Prove that if $A \sim I_n$ and $A \sim I_m$ then $n=m$ Some definitions:
Definition of equinumerous sets
Two sets are equinumerous if there exists a bijection $f: A \rightarrow B$. We write $A \sim B $ if $A$ is equinumerous to $B$
Definition of finite set and cardinality
A set $A$ is said finite if $A \sim I_n$  where $I_n=\{k \in \mathbb{N} | k \leq n\}$ and $\mathbb{N}=\{1,2...\}$
If $A$ is finite the unique number $n$ such that $A \sim I_n$ is called the cardinality of $A$
Now the question:
Prove the cardinality of A is well-defined (that is if $A \sim I_n$  and $A \sim I_m$  then $n=m$ )
Suggestion: prove first that if $n \neq m$ then there exist no bijection between $I_n$ and $I_m$.
So, if I assume by absurd that $m \neq n$, let's say $m &lt; n$
There exists an injective but not surjective mapping  $i:I_m \rightarrow I_n$.
Now since this is so obvious I am having trouble going forward. Actually, I don't know if the previous statement, provided it is needed, is too far a step into the proof, given that what I want to prove is as much as obvious.
Can someone enlighten me?
Proving the statement in the suggestion is not trivial. What you’ve done isn’t nearly enough: the existence of an injection from one set to another that is not a surjection does not prove that there is no bijection between the two sets. For instance, the map $f:\Bbb N\to\Bbb N:n\mapsto n+1$ is a non-surjective injection from $\Bbb N$ to itself, but obviously there are bijections from $\Bbb N$ to $\Bbb N$! It’s only for finite sets that the existence of a non-surjective injection implies that there is no bijection, and that’s essentially what you’re supposed to be proving here.
Let $B$ be the set of $n\in\Bbb N$ such that there is a bijection from $I_n$ to some $I_m$ with $m&lt;n$. Suppose that $B\ne\varnothing$; by the well-ordering principle we can let $n=\min B$. Let $f:I_n\to I_m$ be a bijection, where $m&lt;n$, and let $k=f(n)$. Let $g$ be the restriction of $f$ to $I_{n-1}$; $g$ is a bijection from $I_{n-1}$ to $I_m\setminus\{k\}$.
Now define a function $h:I_m\setminus\{k\}\to I_{m-1}$ as follows:
$$h(i)=\begin{cases}
i,&amp;\text{if }1\le i&lt;k\\
i-1,&amp;\text{if }k&lt;i\le m\;.
\end{cases}$$
It’s easy to verify that $h$ is a bijection. But then $h\circ g$ is a bijection from $I_{n-1}$ to $I_{m-1}$, so $n-1\in B$, contradicting the choice of $n$ as the smallest member of $B$. This contradiction shows that $B$ must be empty and hence that no $I_n$ can be mapped bijectively to an $I_m$ with $m&lt;n$.
Let $m&lt;n$. Suppose that there is a bijection $f: I_m\to I_n$. Write
$$ I_n=f(I_m)\cup A, f(I_m)\cap A=\emptyset, A\neq\emptyset. $$
For $y\in A$, since $f: I_m\to I_n$ is bijection, there is $x\in I_m$ such that $f(x)=y\in f(I_m)$. From this one can see that $y\in f(I_m)\cap A$ which is absurd since $f(I_m)\cap A=\emptyset$.