Math Problem Statement

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.

Solution

To prove that every integer n1n \geq 1 can be expressed in the form n=k=1mckk!n = \sum_{k=1}^{m} c_k \cdot k! where each ckc_k is an integer and 0ckk0 \leq c_k \leq k, we'll use mathematical induction and the division algorithm.

Base Case

For n=1n = 1: We need to express 11 in the form k=1mckk!\sum_{k=1}^{m} c_k \cdot k! where 0ckk0 \leq c_k \leq k for each kk.

Since 1=11!1 = 1 \cdot 1!, we can take c1=1c_1 = 1. Thus, the base case holds with m=1m = 1 and c1=1c_1 = 1.

Inductive Hypothesis

Assume that for some integer n1n \geq 1, we can express nn in the form n=k=1mckk!n = \sum_{k=1}^{m} c_k \cdot k! where 0ckk0 \leq c_k \leq k for all kmk \leq m.

Inductive Step

We need to show that n+1n + 1 can also be expressed in the same form.

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

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

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

    Thus, we have 0qm0 \leq q \leq m.

  4. Construct the Representation for n+1n + 1: Write n+1=qm!+r.n + 1 = q \cdot m! + r. Since r<m!r < m!, by the inductive hypothesis, rr can be written as a sum of factorial terms in the form r=k=1m1ckk!r = \sum_{k=1}^{m-1} c_k \cdot k! where 0ckk0 \leq c_k \leq k for each km1k \leq m-1.

  5. Combine Terms: Substitute the expression for rr back into the equation for n+1n + 1: n+1=qm!+k=1m1ckk!.n + 1 = q \cdot m! + \sum_{k=1}^{m-1} c_k \cdot k!. This expression represents n+1n + 1 as a sum of factorial terms, with coefficients ckc_k for k<mk < m and cm=qc_m = q, satisfying 0ckk0 \leq c_k \leq k for all kmk \leq m.

Therefore, by induction, every integer n1n \geq 1 can be expressed in the form n=k=1mckk!n = \sum_{k=1}^{m} c_k \cdot k! where 0ckk0 \leq c_k \leq k for each kk.

Conclusion

This completes the proof by induction. The result follows that any integer n1n \geq 1 can be expressed as a sum of factorials with the given constraints on the coefficients ckc_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 mm such that m!n+1<(m+1)!m! \leq n + 1 < (m+1)!?
  3. What is the significance of having 0ckk0 \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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Mathematical Induction
Factorials
Division Algorithm

Formulas

n = Σc_k * k!

Theorems

Inductive Step
Division Algorithm

Suitable Grade Level

Grades 10-12