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I know how to prove if a single variable function is injective or surjective but I've been quite confused by an exercise where i need to determine if:

$f :\mathbb{Q}^{2} \to\mathbb{Q}^{2}$
$f(x,y) = (x+y, x-y)$

is injective and/or subjective.

Is it possible to say that this function is not injective because:
$Let \;x_{1} = \frac{1}{2} and \;y_{1} = \frac{0}{2}, f(x_{1},y_{1}) = \frac{1}{2}$
$Let \;x_{2} = \frac{1}{2} and \;y_{2} = \frac{0}{4}, f(x_{2},y_{2}) = \frac{1}{2}$
$f(x_{1},y_{2}) = f(x_{2},y_{2}) \land (x_{1}, y{1}) \neq(x_{2}, y{2})$

Also, I assume this function would be surjective but how to prove with couples?

Anyone could point me in the right direction? Thanks!

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    When you write $f(x_1, y_1) = 1/2$ you're implicitly assuming that the codomain of the function is one dimensional, whereas the codomain is actually two dimensional. The map is in fact both injective and surjective. A hint: the question is the same as that of whether given rational $a$ and $b$ there exist a unique pair of rational $x$ and $y$ such that $a = x + y$ and $b = x - y$.2012-10-31

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