1. ## subspace

hello.
$(a);\left \{ (x,y)\in \mathbb{R}^{2},\exp x=\exp y\right \}$
i have to prove that $a$ is a subspace.
i took two elements from (a), $(x,y)$ and $(x',y')$ and i must have $(x+x',y+y')\in a$.
so i wrote $\exp (x+x')-\exp (y+y')=\exp x \exp x'-\exp y\exp y'$
where i went wrong ?
i know that $a$ can be written as $(a);\left \{ (x,y)\in \mathbb{R}^{2}, x= y\right \}$,but i want to know where was my error.
thanks.

2. Originally Posted by Raoh
hello.
$(a);\left \{ (x,y)\in \mathbb{R}^{2},\exp x=\exp y\right \}$
i have to prove that $a$ is a subspace.
i took two elements from (a), $(x,y)$ and $(x',y')$ and i must have $(x+x',y+y')\in a$.
so i wrote $\exp (x+x')-\exp (y+y')=\exp x \exp x'-\exp y\exp y'$
where i went wrong ?
i know that $a$ can be written as $(a);\left \{ (x,y)\in \mathbb{R}^{2}, x= y\right \}$,but i want to know where was my error.
thanks.
No where went you wrong: $e^xe^{x'}=e^{y}e^{y'}$...why you tried to make a substraction there I can't say.

Tonio

3. Originally Posted by tonio
No where went you wrong: $e^xe^{x'}=e^{y}e^{y'}$...why you tried to make a substraction there I can't say.

Tonio
i tried to show a zero.

4. Since exp is an invertible function, isn't $\{(x,y)|exp(x)= exp(y)\}$ exactly the same as $\{(x,y)| x= y\}$?

5. God ! i forgot the fact that $\exp(x+x')=\exp(x)\exp(x')=\exp(y)\exp(y')=\exp(y+ y')$ since
$\exp(x)=\exp(y)$ and $\exp(x')=\exp(y')$
thanks everyone.

6. You don't have to call me "God"- you can use my webname!

7. Originally Posted by HallsofIvy
You don't have to call me "God"- you can use my webname!
i meant "oh God !"