From How to Prove It:

Prove that there is a unique real number $x$ such that for any real number $y,$ $$xy+x-4=4y.$$

I am having issue with this proof, since the theorem itself seems incorrect: if $y=-1,$ then we have $-x+x-4=-4,$ that is, $ x=x,$ which is true for any $x$ whatsoever. So, how is it true that there is a unique $x$ that works for any $y?$ I can see why $x$ exists, but not why it is unique.

  • $\begingroup$ This is a really great question! thanks for citing the book. $\endgroup$
    – ktm5124
    Apr 1 at 20:07

3 Answers 3


We can factor this polynomial and get the solutions.

(We can find the solutions by inspection. But factoring helps us find all of the solutions.)

\begin{align*} xy + x - 4 &= 4y \\ x(y + 1) - 4 - 4y &= 0 \\ x(y + 1) - 4(y + 1) &= 0 \\ (x - 4)(y + 1) &= 0 \end{align*}

The solutions are $x = 4$ and $y = -1$.

Our solution $x = 4$ works for any real number $y$.


I believe your confusion is caused by an unsound swap of logical quantifiers.

Consider this scenario. There are two doors:

  • door number 1 can be opened with the red key or the blue key;
  • door number 2 can be opened with the red key or the green key.

Now, focus on the difference between the following statements:

  • there is a unique key that opens all the doors (true, only the red key opens both doors)
  • all the doors are opened by a unique key (false, for instance door 1 can be opened by more than one key)

In your own example, the statement is $\exists !x\; \forall y$ but you mentally swapped the quantifiers reading it as $\forall y\; \exists !x $. As the door example shows, you can not swap the quantifiers, since that will change the meaning of the statement, in general.


For any given $~y \in \Bbb{R},~$ you have that

$$xy + x - 4 = 4y \iff $$

$$x(y+1) = xy + x = 4y + 4 = 4(y+1). \tag1 $$

So, the original assertion translates into proving that there exists a unique real number $~x~$ such that

$$\forall ~y ~\in \Bbb{R}, ~x(y+1) = 4(y+1). \tag2 $$

In fact the assertion is true, and the corresponding unique value of $~x~$ is $~x = 4.~$

That is:

  • When $~x = 4,~$ the assertion in (2) will be true, regardless of the value $~y.~$

  • When $~x \neq 4,~$ you will always be able to find at least one value of $~y,~$ (e.g. $~y = 0,~$ for example) such that the assertion is not true for that particular value of $~y.$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .