Multiplicative Inverse Manipulation Valid?

[snipez90]

Homework Statement

Let S be a field containing elements x and y. If a \neq 0 and b \neq 0 then ab \neq 0

Homework Equations

Field Axioms:

Associative law for addition
Existence of additive identity
Existence of additive inverse
Commutative law for addition
Associative law for multiplication
Existence of multiplicative identity
Existence of multiplicative inverses
Commutative law for multiplication
Distributive law

The Attempt at a Solution

The proof I want to implement is as follows:

Suppose ab = 0. Then,

(1)1 = 1 \cdot 1 (Existence of additive identity)
(2)= (a*a^{-1})(b*b^{-1}) (Existence of multiplicative inverses)
(3)= (a^{-1})(b^{-1})ab (Associative law for multiplication)
(4)= (a^{-1})(b^{-1})0 = 0 (A lemma that a*0 = 0, if a is a real)

which is a contradiction.

I have two primary concerns. The both involve the idea of substitution, which the professor didn't mention if we are allowed to use or not. Substitution is of course not an axiom (or we weren't given it).

Specifically I am focused on lines (2) and (4). In line (2), it seems the substitution is more valid in a sense that the product of a real and its multiplicative identity defines 1. But on line (4), it is simply substituting 0 in place for ab. We are allowed to add and multiply both sides by the same quantity and that's really all the algebra that were explicitly mentioned for use.

 

[snipez90]

OK sorry for the bump. The focus of my question is simply this:

We have a field that contain the 9 axioms listed above. These axioms involve binary operations. Adding or multiplying the same thing seems justified under the binary system we have. But what about a = b? Can we logically deduce that a and b are interchangeable? It's an equivalence statement so I can't think of a possible argument against using this in a proof. Nonetheless, it's not an axiom, do I have to prove it or does it follow logically.

 

[logarithmic]

"=" is an equivalence relation, so it is transitive. So substitution is justified.

 

[statdad]

Proof looks good (classical approach) - but I would eliminate the reference to 'a is real' in your final line, as it seems to me that you are referring to fields in general, not to the reals.

 

[snipez90]

Sorry I was aware of that as well, that's what I meant but I didn't want to define the field again (laziness on my part). Thanks for the replies.