To prove that every element of $X$ is even, let's proceed with mathematical induction based on the recursive definition of $X$.

Definition Recap

1. Base case (B): $2 \in X$.
2. Recursive rule (R): If $x \in X$, then $x + 10 \in X$.

We need to prove that every element $x \in X$ is an even number.

Proof by Induction

Step 1: Base Case

The base element $2 \in X$ is clearly even.

Step 2: Inductive Hypothesis

Assume that for some $k \in X$, $k$ is even. In other words, $k = 2n$ for some integer $n$.

Step 3: Inductive Step

By the recursive rule (R), since $k \in X$, we have $k + 10 \in X$. Now, let's show that $k + 10$ is also even: $k + 10 = 2n + 10 = 2(n + 5),$ which is clearly an even number (as it is divisible by 2).

Conclusion

By induction, every element of $X$ is even.

This completes the proof.

Do you have any questions, or would you like more details?

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Here are some related questions to explore:

1. How would the proof change if the recursive rule was $x + 5$ instead of $x + 10$?
2. What would happen if the initial element in $X$ was an odd number?
3. Can we use induction to show all elements of $X$ are divisible by any constant multiple?
4. How would we modify the proof if the base case was $4 \in X$ instead of $2$?
5. What if there were two base cases, like $2 \in X$ and $4 \in X$?

Tip: In recursive proofs, always verify if each added element maintains the required property to simplify induction.