Hi everyone.

I've got a one task to do i need to show if the function g: R -> R^2 by g(x) = (x+1, x+2) is g onto?

I think is not but i do not really know how to show this understanding.

I'll be appreciate for any help.

Regards

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- Aug 23rd 2009, 12:17 PMSnowboarderDefine function
Hi everyone.

I've got a one task to do i need to show if the function g: R -> R^2 by g(x) = (x+1, x+2) is g onto?

I think is not but i do not really know how to show this understanding.

I'll be appreciate for any help.

Regards - Aug 23rd 2009, 12:29 PMPlato
- Aug 23rd 2009, 02:13 PMSnowboarder
yes than it's not but how can i proof that mathematicly?

- Aug 23rd 2009, 02:22 PMPlato
- Aug 24th 2009, 06:17 AMnirax
ok if you see the diagram you may better understand. the domain is a line. the codomain is a plain and the range is a straight line in that plane. now is that range line the same as full plane ?? how can that be ... this is a good way of illuminating things. keep track of pictures.

- Aug 24th 2009, 06:36 AMDefunkt
Simply assume that there exists $\displaystyle x \in {R}$ such that $\displaystyle g(x) = (0,0)$. This gives us:

(I) $\displaystyle x+1 = 0$

(II) $\displaystyle x+2 = 0$

$\displaystyle \Rightarrow x+1 = x+2 \Rightarrow 1 = 2 \Rightarrow$ contradiction.

So there is no $\displaystyle x \in {R}$ such that $\displaystyle g(x) = (0,0)$ and thus g is not onto