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 [math]y' = 2x_{0}[/math] come from? Just pulled it out of a hat? Why not [math]y' = 7x_{0}[/math]?

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

Have you accounted for [math]z = 0[/math]?

 

[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:
\(\displaystyle \begin{align*}(x+z)^2-(x^2+z^2)&=(x^2+2xz+z^2)-(x^2+z^2) \\&=x^2+2xz+z^2-x^2-z^2\\&=2xz\\&=yz \end{align*}\)
\(\displaystyle Q.E.D.\)