Why doesn't mathematical induction work backwards or with increments other than $1$? From my understanding of my topic, if a statement is true for $n=1$, and you assume a statement is true for arbitrary integer $k$ and show that the statement is also true for $k+1,$ then you prove that the statement's true for all $n\geq 1$. Makes sense.
However - why can't I do this backwards? If I show the statement is true for $k-1,$ aren't I showing that if the statement is true for $n=1,$ it's likewise true for $n=0,n=-1,n=-2,\ldots$?
Also, why can't I prove the statement is true for $k+0.1$, and prove the statement true for $n=1.1,1.2,1.3,\ldots$? Both of these scenarios, in my mind, seem to follow the same logic as the "proper" definition of mathematical induction - but apparently they're no-go. Can someone please explain why?
Thank you!
Edit: The consensus seems to be that yes, even though it's abnormal, induction as I've stated above it is logically sound. Which raises the question - why has my math teacher said this is wrong? Is it as I suspect, where she didn't want me straying from the proper definition of $k+1$ induction and possibly confusing myself (or losing points on the test), or is there something else that makes the above fundamentally flawed?
Thanks!

Consider your first example of counting down:
Suppose property $Q(n)$ holds for $n=1$.
Suppose also that $Q(n)$ implies $Q(n-1)$.
Define a new property $P$ by $P(n)$ if and only if $Q(2-n)$.
We supposed $Q(1)$ so we know $P(1)$.
Suppose $P(n)$. That implies $Q(2-n)$. We deduce $Q(2-n-1)$ which is $Q(2-(n+1))$. And that implies $P(n+1)$.
So we can use induction to find that $P(n)$ holds for all $n\ge 1$.
And therefore Q(n) holds for all $n\le 1$.
So counting down works fine. But if you haven't yet proved it works you need to prove it at least once from the usual induction argument.
A similar line of reasoning shows that your second example works fine too.
