Unable to prove a simple inequality Let the function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be strictly increasing. That is: $x_2>x_1 \Rightarrow f(x_2)>f(x_1)$.
In the segment $[0;a] \subset \mathbb{Z}$, we have $f(a)<a$ 
I want to prove that $\forall i \in [0;a] , f(i)<i$ .
I am getting a bit frustrated because it looks so obvious.
One proof I came up with is using reverse induction:
$\bullet$ Base Case: $k=a$, because it is given that $f(a)<a$
$\bullet$ Inductive Case: 
Assume $f(k)<k$. $f$ is increasing so $f(k-1)<f(k)$. (1)
$f(k)<k \Rightarrow f(k) \leq k-1$ . (2)
Combining (1) and (2) we have: $f(k-1)<k-1$.
Thus $\forall k>0 , f(k)<k \Rightarrow f(k-1)<k-1$.
So the conclusion is that starting from $a$ and moving backwards we have $\forall i \leq a$  , $f(i)<i$. Of course we are only interested in the segment $[0;a]$. 
However I am sure there is a simpler proof.

Suppose on the contrary, that there exists $k<a$ such that $f(k)\geq k$, and pick up the biggest such $k< a$. Then $f(k+1)\leq k$, $f(k)\geq k\implies f(k)\geq f(k+1)$, a contradiction.
