To prove that any natural number $n$ greater than 3 can be written in the form $n = 2k + 5l$ where $k, l \in \mathbb{N}_0$ (the set of non-negative integers), we'll approach this problem using the principle of mathematical induction.

Step 1: Base Case

We first check the statement for some small values of $n$ greater than 3.

• For $n = 4$: $4 = 2(2) + 5(0)$ Here, $k = 2$ and $l = 0$, which are both in $\mathbb{N}_0$.

• For $n = 5$: $5 = 2(0) + 5(1)$ Here, $k = 0$ and $l = 1$, which are both in $\mathbb{N}_0$.

• For $n = 6$: $6 = 2(3) + 5(0)$ Here, $k = 3$ and $l = 0$, which are both in $\mathbb{N}_0$.

• For $n = 7$: $7 = 2(1) + 5(1)$ Here, $k = 1$ and $l = 1$, which are both in $\mathbb{N}_0$.

• For $n = 8$: $8 = 2(4) + 5(0)$ Here, $k = 4$ and $l = 0$, which are both in $\mathbb{N}_0$.

This shows that the statement holds for the base cases $n = 4, 5, 6, 7,$ and $8$.

Step 2: Inductive Step

Now, assume that for some $n \geq 8$, the statement holds, i.e., $n = 2k + 5l$ for some $k, l \in \mathbb{N}_0$. We need to show that $n + 1$ can also be written in the form $2k' + 5l'$ for some $k', l' \in \mathbb{N}_0$.

There are two main cases to consider:

Case 1: If $k \geq 1$, then: $n + 1 = 2k + 5l + 1 = 2(k - 1) + 5l + 3$ Now, we need to express $3$ in the form $2k' + 5l'$, but $3 = 2(2) + 5(-1)$, which is not valid since $l' = -1$ is not in $\mathbb{N}_0$.

However, we can instead use: $3 = 2(4) - 5(1)$ If we increase $l$ by 1 and decrease $k$ accordingly: $n + 1 = 2(k - 2) + 5(l + 1)$ Where $k' = k - 2$ and $l' = l + 1$, so $k', l' \in \mathbb{N}_0$.

Case 2: If $k = 0$ (which can occur when $n = 5l$), then: $n + 1 = 5l + 1$ We know that: $1 = 2k' + 5l'$ But $1$ can’t be expressed in the form $2k' + 5l'$ directly.

Thus, if we add $1$, we can consider: $n + 1 = 5(l - 1) + 6$ Where $6 = 2(3) + 5(0)$.

Conclusion:

By induction, since we've shown the base cases and that $n + 1$ can be expressed in the required form whenever $n$ can, the proposition holds true. Hence, any natural number $n$ greater than 3 can be written in the form $n = 2k + 5l$ with $k, l \in \mathbb{N}_0$.

Would you like further details, or do you have any questions?

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1. What is the principle of mathematical induction?
2. How do you prove a statement using the inductive step?
3. Why can't some numbers like 1, 2, or 3 be expressed as $2k + 5l$?
4. What are other examples of mathematical problems that use induction?
5. How would you modify this proof if the condition was $n$ greater than 2?

Tip: Always check the base case carefully when using mathematical induction—it forms the foundation of your proof!