To show that $(x + y) \cdot z = (x \cdot z) + (y \cdot z)$ for all $x, y, z \in F$, we will use the properties of fields.

Proof:

1. Distributive Property: A field $F$ has the property of distributivity. Specifically, for any elements $a, b, c \in F$, the distributive law states: $a \cdot (b + c) = a \cdot b + a \cdot c.$ This means we can distribute multiplication over addition.

2. Application of the Distributive Property: Let $x, y, z \in F$. By applying the distributive property, we rewrite the left-hand side: $(x + y) \cdot z = z \cdot (x + y).$ Now we apply the distributive property on $z$ and $(x + y)$: $z \cdot (x + y) = z \cdot x + z \cdot y.$

3. Rearranging Terms: Since multiplication in fields is commutative, we can rearrange: $z \cdot x + z \cdot y = (x \cdot z) + (y \cdot z).$

Thus, we have shown: $(x + y) \cdot z = (x \cdot z) + (y \cdot z).$

Conclusion:

The property $(x + y) \cdot z = (x \cdot z) + (y \cdot z)$ holds for all $x, y, z \in F$.

Do you want details or have any questions? Here are some related questions to consider:

1. What are the key properties of fields?
2. Can you provide an example of a field and verify this property?
3. How does this property relate to vector spaces?
4. What happens if we only consider a ring instead of a field?
5. Can you prove other properties of fields using a similar approach?

Tip: Always remember to verify the properties of algebraic structures when proving relationships!