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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:)

Mauro ALLEGRANZA
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Geeniee
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  • Oh alright thanks! Are the "cases" simply overkill or can it be justified? @BillDubuque – Geeniee Oct 14 '20 at 22:10
  • The existence proof is correct (but you need only $1$ case since $,ax = xa).,$ The uniqueness proof is not correct. To prove that you need to show that if $,x_1$ and $x_2$ are both roots of $,ax = b,$ then $,x_1 = x_2,$ This follows immediately from your proof, which shows every solution $ = b a^{-1}$, see the dupes. – Bill Dubuque Oct 14 '20 at 22:12
  • Great thank you :)@BillDubuque – Geeniee Oct 15 '20 at 07:28

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