Answer
J is one-to-one but not onto
Work Step by Step
$J: \mathbb{Q}\times \mathbb{Q} \rightarrow \mathbb{R} \\
J(a,b)=a+\sqrt{2}b ,(a,b)\in \mathbb{Q}\times \mathbb{Q} \\
A\,function\,\,J: \mathbb{Q}\times \mathbb{Q} \rightarrow \mathbb{R} \,is\,\,one-to-one\,\Leftrightarrow \\
\forall \,\,x_{1}\,and\,x_{2}\,\,in\,\,\mathbb{Q}\times \mathbb{Q} \,\,if\,
J(x_{1}) = J(x_{2})\,\,then\,x_{1} = x_{2}.\\
let \,x_{1}=\left ( a,b \right ), \,x_{2}=\left ( c,d \right )\\
if\,\,J(a,b)=J(c,d) \\
\Rightarrow a+\sqrt{2}b =c+\sqrt{2}d \\
\Rightarrow a=c \,\,,\sqrt{2}b=\sqrt{2}d \\
\Rightarrow a=c , b=d \\
\Rightarrow (a,b)=(c,d)\\
\because J(a,b)=J(c,d)\Rightarrow (a,b)=(c,d)\\
\therefore J \,\,is\,one-to-one \\
J: \mathbb{Q}\times \mathbb{Q} \rightarrow \mathbb{R} is\,\,not\,\,onto\,\Leftrightarrow \,\\
\exists y\,in\,\mathbb{R} such\,that\,\forall x \in \mathbb{Q}\times \mathbb{Q}, J(x) \neq y.\\
since\,\sqrt{5}\in\mathbb{R} \\
and\,\sqrt{5}\neq a+\sqrt{2}b(as\,a,b\in \mathbb{Q})
so\,\,J\,is\,not\,onto
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