Bijection for two sets. 
Give a bijection between the set of odd numbers and the set of even numbers and provide proof that it is a bijection.

Would this be a feasible bijection:
If $a$ is odd, then $a-1$ is even.
How would I provide a proof, that this is bijective? I understand that this is a bijection in that it is surjective and injective as each element only maps to one.  
Thanks in advance
 A: So you're saying that your function $f : \{ \text{odds} \} \to \{ \text{evens} \}$ is given by $f(a)=a-1$.
This function certainly works. To show $f$ is bijective you need to show that:


*

*$f$ is well-defined, i.e. given any odd number $a$, $f(a)$ really is even;

*$f$ is injective, i.e. if $f(a)=f(b)$ then $a=b$;

*$f$ is surjective, i.e. given any even number $n$ there is an odd number $a$ such that $f(a)=n$.


When you've proved that $f$ is well-defined, injective and surjective then, by definition of what it means to be bijective, you've proved that $f$ is a bijection.
A: Yes, the mapping $\phi:a\mapsto a-1$ is indeed a bijection from the set of odd integers to the set of even integers (I assume, negative integers are included, but it doesn't really make any difference).
I think, the easiest argument now is that the mapping $\psi:b\mapsto b+1$ is an inverse of $\phi$, in that
$$\phi(\psi(b))=b\quad\quad\text{and}\quad\quad \psi(\phi(a))=a$$
for all odd $a$ and even $b$.
A: How do provide a proof in general in mathematics? You have to show that the definition required in the problem holds.
In this case, you are asked to come up with a bijection. So you came up with a function, $f(n)=n-1$ defined for the odd numbers (I'm assuming integers, or natural numbers). This is of course a function, otherwise you'd have to verify that this is indeed a function.
Next to verify that the definition of a bijection holds. So we need to verify that the definition of "injective" is true for this $f$, as the definition of surjective.
Recall that a function is injective if and only if for different inputs it gives different outputs. Equivalently, if the output is equal, the input was equal. Assume that $n$ and $k$ are two odd integers. We have that $$f(n)=f(k)\iff f(n)+1=f(k)+1\iff n=k.$$
Therefore $f$ is injective. To show that $f$ is surjective we have to show that given an even number, $m$ there exists an odd number $n$ such that $f(n)=m$. I will leave this to you to verify.
