Proving that there is a unique real number x such that for every real number $y$, $xy+x-4=4y$

Question

The question is from problem 2 of chapter 3.6 of Velleman's How To Prove It (2nd edition). The question has been asked by another user before, but I couldn't find any thread that addresses the specific problem I have with my proof/problem.

My proof:

Existence Let $x = 4$, then $xy+x-4 = 4y+4-4 = 4y$

Uniqueness Let a and y be arbitrary and suppose $∀(+−4=4)$ and $∀(b+b−4=4)$.
From $∀(+−4=4)$, Suppose $y = b$, then $ab + a -4 = 4b$.
Then from $∀(b+b−4=4)$, suppose y = a, thus $ba+b-4 = 4a$.
Subtracting $ba+b-4 = 4a$ from $ab + a -4 = 4b$, we get $4b-4a = 0$, thus, $a = b$.
Since $a$ and $b$ are arbitrary, we can conclude that $a$ = $b$ = $x$ = $4$. Thus, $x$ is unique.
Q.E.D.

My issue with the proof:
I think this is correct, but something about the existence part of the proof is bothering me. Specifically the fact that we infer $x = 4$ by solving $xy+x-4=4y$ for $x$ like so:
$xy+x-4=4y$ iff
$x(y+1)=4y+4$ iff
$x = 4(y+1)/(y+1)$.

The $(y+1)$ factors cancel out and we are left with $4$. However, isn't $x$ undefined for $y=-1$?
So how is the statement "There exists a real number $x$ for all $y$ such that $xy+x-4=4y$" actually true? (As $x$ is undefined for $y = -1$?) Plugging in $-1$ for $y$ and $4$ for $x$ still makes the equation work, so I think the existence part of my proof is possibly correct, but how?

Answer 1

You have an algebraic error in your uniqueness proof: subtracting $ba+b-4=4a$ from $ab+a-4=4b$ yields $a-b=4b-4a$, not $4b-4a=0$. Of course then you get $5b-5a=0$, which also leads to the desired conclusion that $a=b$.

Solving for $x$ as you did is harmless provided that you then check the case $y=-1$ to be sure that the desired identity still holds, though your last iff is not actually correct, precisely because the fraction is undefined for $y=-1$. However, you can avoid the problem altogether and get the uniqueness at the same time: $xy+x-4=4y$ if and only if $(x-4)y+(x-4)=0$, i.e., if and only if $(x-4)(y+1)=0$. This in turn is the case if and only if $x=4$ or $y=-1$. In particular, if $x=4$ it is true for all $y$. If, on the other hand, $x\ne 4$, then it is true if and only if $y=-1$, so it is not true for all $y$.

Answer 2

All of the other answers, while not incorrect, have failed to address the fundamental misunderstanding in your question.

Under the heading "My issue with the proof," you write that in the existence part of the proof, "… we infer $x=4$ by solving …." But that's not true. Go back and reread your proof. In the existence part of the proof, you do not infer that $x=4$. Rather, you define $x$ to be 4, and then you do the algebra to show that this value of $x$ works. That algebra is correct for all values of $y$, including $y = -1$. So there is no problem with the proof.

What is worrying you is not the proof, but the scratch work you did in figuring out the proof. In that scratch work you did a step that would not be correct for $y = -1$. But that is irrelevant for judging the correctness of the proof. You could have come up with this proof by simply having a flash of insight that $x=4$ would work, or by making a lucky guess. In judging the correctness of the proof, it doesn't matter how you thought of the choice $x=4$; all that matters is whether your proof that this choice works is correct, and it is.

In How To Prove It, go back and reread the paragraph before Example 3.3.3.

Answer 3

Actually, when you try to find $x$ from the given property, you are more in the uniqueness part than the existence part ("If the property holds, then $x$ must be ..."); and indeed your second "iff" in that derivation is not correct as it works only in one direction unless we poystulate $y\ne -1$. Ultimately, you show existence just as you did: you plug in the value $x=4$ and win. How you arrived at that guess, doesn't matter for the existence proof. You could have derived it "legally", or with a dirty trick of ignoring a sometimes undefined expression, or played around with a few values until things work out, our have dreamt about it, or whatnot.

Likewise, you could simplify your uniqueness part by testing $a$ vs. $b$ only at $y=0$, which means $a-4=0$ and $b-4=0$, so $a=b$ ($=4$, but we don't care here). This ignores whether $a$ and $b$ are also good for $y=42$, but that gap doesn't matter for the uniqueness part.

Answer 4

You wrote:

$$x(y + 1) = 4y + 4 \iff x = 4(y + 1) / (y + 1).$$

I disagree. While $x = 4(y + 1) / (y + 1) \implies x(y + 1) = 4y + y$ by multiplying both sides by $y + 1$, the reverse implication is false in the particular case where $y = -1$. This is why you're getting this small uncertainty regarding $y = -1$.

Now, obviously when $y = -1$, there is no way to determine a specific value of $x$. Indeed, if $y = -1$, then any value of $x$ will satisfy the equation. But, the question specifically asks you to show there is a unique $x$ so that any $y$ satisfies the equation. Certainly $y = -1$ is the easiest case, but there will be only one $x$ that satisfies the equation for, say, $y = 3$ (or indeed any other value of $y$ except $y = -1$).