Can there exist a non-associative operation with an identity element,such that every element has an inverse? As far as my understanding goes, the inverse of each element need not be even unique based on what is given. But it looks like they are asking for a where it is actually unique despite it needing not to be. And I am not being able to come up with an example even after trying trying. Can you guys suggest a few examples. And also, can you please share the thought process behind it, like how you came up with it?
 A: Multiplication of (nonzero) octonions is non-associative but has inverses.
For a simpler if less natural example, consider $\mathbb{R}\cup\{\infty\}$ and let $*$ be the operation given as follows:

*

*$\infty*a=a*\infty=a$ for all $a$.


*For $a\not=b$ in $\mathbb{R}$, we set $a*b={a+b\over 2}$.


*For each $a\in\mathbb{R}$ we set $a*a=\infty$.
In this structure $\infty$ functions as the identity element and each element is its own inverse, but $*$ is nonassociative.
A: Well, if you don't have any constraint on the set and the operation it's not hard to create one by just adding new element to a set with a non associative operator and then extending the operator.
For instance take $S=\mathbb{R}^3$ with the vector product $\wedge$, which is a not associative operator which has no inverse.
Take a copy of $S$ and call it $\bar S$. Assume there is a bijection $S\to \bar S$ with maps a element $s$ in $\bar s$. Define a new set $S'$ as
$$S' = S\cup \bar S\cup\{E,F\}\,.$$
You can extend $\wedge$ in the following way. For $v,w\in S$ just the usual $v\wedge w$. For $\bar v,\bar w\in\bar S$ you can define $\bar v\wedge \bar w = \overline{v\wedge w}$. For $v\neq w$ define $v\wedge \bar w = \bar v \wedge w = F$. Then $v\wedge \bar v=\bar v \wedge v = E$. Finally define $F\wedge v = v\wedge F =F\wedge \bar v=\bar v\wedge F = F$. Then $F\wedge F = E\wedge E = E$. And finally make $E$ to be the unit element, i.e. for all $s\in S'$, $s\wedge E = E\wedge s = E$.
It seems very complicated but the idea is just that if you have a set with an associative operation you can add a copy of being the inverse, together with a neutral element. Then I introduced $F$ just to make it easier to extend the operation.
