a⊗b = ab + a + b = ab + b + a (because ORDINARY addition is commutative)

= ba + b + a (because ORDINARY multiplication is commutative)

= b⊗a. you're over-thinking it.

if A has a unity, we should be able to FIND it, ya?

such a unity has the property that

a⊗x = x⊗a = a for all a in A. so let's write this in terms of operations we understand better:

ax + a + x = a

ax + x = 0

x(a+1) = 0

now if a is not -1, this means x= 0. so the multiplicative identity is 0:

a⊗0 = a0 + 0 + a = a.

now to prove any non-zero element of A is invertible, we need to know what "zero" (the additive identity) means.

it should be clear that the additive identity is -1:

a⊕(-1) = a + -1 + 1 = a.

so in A, "non-zero" means A-{-1}. so we have to show that if a is not -1, there is some b in A with a⊗b = 0 (the multiplicative identity).

again, let's try to FIND b, by solving for b in terms of the ordinary operations:

a⊗b = 0

ab + a + b = 0

ab + b = -a

b(a+1) = -a since a is not -1, a+1 is not 0, so we can divide by it:

b = -a/(a+1).

verifying:

a⊗(-a/(a+1)) = -a^{2}/(a+1) + a - a/(a+1) = (-a[SUP]2 - a)/(a+1) + a(a+1)/(a+1) = (-a^{2}- a + a^{2}+ a)/(a+1) = 0/(a+1) = 0.

***************

a faster way, given that A is already a ring:

define φ:Q-->A by φ(q) = q-1.

a) φ is injective:

φ(q') = φ(q) => q'-1 = q-1 => q' = q

b) φ is surjective:

let a be any element of A, then for q = a+1, φ(q) = (a+1) -1 = a.

c) φ is a ring homomorphism:

φ(q+q') = q+q'-1 = (q-1) + (q'-1) + 1 = (q-1)⊕(q'-1) = φ(q)⊕φ(q')

φ(qq') = qq' - 1 = qq' - q - q' + 1 + q - 1 + q' - 1 = (q-1)(q'-1) + (q-1) + (q'-1) = (q-1)⊗(q'-1) = φ(q)⊗φ(q')

together, (a) through (c) say φ is a ring isomorphism, thus φ(Q) = A is a field.