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For context:

Theorem. There is a unique real number $x$ such that for every real number $y$, $xy+x-4=4y$.

Proof. We show the existence by choosing $x=4$. Substituting it in gives. $xy+x-4=4y+4-4=4y$. Suppose there is another real number $z$ such that for every real number $y$, $zy+z-4=4y$. Add $4$ to both sides gives $z(y+1)=4(y+1)$. From this we see, that $z=4=y$. Thus $x=4$ is a unique solution.

By cases, it would be:

  1. Case: $y \ne -1$: We divide both sides by $(x+1)$ and get $z=4$
  2. Case: $y = -1$: Leads to $0=0$ which is true, but does it say anything?

The $y=-1$ case bugs me. I can see, that the solution must be $z=4$ but I am somehow not convinced by the proof by cases as the second case only leads to $0=0$ and says nothing about $z$. How could I show rigorously what I see?

Source: It is an exercise from "How to prove it" by Daniel J. Velleman. The exercise 3.6.2 is "Prove that there is a unique real number $x$ such that for every real number $y$, $xy+x-4=4y$."

Thomas Deniffel
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2 Answers2

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It is not true that $xy+x-4=4y$ has a unique solution $x$ for every $y\in \Bbb R$. Indeed, for $y=-1$, every $x\in \Bbb R$ is a solution.

However, the Theorem is meant to say that there is a unique $x\in \Bbb R$ such that $xy+x-4=4y$ is true for all $y\in \Bbb R$. And this $x$ must be $4$, by considering $y=0$. Any other $x$ would not satisfy the equation for $y=0$, i.e., would not satisfy it for all $y$. This shows it "rigorously".

Dietrich Burde
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  • So you are saying that the theorem is meant to say this: $\exists ! x \in \mathbb{R} \forall y \in \mathbb{R} (xy+x-4=4y)$ – Thomas Deniffel Apr 29 '21 at 09:24
  • I am sorry, could you use words? I think that theorems in books are still formulated in words. Let me try. Theorem: There exists a unique real number $x$ such that the equation $xy+x-4=4y$ holds for all real numbers $y$. – Dietrich Burde Apr 29 '21 at 09:26
  • Sorry - you are right. I guess in it is nearly the theorem: There is a unique real number $x$ so that for every real number $y$, $xy+x-4=4y$ is true. But then I have trouble distinguishing how the theorem differed from the corrected version, proposed by you. – Thomas Deniffel Apr 29 '21 at 09:33
  • I think it is clear now. There is no difference. May I ask in which language the Theorem is originally formulated? Do you have a link? – Dietrich Burde Apr 29 '21 at 09:35
  • It is an exercise from "How to prove it" by Daniel J. Velleman. The exercise 3.6.2 is "Prove that there is a unique real number $x$ such that for every real number $y$, $xy+x-4=4y$." I have forgotten to add it to the question - I add it now... – Thomas Deniffel Apr 29 '21 at 09:37
  • "To see, that $x=4$ is unique, suppose $zy+z-4=4y$. This $z \in \mathbb{R}$ has also to be $4$, because any other $z$ would violate the case $y=0$." - The solution of @dietrich-burde in my own words (as I understand it). Replacing the uniqueness proof from the original post. – Thomas Deniffel Apr 29 '21 at 10:03
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You started your proof of uniqueness correctly, by writing "Suppose there is another real number $z$ such that for every real number $y$, $zy+z-4 = 4y$." I think you then confused yourself by ignoring the phrase "for every real number $y$" in this sentence. In your next step, you concluded $z(y+1) = 4(y+1)$, as if $y$ were a fixed number. The correct statement would be that for every real number $y$, $z(y+1) = 4(y+1)$. If something is true for every real number $y$, then you are free to plug in any value you please for $y$. (See the explanation of using a given of the form $\forall xP(x)$ in Section 3.3 of How To Prove It.) So you can simply set $y=0$ and conclude that $z = 4$.

Dan Velleman
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