To prove that in any field $F$, for all $a, b \in F$, the equation $(-a) \cdot (-b) = a \cdot b$ holds, we will use the properties of fields, specifically the distributive property, the existence of additive inverses, and the identity elements for addition and multiplication.

Proof:

Step 1: Consider the expression $(-a) \cdot b$.

Using the distributive property, we have: $(-a) \cdot b = (-a) \cdot (b + 0)$ Since $0$ is the additive identity in $F$, we can write: $(-a) \cdot b = (-a) \cdot (b + (0)) = (-a) \cdot b + (-a) \cdot 0$ By the property of multiplication with zero (in any field), $(-a) \cdot 0 = 0$. Therefore: $(-a) \cdot b = (-a) \cdot b + 0 = (-a) \cdot b$ So, this equation is trivially true.

Now consider: $(-a) \cdot b + a \cdot b = ((-a) + a) \cdot b$ Since $(-a) + a = 0$ (the definition of the additive inverse), we have: $0 \cdot b = 0$ So: $(-a) \cdot b + a \cdot b = 0$ This implies that: $(-a) \cdot b = -(a \cdot b)$ This shows that the product of $-a$ and $b$ is the additive inverse of $a \cdot b$.

Step 2: Consider the expression $(-a) \cdot (-b)$.

Using the result from Step 1 with $b$ replaced by $-b$, we have: $(-a) \cdot (-b) = -(a \cdot (-b))$ From Step 1 again, we also know that: $a \cdot (-b) = -(a \cdot b)$ Therefore: $(-a) \cdot (-b) = -(-(a \cdot b))$ In a field $F$, the additive inverse of the additive inverse of any element $x \in F$ is $x$ itself. That is: $-(-(a \cdot b)) = a \cdot b$ Thus, we have shown that: $(-a) \cdot (-b) = a \cdot b$ This completes the proof.

Conclusion:

In any field $F$, for all $a, b \in F$, $(-a) \cdot (-b) = a \cdot b$. This result follows directly from the field properties, particularly the distributive property and the behavior of additive inverses.

Would you like to explore any particular aspect of this proof further, or do you have other questions?

Related Questions:

1. How does the property $(-x) \cdot (-y) = x \cdot y$ extend to non-commutative rings?
2. Can this property be used to derive any other identities in a field?
3. What happens if we consider elements of a ring that is not a field?
4. How does the distributive property play a critical role in this proof?
5. How would you prove that the additive inverse is unique in any field?

Tip:

When working with fields, always rely on the fundamental properties (associativity, commutativity, distributivity, identity elements, and inverses) to construct rigorous proofs. These properties are key to understanding how elements in fields interact.