To prove that $1x = x$ for all $x$ in a finite field $\mathbb{F}_n$, we can use the basic properties of fields. Here's a step-by-step proof:

Proof:

1. Field Axioms: By definition, a field $\mathbb{F}_n$ is a set equipped with two operations, addition (+) and multiplication (·), that satisfy certain axioms (e.g., commutativity, associativity, distributivity, existence of identity elements, and existence of inverses for all elements except zero for multiplication).

2. Multiplicative Identity: One of the axioms of a field is the existence of a multiplicative identity, denoted as 1, such that for any element $x \in \mathbb{F}_n$, $1 \cdot x = x \cdot 1 = x$ This property holds for all $x$ in the field $\mathbb{F}_n$.

3. Applying the Axiom: According to this axiom, by multiplying any element $x$ by 1, the result is $x$ itself. Specifically: $1 \cdot x = x$ This shows that the element 1 acts as the multiplicative identity, meaning it leaves any element $x$ unchanged when multiplied by it.

4. Conclusion: Therefore, for any $x$ in the field $\mathbb{F}_n$, we have shown that $1x = x$, proving the statement.

This completes the proof that $1x = x$ for all $x$ in $\mathbb{F}_n$.

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Would you like more details or have any questions?

Here are 5 related questions to consider:

1. What is the definition of a finite field $\mathbb{F}_n$?
2. Why is the existence of a multiplicative identity important in a field?
3. How can you prove that $0 \cdot x = 0$ in a field $\mathbb{F}_n$?
4. What are some examples of finite fields?
5. How does the property $1x = x$ relate to the concept of scalar multiplication in vector spaces?

Tip: When dealing with fields, always remember that the identity elements (both additive and multiplicative) play a crucial role in defining the structure and properties of the field.