How do you prove that a/b=ab^-1?

L.H.S.
= a / b
= 1 × (a/b)              (Identity)
= (b⁻¹ × b) × (a/b)   (Inverse)
= b⁻¹ × (a × b/b)     (Associativity)
= b⁻¹ × (a × 1)        (Identity)
= b⁻¹ × a                 (Identity)
= a b⁻¹                    (Commutativity)
= R.H.S.

Hence, a / b = a b⁻¹

by definition b^(–1) = 1/b

Suppose S is a field, then the subset S'=S-{0} forms
a commutative group under multiplication with the neutral element=1.
Let a, b & x be the element of S' &
x=ab^-1
=>
xb=a(b^-1)b
=>
xb=a[(b^-1)b]
=>
xb=a*1=a
=>
xb(1/b)=a(1/b)
=>
x=a/b
(since 1/b is an element in S')
Thus,
ab^-1=a/b

a/b = a(b^-1)

Multiply both sides by (1/a) and we have

1/b = b^-1

Multiply both sides by b

1 = b (b^-1)

Using law of exponents , right hand side becomes

 b ^ (1 - 1)

= b ^ 0

= 1 

Left hand side is now equal to right hand side i.e.

1 = 1

True. 

a/b = a/b = ab^-1
True

Read billrussel42's short answer and think about awarding a favorite there.

There is no division operation in (the usual statement of) the field axioms.  Division is *defined* to mean: x / y = x • y⁻¹.  There is no proof for a definition.

A later answer gives a long-winded "proof", but stumbles at the fifth step where it replaces b/b with 1 without explanation.  That's basically assuming what is to be proven, at least for the case b=a.  If you try to "fix" the missing explanation by first replacing a/b with a • b⁻¹, according to the definition, then you have directly used what is to be proven.