What is an example of a groupoid which is not a semigroup? I know that groupoid refers to an algebraic structure with a binary operation. The only necessary condition is closure.
However, I couldn't find any easy-to-understand example of a groupoid which is not a semigroup. I did come across some examples of (certain type of) matrices but then matrix multiplication is always associative (thus making it a semi-group).
So, could someone please provide me an example of a groupoid which isn't a semigroup? 
 A: These are called magmas, not groupoids.
The ``midpoint'' operation $s\ast t=\frac{s+t}{2}$ on $\mathbb{R}$ makes it a magma which is not a semigroup.
A: Here's three different examples.


*

*Take an abelian group $(A,+)$ and define a new binary operation $\circ$ on $A$ by $x\circ y=x+(-y)$. This is an example of a quasigroup.

*Take a group $(G,\cdot)$ and define a new binary operation $\triangleleft$ on $G$ by $x\triangleleft y=x\cdot y \cdot x^{-1}$. This is an example of a quandle.

*Take a digraph $(V,A)$ with the property that for any two distinct vertices $v,w\in V$, exactly one of the arcs $vw$ or $wv$ is in $A$. Define a commutative binary operation $\cdot$ on $V$ by $v\cdot v=v$ and $v\cdot w=w$ if and only if $vw\in A$. This is an example of a tournament.
A quasigroup is associative if and only if it is a group, a quandle is associative if and only if it is trivial, and a tournament is associative if and only if it is a commutative idempotent semigroup (aka a semilattice).
A: $( \Bbb Z , -)$
This is a groupoid and not a semi-group. 
A: First, the term 'groupoid' recently rather means primarily a category with invertible arrows, and the term 'magma' is arising for an algebraic structure with a binary operation. 
For a familiar example, consider $\Bbb Z$ (or almost any Abelian group) with the subtraction. 
Or, define $a*b:=a+b+1$ (or whatever..)
Other examples arise e.g. from finite quasigroups whose multiplication table is a latin square: having each element  once in every row and in every column. 
A: Let $a,b,c$ be distinct members of a three element set and $ab=c=cc $, $bc=a=aa$ and define $ac, bb $ however you like (but in $\{a,b,c\}$.) You have a nonassociative binary operation.
A: Let $\{a,b\}$ be a set with two distinct elements.
Define a partial multiplication by $a\times a=a$ and $b\times b=b$ and nothing more (so $a\times b$ and $b\times a$ are not defined). 
Then $\{a,b\}$ equipped with this multiplication is a groupoid, but not a semigroup. 
I suspect you use another definition of groupoid. If it is what I would call a magma then see the other answers.
A: What about the set $\mathbb{R}^+$ of positive real numbers with exponentiation?
It is not associative:
$$2^{(1^3)}=2\ne 8=(2^1)^3$$
and not commutative:
$$2^3=8\ne 9=3^2.$$
It neither has identity element. Actually, $1$ is a left-absorbing and right identity element:
$${1^n=1\ \forall\,n}\quad\text{and}\quad{n^1=n\ \forall\,n}.$$
A: A non-numerical example of non-associative magma is the Jankenpon game. Let's assume the dyadic operation "*" determines the winner of a match:
(rock * paper) * scissors = paper * scissors = scissors
rock * (paper * scissors) = rock * scissors = rock
Therefore,
(rock * paper) * scissors $\neq$ rock * (paper * scissors).
Incidentally, this magma happens to be commutative and idempotent.
