# There is a unique real $x$ such that for any real $y,\,xy+x-4=4y$

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.

• This is a really great question! thanks for citing the book. Apr 1 at 20:07

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 $$~y.$$