    by pos 1 we conclude
    1 is positive
     
  11. Jul 15, 2013 #10
    How does (-1)(-1) being positive violate POS 2?
     
  12. Jul 15, 2013 #11

    D H

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    Contradiction. The assumption that -1 is positive means that 1 is positive. A number and its additive inverse cannot both be positive by POS2.
     
  13. Jul 15, 2013 #12

    SteamKing

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    If a = -1 and b = -1, then by Postulate 1, in order to conclude that a and b are both positive, we must calculate the product ab and the sum a+b.

    Now ab = (-1)(-1) = 1 : so far so good.
    But a + b = -1 + -1 = -2 : there is a different result here which is not positive

    We can conclude thus that -1 is not positive. It's the word 'and' in the postulate which is key, where it specifies that the product AND the sum of a and b must be positive.
     
  14. Jul 15, 2013 #13

    HallsofIvy

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    I suspect that it stands for "positive" since these are axioms or postulates describing "positive" numbers.

    Yes, both (1)(1) and (-1)(-1) are positive. However, we can from "Pos 2" (the "trichotomy" axiom) conclude that (since 1 is not the additive identity it is NOT 0) 1 is either positive or negative. If 1 is positive, we are done. So assume that 1 is negative and derive a contradiction. That is, assume that -1 is positive.

    HOW do you say that? Of course, (-1)+ (-1)= -2 but how do you know that is not positive?

     
  15. Jul 15, 2013 #14
    I'm also beginner like wifi, but maybe I can rephrase this informally. (Please correct me also in case of fallacy)

    I want to know whether -1 or 1 is positive. The clear thing is 1 is not 0 because of the existence of multiplicative identity. Any numbers squared ##a^2## will always be positive. If we multiply (-1)(-1), then it will be positive. If we multiply (1)(1) it will also be positive. But what is (1)(1)?

    We also know for certain that (1)(1) = 1 because of the identity axiom, not only because of pos1 and pos2. (a)(1) = (a)

    If -1 is positive then (-1)(-1) should be (-1), and consequently (1)(1) = -1. Surely this is not right.

    The important point I think is the special case that ##a^2## will always be positive, where a is not 0. I think if we only know that (1)(1) = (1) and the special case of number squared \, it's already enough to prove that 1 > 0. (Is this right?)

    I also think it might be possible to agree on the fundamental definition to switch sign of positive and negative. So we treat 1 as negative and -1 as positive.

    I know (1)(1) = 1. Still, I'm not sure whether it's positive or negative but I know ##a^2## must be positive for any non-zero number. In that case 1 is positive, and so (1)(1). If it's positive it's not negative. Hopefully this is not a circular argument and you can understand it.
     
    Last edited: Jul 15, 2013
  16. Jul 15, 2013 #15

    D H

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    You do not have to calculate a+b. Calculating ab is all that is needed.

    There is two items left out of the proof, one of which is showing that (-1)(-1)=1. The supplied proof merely says "But this product is equal to 1." Proving that (-1)(-1) is indeed 1 is trivial.

    If -1 is positive, then so is 1 by POS1 and the fact that (-1)(-1) is 1. But this contradicts POS2. That's all that is needed to show that -1 is not positive.

    What about 1? Assuming 1 is positive does not lead to a contradiction. Lack of a contradiction is not sufficient, because POS1 says nothing about the possibility of 1 being zero. Dealing with that possibility is the other item left omitted from the proof.
     
  17. Jul 15, 2013 #16
    This is exactly how I see it!! How is this not correct?
     
  18. Jul 15, 2013 #17

    D H

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    You don't know whether -2 is positive or negative. Using the sum doesn't accomplish a thing.
     
  19. Jul 15, 2013 #18
    Hm. I think I'm getting closer to understanding this.

    So, at a certain point in the proof we've basically proved that -1 and 1 are both positive. This certainly isn't the case, because by POS 2 we know only one of them can be positive. But how we do we know that it's 1 that's positive?
     
  20. Jul 15, 2013 #19

    D H

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    You know that -1 is not positive, so by POS2 -1 can only be zero or negative. Thus 1 is either zero or positive. All that's left is to show that 1 is not zero.
     
  21. Jul 15, 2013 #20
    I wonder if it would be possible to leave the contradiction argument entirely by deciding with certainty that 1 is in fact positive and not 0, i.e knowing (1)(1) = 1 and squaring any non-zero number will produce a positive number.
     
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