5. Prove that any natural number n greater than 3 can be written in the form n = 2k + 5l with k, l ∈ N0.

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?

---

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!

Learn how to prove that any natural number n greater than 3 can be expressed as n = 2k + 5l, where k and l are non-negative integers. This detailed proof uses mathematical induction, covering key concepts in number theory and induction. Suitable for advanced high school students and college undergraduates.

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