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Existence/Uniqueness of x?

  1. Aug 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that there is a unique real number x such that for every real number y, xy + x -4 = 4y

    2. Relevant equations



    3. The attempt at a solution
    Givens:
    x, y are real numbers

    The goal has the form
    ∃!x(∀y(xy+x-4=4y))

    Existence:
    We want to prove that ∃x(∀y(xy+x-4=4y))
    First, we let y be an arbitrary real number, then we solve the equation xy+x-4 = 4y for x, to get

    x = 4y+4/(y+1)
    However, this equation has two special cases depending on if y does or doesn't equal -1, therefore I split up the Existence proof into two cases

    Case 1a) y = -1,
    then x(-1+1) = 4(-1) + 4 which turns into x(0) = 4(0) + 4 = 0
    The first and last parts of this equation tell us that x(0) = 0, but this is true for ALL X, therefore for any value of x this case where y = - 1 is true, therefore existence is proved in this case.

    Case 1b) y=/=-1
    then x = 4y+4/(y+1)

    I have proved existence, now I want to prove Uniqueness: that is, if there exists some other real number, say z, such that
    zy + z - 4 = 4y, that z = x.

    So we let z be an arbitrary real number and we make the assumption that zy + z - 4 = 4y and prove that z = x:

    but this just brings up two cases again:

    case 2a) y = -1, then z can be anything (just like in case 1a)
    case 2b) y=/= -1, then z = 4y+4/(y+1) which implies that z=x

    In this case Uniqueness is impossible since when y=-1, z could be any real number and x can be any real number, for example, z = 3/4 and x = 1000000000
    ------------------------------------------------------------------------------------------



    Was there some sort of mistake in my reasoning? The text never explicitly states in the question that y=/=-1 so I have to consider the case where it is -1 since the equation 'breaks down' since I can't divide by 0. I'm sort of stuck on this, is this a mistake on my part or should the text have specified that y=/=-1? Any help would be much appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 22, 2014 #2

    Orodruin

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    You are missing some parentheses here ...

    xy + x - 4 = 4y
    x(y+1) = 4y + 4
    x(y+1) = 4(y+1)
    x = ...
     
  4. Aug 22, 2014 #3

    haruspex

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    No you haven't. You have expressed x as a function of y, but you have to show there's a single x that works for all y.
    Hint: the way you have written that equation is wrong. Get the parentheses right.

    Edit: Orodruin pipped me at the 'post'.
     
  5. Aug 22, 2014 #4
    The business with cases is entirely unnecessary. You're misinterpreting the question I think. Find an explicit x (that does not depend on y!) such that xy + x -4 = 4y for all y. Yes, including y=-1.

    Don't solve for x in terms of y. Don't do anything to the equation! Pick a couple x to see what happens. Try x=1, 2, -50, pi, etc. There's a particular x such that this is true, regardless of y.
     
  6. Aug 22, 2014 #5

    Orodruin

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    You may also be misinterpreting the question, your x should be the same regardless of the value of y, i.e., there is a fixed x for which the equality holds regardless of what y is.
     
  7. Aug 22, 2014 #6
    Ah, yes. That's exactly what I meant, it was a slip-up on my part.

    So in other words I should just be able to say x=4 works for all possible values of y. Woops. I should probably take a break.
    Thanks for the clarification.
     
  8. Aug 22, 2014 #7
    Are you talking to me? Because that's exactly what I said.
     
  9. Aug 22, 2014 #8
    Yup! 4 works. Then you must prove uniqueness.
     
  10. Aug 22, 2014 #9

    Orodruin

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    When I started writing that I was still had the lone reply in this thread :tongue:
     
  11. Aug 22, 2014 #10
    Haha, I hate that. I forgive you :)
     
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