Answer
a-G is one-to-one
b-G is onto
Work Step by Step
$G:\mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}\times \mathbb{R} \\
G(x,y)=(2y,-x)\,for\,all\,(x,y)\in\mathbb{R}\times \mathbb{R} \\
A\,function\,\,G:\mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}\times \mathbb{R}\,is\,\,one-to-one\,\Leftrightarrow \\
\forall \,\,x_{1}\,and\,x_{2}\,\,in\,\,\mathbb{R}\times \mathbb{R}\,\,if\,
G(x_{1}) = G(x_{2})\,\,then\,x_{1} = x_{2}.\\
proof:\\
let\,x_{1}=(a,b),x_{2}=(c,d)\\
if\,G(a,b)=G(c,d)\\
G(a,b)=(2b,-a)=G(c,d)=(2d,-c)\Rightarrow \\
2b=2d \,and\,-a=-c \Rightarrow \\
b=d \,and\,a=c \Rightarrow (a,b)=(c,d)\\
\because G(a,b)=G(c,d)\Rightarrow (a,b)=(c,d) \\
\therefore G\,\,is\,one-to-one \\
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$G:\mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}\times \mathbb{R} \\
G(x,y)=(2y,-x)\,for\,all\,(x,y)\in\mathbb{R}\times \mathbb{R} \\
G: \mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}\times \mathbb{R} is\,\,onto\,\Leftrightarrow \,\\
\forall y\,in\,\mathbb{R}\times \mathbb{R} ,\exists x \in \mathbb{R}\times \mathbb{R}\,such\,that\, G(x) = y.\\
let\,y=(f,g) ,\,\,\,x=(w,z)\\
suppose\,(w,z)=(-g,\frac{f}{2})\\
G(w,z)=(2.\frac{f}{2},-(-g))=(f,g)\\
G\,\,is\,\,onto \\
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