How to write a definition of less than $<$? I'm learning the fundamentals of discrete mathematics, and I have been requested to solve this problem:
According to the set of natural numbers
$$
\mathbb{N} = {0, 1, 2, 3, ...} 
$$
write a definition for the less than relation.
I wrote this:
$a < b$ if $a + 1 < b + 1$
Is it correct?
 A: Regarding to this particular set, you can define $<$ as $a < b$ if $b - a \in \mathbb{N}$ and $b - a \neq 0$.
A: $a<b  \iff \exists p \in \mathbb{N_{>0}}$: $b=a+p$.
A: How can you decide if $3<5$ using your definition?
You can say $3<5$ if $4<6$ if $5<7$ and so on, but this sequence will never end.
It works the other way round:  


*

*if $b \ne 0$: $0 \lt b$  

*if $a \lt b$: $a+1 \lt b+1 $  


$2 \ne 0$ , so $0 \lt 2$, therefore $1 \lt 3$, therefore $ 2 \lt 4$ , and finally $3 \lt 5$
A: A way to think about the natural numbers is in terms of the Peano Axioms.
There exists a "successor" map 
$$ S: \mathbb{N}\rightarrow \mathbb{N} $$
such that in particular


*

*$S(0) = 1 $

*$0\notin S(\mathbb{N}) $


The action of $S$ is usually written as $S(n) =: n+1$ 
The ordering of $\mathbb{N}$ may then be defined as
$$ a\leq b :\Longleftrightarrow \exists k\in\mathbb{N}: S^k(a) = b$$
where the $k$-th power is understood as $k$ fold application of $S$.
This is essentially the same answer already given by Solitary.
A: You can either have a direct definition or a recursive definition.  If you have a recursive definition you need a base case from which all cases arrive.
Your function appears to be recursive but it has no base case.
a < b if a + 1< b + 1 which raises the question what is the definition of a + 1 < b + 1 to which a + 1 < b+a if a + 2 < b +2, and final verification is pushed further and further away.
So if you are going to do recursion, you need a base case involving 0


*

*$0 < b$ if $b \ne 0$


Now your definition $a < b$ if $a + 1 < b + 1$ ... isn't good because it is taking you away from the base case.  We need a definition that either a) takes you from the base case to $a < b$ or b) takes you from $a < b$ to the base case.
Either
2a. $a < b $ if $a - 1 < b -1$ (allows the user to start at $a<b$ and work down to $0 < b'$)
Or 
2b. if $a < b$ then $a + 1 < b + 1$ (allows the user to start at $0<b'$ annd work up to $a < b$)
will do.  Which one you like is a matter of taste.
====
Then there is a direct definition.  This is less obvious to see but more "powerful" and ,ahem, direct to use.  When is $a < b$ true?  It's true when $0 < b- a$ which, as these are natural numbers rather than integers, is true whenever $b - a \ne 0$ and $b - a$ is a legitimate natural number.
So


*

*$a < b$ if $b - a \ne 0$ and $b - a \in \mathbb N$.

