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Implication math problem

  1. Sep 11, 2007 #1
    A chapter I am reading says that with "[tex]\frac{1}{x}[/tex]=[tex]\frac{1}{2x+1}[/tex][tex]\Rightarrow[/tex]2x+1=x", the [tex]\Rightarrow[/tex]cannot be replaced by [tex]\Leftrightarrow[/tex], but if a = b, does not 1/a necessarily = 1/b? Is this a misprint or are they right? If they are right, could you illustrate this with an example please?
  2. jcsd
  3. Sep 11, 2007 #2
    a=b implies 1/a=1/b iff [itex]a=b\neq 0[/itex]. In this case, x can't be 0, so it won't be the case, but maybe you didn't know that until after that particular step. Once you know that x isn't 0, you can put an equivalence arrow there, but in general you cannot.
  4. Sep 11, 2007 #3
    Many thanks Eighty. In this case it seemd to me that the second equation implied that the denominators were not 0 and thus implied the first equation, so is a particular result (that the denominator in not 0) not allowed in a chain of implication? I'm still a bit puzzled here.
  5. Sep 11, 2007 #4


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    You are right only with the benefit of hindsight (after having solved for x). Before you solve for it, x can be anything.
  6. Sep 12, 2007 #5
    I don't know what preliminary steps the author took in an attempt to ensure the non-replacement claim.
    You haven't provided enough information.
  7. Sep 12, 2007 #6

    There were no preliminary steps; it was just an answer to an exercise. I am aware of the a = b = 0 possibility but 2x+1 = x does not allow this.

    I think what I'm really asking is this: can one statement imply a second by virtue of implying a third?

    Can 2x+1 = x imply that [tex]\frac{1}{2x+1}[/tex] = [tex]\frac{1}{x}[/tex] by virtue of implying that neither x, nor 2x + 1 = 0? and is there any way to write such a thing mathematically in one go?
  8. Sep 12, 2007 #7


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    I don't know how, short of solving for x or assuming x [itex]\ne[/itex] 0 ex ante.
  9. Sep 12, 2007 #8
    Yes, when the third implies the second. Ie [tex]a\Rightarrow b\Rightarrow a[/tex] gives you [tex]a\Leftrightarrow b[/tex]. The chain of implications really is:
    [tex]1/x=1/(2x+1) \Rightarrow x=2x+1 \Leftrightarrow x=-1 \Rightarrow 1/x=1/(2x+1)[/tex]
    Since [tex]1/x=1/(2x+1) \Rightarrow x=2x+1[/tex] and [tex]x=2x+1\Rightarrow 1/x=1/(2x+1)[/tex], we have [tex]1/x=1/(2x+1)\Leftrightarrow x=2x+1[/tex]

    But you could just write merely [tex]1/x=1/(2x+1)\Leftrightarrow x=2x+1[/tex] if you want. It's true. But if you're doing a proof, you may want to use the implication arrows to indicate your train of thought (if the statement isn't trivial).
  10. Sep 13, 2007 #9
    Many thanks Eighty, EnumaElish and fopc. Eighty's answer, then, seems to be saying that the two-way arrow is O.K.here, and that the book is wrong to say that it is not?
    Last edited: Sep 13, 2007
  11. Sep 15, 2007 #10
    Apparently no details are given by the author (even in the exercise statement). So you'll have to make some assumptions.
    It's a hypothetical formula, so I think it's safe to say it's universally quantified. No problem.
    But, the problem domain is another matter. I suspect the author meant for it to be R, because
    if he meant say R+ or R-{0}, then clearly <-> holds and the book is wrong.
    So assume R. But consider:

    Look at the antecedent, and remember that it's universally quantified (for every x in R ...).
    Notice there's an element sitting in R that turns it into nonsense.
    Specifically, 1/0 = 1/(0+1) is nonsense. It is not true, and it is not false.
    This is not the same as saying it has a truth value, but we don't know which one.
    It does not have a truth value. So now you have a bound antecedent (a closed subformula)
    that does not have a truth value for every x in R.

    Last point: Note that the problem says nothing about x = -1.
    If this is introduced, then you have a different problem.
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