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

Question

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

To start this proof I will translate the goal into a logical form.

$\exists !x\forall y(xy + x -4 = 4y)$.

If I let $P(x) = \forall y(xy + x -4 = 4y)$, I would want to solve $\exists x(P(x) \land \forall z(P(z) \rightarrow z = x)$.

If Let $x =4$ I must show $P(4)$ for existence and $\forall z(P(z) \rightarrow z =4)$ for uniqueness.

To prove existence:

If $x =4$ then $4y + 4 - 4 = 4y \iff 4y = 4y$.

To prove uniqueness:

Suppose $z$ is some arbitrary real number such that $P(z)$. Then $zy + z -4 = 4y$. Thus $z =4 = x$. Since $z$ was arbitrary this holds for all $z$.

Is my logic correct or am I missing something? I have seen other sources use cases but I am unsure as to how that approach even crops up.

"Is my logic correct or am I missing something? " Maybe. I quit reading when you said "To start this proof I will translate the goal into a logical form". There's nothing wrong with saying "To start this proof I'll just do it. If we solve for $x...." — fleablood, Jun 21 '20 at 23:42
@fleablood I should have been more clear. I wasn’t intending to just do the proof but show my thought process. — Timmy Chyrklund, Jun 22 '20 at 00:00
Yes, your proof is logical (though tedious). However, I disagree with calling the second half a proof of uniqueness, since it merely proves *conditional uniqueness, that if a solution exists, then there is a unique solution, in other words, that at most one solution exists*. Uniqueness is then finally proven by combining the two halves of your proof. — ryang, Nov 05 '24 at 02:31

score 2 · Answer 1 · answered Jun 21 '20 at 23:33

2

If $xy + x -4 = 4y$ holds for all $y$, then it holds for $y=0$. This implies that $x=4$.

Conversely, $x=4$ works, as is easily checked.

answered Jun 21 '20 at 23:33

lhf

221,500

I think I understand why cases are appropriate. However, is my logic sound? – Timmy Chyrklund Jun 21 '20 at 23:37

Khashayar Baghizadeh · Answer 2 · 2020-07-28T14:08:24.183

0

You can also prove it like this:

Proof.

Existence: Let $x=\frac{4y+4}{y+1}$. Then

$$\frac{4y+4}{y+1}y+\frac{4y+4}{y+1}-4=4y.$$

Uniqueness: Let $a$ and $b$ be arbitrary real numbers and suppose $\forall y(ay+a-4=4y)$ and $\forall y(by+b-4=4y)$. From $\forall y(ay+a-4=4y)$ and $b$ we obtain $ab+a-4=4b$. From $\forall y(by+b-4=4y)$ and $a$ we obtain $ba+b-4=4a$. Subtracting $ab+a-4=4b$ from $ba+b-4=4a$ we get $5a=5b$ and consequently $a=b$. $Q.E.D.$

edited Jul 28 '20 at 14:08

answered Jul 28 '20 at 13:57

Khashayar Baghizadeh

1,531

Can you please elaborate on why we are allowed to assume that we can safely perform division by $y + 1$? Are not we supposed to assume that $y$ is not equal to $-1$? – Vlad Mikheenko Nov 14 '24 at 08:17

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

2 Answers2