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Simple Proof From Loomis and Sternberg's Calculus

  1. Jun 23, 2014 #1
    Hello everyone,

    I am currently working through Loomis and Sternberg's Advanced Calculus, and am having difficulty with rather simple proofs. Despite this, I am NOT having much difficultly with proofs that require more sophisticated mathematics. For instance, I am trying to prove that if ab = 0, then either a = 0 or b = 0. I have tried to add zero in manifold ways, yet have not been successful in proving this claim.

    I have two questions: could someone provide me with some hints; and is this an early indication of my failure as a mathematician? To me, problems of this sort are not truly going to display your ability as a mathematician, as they require merely exhausting all possibilities. Does anyone else share this sentiment?

    Thank you.
     
    Last edited: Jun 23, 2014
  2. jcsd
  3. Jun 23, 2014 #2

    micromass

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    Not at all. I have often found such "easy" proofs to be the hardest.

    Anyway, what axioms can you use?
     
  4. Jun 23, 2014 #3

    pasmith

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    One axiom of the real numbers is that the non-zero reals are closed under multiplication; in other words, if you multiply any two non-zero real numbers together, you get a non-zero real number.
     
  5. Jun 23, 2014 #4

    MarneMath

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    Have you tried to use the contraposition? For me, this makes the proof a bit more mangable to work with. Also, don't feel too bad about 'simple' proofs being difficult. I think a lot of times these 'simple' proofs require a subtleness that is often easy to overlook.
     
  6. Jun 23, 2014 #5
    micromass, that is heartening to know. I attached a picture containing the properties that I am permitted to use.
     

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  7. Jun 23, 2014 #6

    micromass

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    OK, so you are working in a vector space. And you need to prove that if ##\alpha\boldsymbol{v}=\boldsymbol{0}##, then either ##\alpha=0## or ##\boldsymbol{v}=\boldsymbol{0}##.

    Have you tried multiplying with ##\alpha^{-1}##?
     
  8. Jun 23, 2014 #7
    But do I truly know what alpha inverse is? That was my first thought, to multiply by the inverse element, but according to the axioms, I do not know such a thing as of yet.
     
  9. Jun 23, 2014 #8

    verty

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    It took me a while to see it, look carefully at the domain of the scalar multiplication operation. This is rather tricky, I must admit.
     
  10. Jun 23, 2014 #9
    Is it R x V?
     
  11. Jun 23, 2014 #10

    micromass

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    But ##\alpha## is a real number. That structure has different axioms.

    This axiom system defines addition and scalar multiplication on the set ##V## to make a vector space. But in order to have a vector space, you need another structure first. You need ##\mathbb{R}## which already has a addition and multiplication. So you can use this structure.
     
  12. Jun 23, 2014 #11
    Oh, okay. In that case, if we use the properties of R, then it is relatively simple to show that the statement ax = 0 implies a = 0, but, in comparison, it is greatly difficult to show that v = 0.
     
  13. Jun 23, 2014 #12

    micromass

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    But this is false.
     
  14. Jun 23, 2014 #13
    Whoops, I got the two cases mixed up.
     
  15. Jun 23, 2014 #14
    Okay, so I understand how to show that av = 0 implies that v = 0, by multiplying by a^-1; but how might I show the statement av = 0 implies a = 0?
     
  16. Jun 23, 2014 #15

    verty

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    If the scalars were from ##\mathbb{Z}/6\mathbb{Z}##, this wouldn't be provable. We could have 3<2,2> = <0,0> but neither part is zero.
     
    Last edited: Jun 23, 2014
  17. Jun 23, 2014 #16

    micromass

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    No, this is not what you need to show and it isn't true.

    It doesn't and this is not what you need to show.
     
  18. Jun 23, 2014 #17
    Then what am I suppose to show?
     
  19. Jun 23, 2014 #18

    BruceW

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    Is this true? think more about it.

    edit: maybe you are just not explaining fully what you mean. But anyway, it is important to be precise in what you say.
     
  20. Jun 23, 2014 #19

    verty

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    Bashyboy, I saw in an older post what you were using Spivak before, but now you are using this book. Both of these are pretty difficult books. Why are you choosing these tough books? I mean this proof was difficult, it was almost like a review question. You may be biting off more than you can chew here, or that anyone can chew without some preparatory exposure.
     
  21. Jun 23, 2014 #20

    BruceW

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    we are allowed to assume the scalars are the reals. So it is not that difficult.

    edit: although, I don't know if the rest of the book will be too difficult. I have not read either of them.
     
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