Question

[SOLVED] Here's a cute predicate logic problem.

Prove or disprove: \[\forall x \,\exists y \,\forall z\, \exists w \;\;\text{such that} \;\;x+y+z+w=xyzw\]Assume that \(x, y, z, w \in \mathbb{C}\).

Answer 1

is xyzw a combination of digit or multiplication of digit??

Answer 2

multiplication

Answer 3

lol .. didn't realize they were complex

Answer 4

It doesn't really matter. I just put that specification on there so that know one tried to change the problem so that addition and multiplication had different meanings.

Answer 5

Just prove/disprove it for \(\mathbb{R}\) and it will become clear why it doesn't matter if they're complex numbers.

Answer 6

Hint below in white LaTeX\[\color{white}{\text{There is a certain value for y that you should choose to prove it.}}\]

Answer 7

\[ \color{white}{\text{LOL i didn't know that!!.}} \]

Answer 8

Another, probably more helpful hint, below in white LaTeX\[\color{white}{\text{What value of y can you choose so that xyzw is constant?}}\]

Answer 9

Choose y=0, then choose w=-(x+z).

Answer 10

Bingo.

Answer 11

But ∃y doesn't mean that y is rather a variable such that we have to fit constraint ??

Answer 12

\(\exists\, y\) means that we can choose a specific value for y. It's a variable, but one that we have complete control over.

Answer 13

then it must be false ... i guess so!!

Answer 14

The "there exists" only means that at least one possible value of y where this is true. One value that we know works is \(y=0\). This means that there does exists at least one value for \(y\) that makes the statement true, so we have proven that the original statement is true.

Answer 15

does't y=0 make statement untrue??
xyzw=0, for y=0

Answer 16

Right, both the sum and the multiplication end up equaling zero because you set w=-(x+z).

Answer 17

If \(y\) is 0, and \(x=-(x+z)\) we have \[x+0+z+(-(x+z))=x+z-x-z=0\]and \[x\cdot0\cdot z\cdot w=0\]So that means that \[x+y+z+w=xyzw\]

Answer 18

Oh ... stupid me ... i never considered the negative values!!

Answer 19

Hope everyone enjoyed this little (unintuitive) problem. Good night.

Answer 20

thanks!! i'll try to think broader way next time!!

Answer 21

I'll try to get a problem just for you next time then. Any preference about the subject?