why division by zero is not possible

Division, by definition, is multiplying by an inverse element. This means that $\frac{a}{b}$ is defined to be equal to $a\cdot b^{-1}$, and $b^{-1}$ is defined as

$b^{-1}$ is the unique real number that satisfies $b\cdot b^{-1}=1$

The problem with dividing by $0$ is that $0^{-1}$ does not exist, because if $x=0^{-1}$ did exist, then it would satisfy the equation $0\cdot x = 1$, but since $0\cdot x=0$ for all $x$ and $0\neq 1$, this is impossible.

This example shows that multiplication and division are actually not inverse processes. "Partially inverse processes" would be a better description. More precisely, multiplication by zero is not an invertible operation, because we can find many inputs that all give the same result when multiplied by zero.

In fact, it's worth noting that multiplication by 2 is not invertible in the complementary sense, when we're working in the integers: e.g. $2x = 3$ has no solutions for $x$ when we are working in the integers.

The point is that multiplication can only be partially inverted: "multiplication by $m$" can only be inverted if $m$ is a "unit": that is, if we can solve the equation $mx = 1$ for $x$.

Also, multiplication by $m$ can be partially inverted if $m$ is 'cancellable': that is, if the only solution to $mx = 0$ is $x=0$. If we had two solutions $x=a$ and $x=b$ to the solution $mx=c$, then $m(a-b)=ma - mb = c - c = 0$ and therefore $a=b$. Thus in this case, solutions are unique if they exist, although they aren't guaranteed to exist. (e.g. my multiplication by $2$ in the integers example above)

Since $a\cdot 0=0$ and $b \cdot 0=0$, then $\frac{0}{0}=?$ $a$ or $b$?