Let M be any set with multiplication. Given and only using the following axioms
$1. \forall a,b\in M : a\cdot b=b\cdot a \\ 2. \forall a,b,c\in M: (a\cdot b)\cdot c=a\cdot (b\cdot c) \\3. \exists 1\in M\ ~ \forall a \in M:a\cdot 1=a \\4. \forall a \in M \setminus \{0\} \exists a^{-1} \in M : a\cdot a^{-1}=1$
Show that: $\forall a,b \in M: a\neq0$ the equations $a\cdot x=b$ and $x\cdot a=b$ has an unambigous/unique solution x.
My solution:
Separate the equation into two cases and solve for x respectively
case 1:
$a\cdot x=b \\ a^{-1}\cdot (a\cdot x)=(b)\cdot a^{-1}\\ (a^{-1}\cdot a) \cdot x= b\cdot a^{-1}\\1\cdot x=b\cdot a^{-1}\\x=b\cdot a^{-1}\\$
case 2:
$x\cdot a=b\\a^{-1}\cdot (x\cdot a)=(b)\cdot a^{-1}\\x\cdot (a^{-1}\cdot a)=b\cdot a^{-1}\\x\cdot 1=b\cdot a^{-1}\\ x=b\cdot a^{-1}$
Both cases gave the same solution for x, therefore x is an unambigous/unique solution to the equations.
I'm new to latex so feel free to edit or comment on that as well! thanks:)