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Contrapositive of a (if p and q, then r) statement?

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data
    I hope this is the right place to post this.

    For my linear algebra homework, I have to prove that
    "If [itex]\vec{u} \neq \vec{0}[/itex] and [itex]a\vec{u} = b\vec{u}[/itex], then [itex]a = b[/itex]."

    2. Relevant equations

    3. The attempt at a solution
    I'm trying to prove the contrapositive, but I'm not sure how to alter the statement.

    I think it's something like
    "If [itex]a \neq b[/itex], then [itex]\vec{u}=\vec{0}[/itex] or [itex]a\vec{u} \neq b\vec{u}[/itex]."

    But then I don't know where to begin, so I'm guessing a proof by contrapositive isn't the right method to use here. It seems to me that a direct proof might me simpler, but I guess I wanted to challenge myself. Any ideas? Thanks.
     
  2. jcsd
  3. Nov 8, 2012 #2

    Zondrina

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    A direct proof would be simpler here. Basically what you're trying to prove is the cancellation law.

    I'm sure you're in a field here, particularly a vector space. What do you know about the properties of a vector space? Think about axioms.
     
    Last edited: Nov 8, 2012
  4. Nov 8, 2012 #3
    Well we know quite a few axioms, and we also know other properties as a result of earlier proofs.

    My "direct proof" goes like this:
    Let [itex]U[/itex] be a vector space and [itex]\vec{u} \in U[/itex]. Let [itex]a,b[/itex] be arbitrary scalars.

    Given that [itex]a\vec{u} = b\vec{u}[/itex], by substitution we have
    [itex]\frac{1}{a} (a\vec{u}) = \frac{1}{a} (b\vec{u})[/itex].

    By axiom 9 (an associative property),
    [itex](\frac{1}{a} \cdot a) \vec{u} = (\frac{1}{a} \cdot b) \vec{u}[/itex]
    [itex](1) \vec{u} = \frac{b}{a} \vec{u}[/itex]

    So for equality to hold, we must have
    [itex]1 = \frac{b}{a}[/itex], and from this it follows that
    [itex]a = b[/itex]. Q.E.D.

    I wrote this up pretty quickly, so I agree that the direct proof is easier. Is there a way to work with the contrapositive, though? It doesn't look like it to me.
     
  5. Nov 8, 2012 #4

    Zondrina

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    Wait what substitution? You multiplied by the left inverse of a, that's not a substitution. Also, your vector u doesn't automatically cancel out of thin air. The point is, you have to argue as to why u is going to cancel.

    You have that u is non-zero and au = bu. You want to show that a = b.

    Since u is a non-zero vector... hmm ;). Another hint is to abandon the fraction notation and use inverse notation.
     
  6. Nov 8, 2012 #5
    My professor considers "multiplying both sides by the same thing" to be substitution. I think what he means by that is since, for example, for some vectors u and v, if u=v then au = av. Does that clear it up?

    As for getting rid of the vector u, I just matched up the scalar coefficients. I don't how else to describe what it is I did there... I guess I'm saying I don't actually get rid of u. How is this normally shown in a proof knowing that u is non-zero?

    Edit: I think I can answer my own question.
    [itex](1)\vec{u} = \frac{b}{a}\vec{u}[/itex]
    [itex](1)\vec{u} + (-\frac{b}{a}\vec{u}) = \frac{b}{a}\vec{u} + (-\frac{b}{a}\vec{u})[/itex]
    [itex](1)\vec{u} - \frac{b}{a}\vec{u} = \vec{0}[/itex]
    [itex](1 - \frac{b}{a})\vec{u} = \vec{0}[/itex]

    Since u is non-zero, it follows that (1 - b/a) = 0, and then I solve to get a=b. Is this right?
     
    Last edited: Nov 8, 2012
  7. Nov 8, 2012 #6

    Zondrina

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    If u is non-zero, then it has an inverse? What happens if you apply the inverse to what you're trying to show.
     
  8. Nov 8, 2012 #7
    We haven't learned anything about inverse vectors, but I think I get what you're saying here. I figured another way which I added to my last comment.

    Edit: Oops, I misinterpreted "inverse." I was thinking multiplicative for some reason...
     
  9. Nov 8, 2012 #8

    Zondrina

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    Ah you haven't been taught about it. The problem with what you're doing is you're assuming a has an inverse. What if a=0? The whole thing goes down the drain.
     
  10. Nov 8, 2012 #9
    Sorry, I was mistaken. We actually did learn about the inverse of a vector, namely the negative of a vector. I don't know... I knew of the additive inverse but it didn't quite register.

    And it looks like I have to provide a proof for when either a or b are zero. Thanks for reminding me.
     
  11. Nov 8, 2012 #10

    Dick

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    Why are you dividing by anything? You've got (a-b)u=0 where u is nonzero. Isn't that enough?
     
  12. Nov 8, 2012 #11
    I believe the contrapositive here would be what you guessed it to be. As others have said, I think its easier to prove directly. For fun, let's prove the contrapositive, worded better for proving:

    "For any vector [itex]u[/itex], in some vector space [itex]V[/itex], and any scalars [itex]a,b[/itex], in some field [itex]F[/itex], consider the equation [itex]au=bu[/itex], with truth of equality undetermined. If [itex]a \neq b[/itex], then either [itex]u=0[/itex] (so that [itex]au=bu[/itex]), or [itex]au \neq bu[/itex]."

    Let's begin the proof. For the first case, suppose [itex]u\neq 0[/itex]. Because by assumption [itex]a\neq b[/itex], we thus have [itex]a-b\neq 0[/itex]. Therefore, we can conclude that

    [itex](a-b)u\neq 0[/itex]

    From this we can easily conclude by moving terms around that it must be that [itex]au\neq bu[/itex]. (How?) Next, suppose instead (working on the second and final case) that [itex]u=0[/itex]. To be clear, we want to draw from this that [itex]au=bu[/itex]. Of course, that is obvious.

    Note that, because this is a mathematical statement, the "or" component means we could have both parts of the conclusion of the statement, i.e. both that [itex]u=0[/itex] and [itex]au \neq bu[/itex]. However, this cannot be. (Why?) So, I guess for this theorem (problem), it must be that we can only have one or the other condition, i.e. it is an "exclusive-or" conclusion, rather "inclusive-or."

    This concludes the proof.

    My bad if I did anything wrong. I'm a bit tired, but I feel like there's enough good info there to give reason for me to submit this reply, anyway
     
  13. Nov 9, 2012 #12
    I looked over the proof last night and I realized that now (that I have (a - b)u = 0). Thanks!

    And to 5hassay, I think I follow. Since it's an exclusive-or statement, would that mean I would only have to show one for the statement to hold?
     
  14. Nov 9, 2012 #13

    If I understand correctly, since our conclusion is a "result 1 OR result 2" conclusion, and it appears we can only have one or the other, but not both, i.e. the "OR" as an "exclusive-or," then we must show both that result 1 can occur while result 2 does not occur, and, likewise, that result 2 can occur while result 1 does not occur
     
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