Let n >= 1 be an integer. Prove that n can be expressed in the form :
n = Σc_k * k! , from k = 1 to m
where each c_k is an integer and 0 <= c_k <= k. For example, 44 = 1*(4!) + 3*(3!) + 1*(2!) + 0*(1!).
Hint: Use induction and apply the division algorithm to divide n+1 by m! for a suitable m.  

To prove that every integer \( n \geq 1 \) can be expressed in the form
\[
n = \sum_{k=1}^{m} c_k \cdot k!
\]
where each \( c_k \) is an integer and \( 0 \leq c_k \leq k \), we'll use mathematical induction and the division algorithm.

### Base Case
For \( n = 1 \):
We need to express \( 1 \) in the form \( \sum_{k=1}^{m} c_k \cdot k! \) where \( 0 \leq c_k \leq k \) for each \( k \).

Since \( 1 = 1 \cdot 1! \), we can take \( c_1 = 1 \). Thus, the base case holds with \( m = 1 \) and \( c_1 = 1 \).

### Inductive Hypothesis
Assume that for some integer \( n \geq 1 \), we can express \( n \) in the form
\[
n = \sum_{k=1}^{m} c_k \cdot k!
\]
where \( 0 \leq c_k \leq k \) for all \( k \leq m \).

### Inductive Step
We need to show that \( n + 1 \) can also be expressed in the same form.

1. **Choose \( m \)**: Let \( m \) be the largest integer such that \( m! \leq n + 1 \). According to this choice, we have:
   \[
   m! \leq n + 1 < (m+1)!
   \]

2. **Apply the Division Algorithm**: Divide \( n + 1 \) by \( m! \) using the division algorithm. This gives:
   \[
   n + 1 = q \cdot m! + r
   \]
   where \( 0 \leq r < m! \) and \( q \) is an integer.

3. **Check Bounds for \( q \)**: Since \( n + 1 < (m+1)! = (m+1) \cdot m! \), it follows that \( q \leq m \).

   Thus, we have \( 0 \leq q \leq m \).

4. **Construct the Representation for \( n + 1 \)**: Write
   \[
   n + 1 = q \cdot m! + r.
   \]
   Since \( r < m! \), by the inductive hypothesis, \( r \) can be written as a sum of factorial terms in the form
   \[
   r = \sum_{k=1}^{m-1} c_k \cdot k!
   \]
   where \( 0 \leq c_k \leq k \) for each \( k \leq m-1 \).

5. **Combine Terms**: Substitute the expression for \( r \) back into the equation for \( n + 1 \):
   \[
   n + 1 = q \cdot m! + \sum_{k=1}^{m-1} c_k \cdot k!.
   \]
   This expression represents \( n + 1 \) as a sum of factorial terms, with coefficients \( c_k \) for \( k < m \) and \( c_m = q \), satisfying \( 0 \leq c_k \leq k \) for all \( k \leq m \).

Therefore, by induction, every integer \( n \geq 1 \) can be expressed in the form
\[
n = \sum_{k=1}^{m} c_k \cdot k!
\]
where \( 0 \leq c_k \leq k \) for each \( k \).

### Conclusion
This completes the proof by induction. The result follows that any integer \( n \geq 1 \) can be expressed as a sum of factorials with the given constraints on the coefficients \( c_k \).

Would you like further details on any part of this proof or have additional questions?

1. How does the division algorithm help in breaking down numbers?
2. Why do we choose \( m \) such that \( m! \leq n + 1 < (m+1)! \)?
3. What is the significance of having \( 0 \leq c_k \leq k \)?
4. How does this approach ensure uniqueness in the representation?
5. Can this factorial representation be applied to negative numbers?

**Tip:** Inductive proofs often simplify complex problems by establishing a base and building incrementally from it. This can help in breaking down abstract concepts into manageable parts.

Let n >= 1 be an integer. Prove that n can be expressed in the form: n = Σc_k * k!, from k = 1 to m where each c_k is an integer and 0 <= c_k <= k. For example, 44 = 1*(4!) + 3*(3!) + 1*(2!) + 0*(1!). Hint: Use induction and apply the division algorithm to divide n+1 by m! for a suitable m.

Learn how to express any integer n ≥ 1 as a sum of factorials with integer coefficients using mathematical induction and the division algorithm. This proof covers key concepts in factorials and division algorithms. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation