Help with simple proof

[Aion]

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

Can anyone give feedback on my proof thanks!

Suppose $\displaystyle x_0$ is an arbitrary element of R.
Let $\displaystyle y' = 2x_0$ and $\displaystyle y' \in R$
Suppose $\displaystyle z_0$ is an arbitrary element of R.
Then $\displaystyle y'z_0 = 2x_0z_0 = (x_0 + z_0)^2 - (x_0^2 + z_0^2)$
Since $\displaystyle z_0$ was arbitrary we conclude
$\displaystyle \forall z \in R (y'z= (x_0 + z )^2 - (x_0^2 + z^2))$
Therefore $\displaystyle \exists y \in R \forall z \in R (yz= (x_0 + z )^2 - (x_0^2 + z^2))$
Since $\displaystyle x_0$ was arbitrary,
$\displaystyle \forall x \in R \exists y \in R \forall z \in R (yz= (x + z )^2 - (x^2 + z^2))$

 

[tkhunny]

Where did $$y' = 2x_{0}$$ come from? Just pulled it out of a hat? Why not $$y' = 7x_{0}$$?

Seems like a lot of trouble. Can't you expand and simplify the RHS then simply construct y?

Have you accounted for $$z = 0$$?

 

[Aion]

I'm not sure I understand you. How would you have proven it? Thanks for the help.

I thought I wrote:

Let x be an arbitray real number. Let y = 2x. Now let z be an arbitrary real number.

Simplyfying [MATH](x+z)^2 - (x^2+z^2) = x^2 + z^2 +2xz - x^2 - z^2 = 2xz[/MATH]
Now [MATH] y=2x[/MATH], so we have [MATH] yz =2xz[/MATH]Thus we conclude that [MATH] yz = (x+z)^2 - (x^2+z^2)[/MATH]
And if [MATH]z=0[/MATH] then [MATH]0=0?[/MATH]

 

[pka]

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

I will write the question as if it were given in a symbolic logic text.
$\displaystyle (\forall x\in \mathcal{R})(\exists y\in \mathcal{R})(\forall z\in \mathcal{R})[yz=(x+z)^2-(x^2+z^2)[/tex]$
That tells us the order in which things must be done.
Suppose that $\displaystyle x\in \mathcal{R}$ then let $\displaystyle y=2x$ so that $\displaystyle y\in \mathcal{R}$.
Now if $\displaystyle z\in\mathcal{R}$ it follows that:
\(\displaystyle \begin{align*}(x+z)^2-(x^2+z^2)&=(x^2+2xz+z^2)-(x^2+z^2) \\&=x^2+2xz+z^2-x^2-z^2\\&=2xz\\&=yz \end{align*}\)
$\displaystyle Q.E.D.$