- #1
pandaBee
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Homework Statement
Prove that there is a unique real number x such that for every real number y, xy + x -4 = 4y
Homework Equations
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.