Prove that for every real number x, there is a real number y such that for every real number z, yz = $(x + z)^2-(x^2+z^2)$

Question

I don't understand how to prove this:

Prove that for every real number x, there is a real number y such that for every real number z, yz = $(x + z)^2-(x^2+z^2)$

In my understanding, I first have to convert this question into logical operators and I have this: $\exists y\forall z[(x + z)^2-(x^2+z^2)]$

After, I solved for $y$ and I get $y=2x$ but I don't understand what I have to do after it.

Thanks

Answer 1

If you want olny to have the solution then y=2x is correct, but if you want to write with a logic formalism so you need to use the logic operator for argument in First-order logic, more specifically the Existential Generalization (EG) and the Universal Generalization (UG). Following

1. $(r+s)^2=r^2+s^2+2rs$
2. $2rs=(r+s)^2-(r^2+s^2)$
3. $(\forall x)(2xs=(x+s)^2-(x^2+s^2))$ (2-UG)
4. $(\forall x)(\exists y=2x)(ys=(x+s)^2-(x^2+s^2))$ (3-EG)
5. $(\forall x)(\exists y=2x)(\forall z)(yz=(x+z)^2-(x^2+z^2))$ (4-UG)