$ \ f: \mathbb{N} \times \mathbb{N} \to \mathbb{R}$ via $ \ f(a,b) = a + b. \sqrt{11}$ Question:
Let $ \ f: \mathbb{N} \times \mathbb{N} \to \mathbb{R}$ via $ \ f(a,b) = a + b. \sqrt{11}$
Is $ \ f$ an injection? Is it a surjection?
My attempt:
It is injective. 
$ f(a,b) = f(c,d) \implies a + b. \sqrt{11} = c + d. \sqrt{11} \implies (a-c) + \sqrt{11}(b-d) = 0 \implies a -c = 0$ and $ \ \sqrt{11}(b-d) = 0 \implies a = c$ and $ \ b = d$.
It is not surjective. Notice that $ \ f(a,b) \neq 1\  \forall \ a,b \in \mathbb{N}$. That is $ \ a+ b.\sqrt{11} \neq 1\  \forall \ a,b \in \mathbb{N}$. So $ \ 1 \notin $ image(f). Hence not surjective.  
Is my approach and reasoning correct?
 A: Your answers are correct, but your proofs need some improvement. You jump too quickly to unjustified conclusions, which immediately invalidates the whole "proof".
Injectivity. Up to $\color{blue}{(a−c)+\sqrt{11}(b−d)=0}$ everything was fine, but how do you know that $\color{red}{a−c=0}$ and $\color{red}{\sqrt{11}(b−d)=0}$ from there? Two numbers that add up to zero don't have to be zeroes themselves. Yes, it has a lot to do with $a,b,c,d$ being integers and $\sqrt{11}$ being, hmmm… a different kind of number. But you still have to justify each conclusion. Hint: go back to the blue equation and solve it for $\sqrt{11}$.
Surjectivity. Same thing: how do you know that $\color{red}{f(a,b)\neq1,\;\forall a,b\in\mathbb{N}}$, i.e. that $\color{red}{a+b\sqrt{11}\neq1,\;\forall a,b\in\mathbb{N}}$? I'm not saying that it's false, because it's actually true. The question is: how do you know that? You have to prove this claim. You can use the same idea: solve for $\sqrt{11}$.
By the way, you can show that this map isn't surjective by a completely different line of reasoning: counting. How many elements are there in the domain $\mathbb{N}\times\mathbb{N}$?
