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Pretty easy when its only a function of the form $y = ax + b$, however I'm trying to find if this function:

$f(x,y) = 2x + y$

is onto and one-to-one. Is there a procedure to follow to check this? Thanks!

EDIT: My reasoning.

onto:

$z = 2x + y$

$x = (z-y)/ 2$

$y = z - 2x$

Seeing as both $x$ and $y$ can be represented in "safe" ways (no dividing by zero), I'm assuming that the function is onto.

One-to-One:

$2x_1 + y_1 = 2x_2 + y_2$

$2(x_1-x_2) = (y_2-y_1)$

Therefore if $y_2 = y_1$, $x_2 = x_1$.

I guess my reasoning is wrong :/

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    Yes: check to see if it's onto, then check to see if it's one-to-one. (Note that it's generally easier to check if either of these properties _fails_, since all you have to do is find some points at which the corresponding definitions don't hold.)2011-03-09
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    You'll have to define your function a bit better first before the question even makes sense. Try to write it as $f\colon A\to B \colon x\mapsto f(x)$, for instance, like $f\colon \mathbf R\to \mathbf R \colon x\mapsto ax+b$.2011-03-09
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    To check if the function is onto, try to solve the equation $2x+y = z$ for arbitrary $z$ (one solution is enough). To check if the function is one-to-one, check if $2x_1+y_1 = 2x_2+y_2$ implies $x_1=x_2$ and $y_1=y_2$.2011-03-09
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    Thanks Yuval Filmus! I've found that it is both 1-1 and onto.2011-03-09
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    @Max: Then you've either made some very odd choices for the domain and codomain, or have got it wrong.2011-03-09
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    @Chris Eagle: I just edited my post2011-03-09
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    @Max: If your function is $\mathbf R^2\to \mathbf R$ then notice that $f(0,x) = f(x/2,0) = x$ so it's definitely not one-to-one...2011-03-09
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    Correct me if I am wrong. For one-one you must have $z_1 = z_2 \Leftrightarrow (y_1 = y_2 \wedge x_1 = x_2)$ however in your solution you only checked $ (y_1 = y_2 \wedge z_1 = z_2) \rightarrow x_1 = x_2$2017-01-21

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