Proof that the identity permutation is an even permutation by induction I think a similar question has been answered before but the part I need was still not answered and I couldn't comment on the answers. So here is the theorem

If the identity permutation ,$\epsilon=\beta_1\beta_2\dots\beta_r$, where $\beta$'s are 2-cycles, then $r$ is even.
Proof: Clearly  $r\neq 1$,  since  a  2-cycle is not the identity . If $r=2$ we are done. So, we suppose that $r\gt2$ ,and we proceed by induction. Suppose that the the rightmost 2-cycle is $(ab)$. Then, since $(ij)=(ji)$, the product $\beta_{r-1}\beta_r$ can be expressed in one of the following forms:
1) $\epsilon=(ab)(ab)$
2) $(ab)(bc)=(ac)(ab)$ 
3) $(ac)(cb)=(bc)(ab)$ 
4) $(ab)(cd)=(cd)(ab)$
If the first case occurs, we may delete $\beta_{r-1}\beta_r$, from the original product to obtain $\epsilon=\beta_1\beta_2\dots\beta_{r-2}$, and therefore by the second principle of mathematical induction, $r-2$ is even.

I don't understand how or why the principle of induction is applied here. I mean shouldn't an inductive set have consecutive integers. Here the values of r are only even numbers_
 A: 
$P(k)$:   The identity permutation can be a product of $E_k =2k$ 2-cycles.

Math is not magical incantations.
Induction is:
a) If something is true for a base case.  b) if we can prove that if is true for any case it is true for the next one after that, then Conclusion) logically it follows that it is true for all cases after the first one.
There is no magic that says the base case has to be the number $1$ and that only way the next case has to be the next consecutive number.
These could be people walking through a door if you can figure out a way to proof that if one person through the door having a condition proves that the next person through the door will also have the condition.
.....
Well, okay, all constructivists are reaching for their keyboards to point out how wrong I am.
Perhaps I should point out more formally instead:
Yes, induction is done on consecutive natural numbers but if there is a bijective mapping from the natural numbers to your cases you can do induction on your cases directly by doing induction on the indexes of your cases.   
In that way may case with people walking through doors is $P_1$ is the first person through the door, and $P_k$ is the $k$th person.
And as $E_1 = 2 = 2*1 $ is the first odd number,  and $E_2 = 4=2*2$ is the second, we can index the odd numbers as $E_k$ and our induction proposition is:

$P(k)$:   The identity permutation can only be a product of $E_k =2k$ 2-cycles.

