(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant 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

3. The attempt at a solution

The proof I want to implement is as follows:

Suppose [tex]ab = 0[/tex]. Then,

(1)[tex]1 = 1 \cdot 1[/tex] (Existence of additive identity)

(2)[tex]= (a*a^{-1})(b*b^{-1})[/tex] (Existence of multiplicative inverses)

(3)[tex]= (a^{-1})(b^{-1})ab[/tex] (Associative law for multiplication)

(4)[tex]= (a^{-1})(b^{-1})0 = 0[/tex] (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.

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# Homework Help: Multiplicative Inverse Manipulation Valid?

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