Prove that whenever $n$ and $m$ are natural numbers such that $m > n$, then we have $a_{m} > a_{n}$ Let $(a_{n})_{n=0}^{\infty}$ be a sequence of real numbers, such that $a_{n+1} > a_{n}$ for each natural number $n$. Prove that whenever $n$ and $m$ are natural numbers such that $m > n$, then we have $a_{m} > a_{n}$. (We refer to these sequences as increasing sequences.)
MY ATTEMPT
Intuitively, one has that $n < n+1 < \ldots < m-1 < m$. Consequently, $a_{n} < a_{n+1} < \ldots < a_{m-1} < a_{m}$.
However it doesn't seem to be rigorous enough. 
Could someone provide a formal proof based on the principle of mathematical induction or any other method?
 A: Let $m=n+k$, where $k\in\mathbb{N}$. We will induct on $k$. For $k=1$, the claim holds via the assumption. Now let $k>1$, and assume the claim holds for $k-1$. This means $a_{n+{k-1}}>a_n$. But $a_{n+k}>a_{n+k-1}$, so $a_{n+k}>a_n$ and we're done.
A: Induction.
The statement is.  For any $m$ so that $m >n$ then $a_m > a_n$.
Base case:  Let $m$ be the very first natural number that is larger than $n$.  That is, let $m = n+1$.  Prove $a_m > a_n$.
Pf: That was given $a_m = a_{n+1} > a_n$.
Induction case:  Suppose that we know that if $k > n$ and $a_k > a_n$ then prove that it follows that $a_{k+1} > a_n$.
Pf:  $a_{k+1} > a_k$ is a property of the sequence.  And $a_k > a_n$ so $a_{k+1} > a_k > a_n$. So $a_{k+1} > a_n$.
So we have proven it by induction.  We have proven in the base case, that $a_m > a_n$ for the very first possible $m > n$, that is to say for $m = n+1$.  Ane wh have proven the inductive step that if for any $m=k>n$ that if $a_k > a_n$ then it is true for the next vale $m=k+1$.  Therefor it is true, by repeated induction, that it is true for all $m > n$.
