Explanation on $1-1$ functions and onto functions $ f : \Bbb{Z} \to \Bbb{Z}$ (integers), $f(n) = 3n + 2$ is $1-1$ 
$ f : \Bbb{R} \to \Bbb{R}$ (real number), $f(x) = 2x - 3$ is onto
can someone explain $1-1$ and onto functions? I don't really understand the idea behind $1-1$ and onto
 A: Definition: A function $f: A \to B$ is said to be $1-1$, whenever $f(x)=f(y)$, then $x=y$. 
So the function which you gave $f: \mathbb{Z} \to \mathbb{Z}$ by $f(n)=3n+2$ is $1-1$ because $f(x)=f(y) \Longrightarrow 3x+2=3y+2$ implies $x=y$.
Definition:  A function $f$ with domain $X$ and codomain $Y$ is onto if for every $y \in Y$ there exists at least one $x \in X$ with $f(x) = y$. Check whether that holds, in your example.
Just put $y=2x-3$ and write $x$ is terms of $y$. So you have $x = \frac{y+3}{2}$, which says that $f$ is onto.
Added. The function $f:\mathbb{Z} \to \mathbb{N}$ defined by $f(x)=x^{2}$ is not $1$-$1$ because, $f(x)=f(y)$ doesn't imply $x=y$. That is from $x^{2}=y^{2}$, we don't get $x=y$. We also get $x=-y$. Hence the function defined by $f(x)=x^{2}$ is not $1$-$1$. But if we change the Co - Domain, from $\mathbb{Z}$ to $\mathbb{N}$, then the function $f(x)=x^{2}$ becomes $1$-$1$. 
A: For definitions, see the other answers. 
Let's start with the opposite of 1 to 1, which is "many to 1". (Not "1 to many" - that would not be a function!) For example: The function $f(x) = x^3 - x$ is "many to 1", since the values $x = -1, \, 0, \, 1$ all are mapped to $y = 0$. A function is "many to 1" if there are values $x_1 \ne x_2$ such that $f(x_1) = f(x_2)$. A function is 1-1 if this is impossible, or with quantifiers
$$ \forall x_1, \, x_2 \in dom(f) \; x_1 \ne x_2 \Longrightarrow \; f(x_1) \ne f(x_2) \, .
$$
You should check that the negation of this statement is precisely what I called informally "many to 1".
A function $f: A \to B$ is onto if all elements of $B$ are "hit" by the function $f$. Put differently, as $a$ attains all possible values in $A$, $f(a)$ attains all possible values in $B$. Differently yet, a function $f$ is onto $B$ if for all $b \in B$ the equation $f(x) = b$ is guaranteed to have a solution $x \in A$. In quantifier notation,
$$\forall b \in B \, \exists x \in A \, \ni \, f(x) = a \, .$$
The function $f(x) = x^3 - x$ is onto $\mathbb{R}$ since the equation $x^3 - x = b$ has a solution for all real numbers $b$.   
Hope this helps, good luck with your final.     
A: 1-1 is better called injective. Onto is also called surjective. Some people use 1-1 to mean bijective, hence the confusion. The wikipedia pages I've linked to contain definitions and examples. If you have concrete questions, please ask again.
