Homework Help: Existence/Uniqueness of x?

1. Aug 22, 2014

pandaBee

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
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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. Aug 22, 2014

Orodruin

Staff Emeritus
You are missing some parentheses here ...

xy + x - 4 = 4y
x(y+1) = 4y + 4
x(y+1) = 4(y+1)
x = ...

3. Aug 22, 2014

haruspex

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'.

4. Aug 22, 2014

johnqwertyful

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.

5. Aug 22, 2014

Orodruin

Staff Emeritus
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.

6. Aug 22, 2014

pandaBee

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.

7. Aug 22, 2014

johnqwertyful

Are you talking to me? Because that's exactly what I said.

8. Aug 22, 2014

johnqwertyful

Yup! 4 works. Then you must prove uniqueness.

9. Aug 22, 2014

Orodruin

Staff Emeritus
When I started writing that I was still had the lone reply in this thread :tongue:

10. Aug 22, 2014

johnqwertyful

Haha, I hate that. I forgive you :)

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