Prove, by induction, that n - n is divisible by 3 for any natural number n.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![**Proof by Induction: Divisibility of \( n^3 - n \) by 3**
**Statement:** Prove, by induction, that \( n^3 - n \) is divisible by 3 for any natural number \( n \).
**Base Case:**
For \( n = 1 \),
\[ n^3 - n = 1^3 - 1 = 0, \]
which is divisible by 3.
**Inductive Step:**
Assume the statement is true for some natural number \( k \), i.e.,
\[ k^3 - k \]
is divisible by 3.
**Inductive Hypothesis:**
Suppose \( k^3 - k = 3m \) for some integer \( m \).
**To Prove for \( n = k + 1 \):**
Consider \( (k+1)^3 - (k+1) \).
Expanding \( (k+1)^3 \):
\[ (k+1)^3 = k^3 + 3k^2 + 3k + 1. \]
So,
\[ (k+1)^3 - (k+1) = (k^3 + 3k^2 + 3k + 1) - (k + 1) \]
\[ = k^3 + 3k^2 + 3k + 1 - k - 1 \]
\[ = k^3 - k + 3k^2 + 3k. \]
By the inductive hypothesis, \( k^3 - k \) is divisible by 3.
Also, \( 3k^2 + 3k \) is clearly divisible by 3.
Thus, \( (k+1)^3 - (k+1) \) is divisible by 3, completing the inductive step.
**Conclusion:**
By mathematical induction, \( n^3 - n \) is divisible by 3 for all natural numbers \( n \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13fe6e57-e6d9-41ab-bcf9-0392508f30a7%2F4aa32f60-cef5-4587-83b6-87f04dcf6ef2%2F02toocp_processed.png&w=3840&q=75)
Transcribed Image Text:**Proof by Induction: Divisibility of \( n^3 - n \) by 3**
**Statement:** Prove, by induction, that \( n^3 - n \) is divisible by 3 for any natural number \( n \).
**Base Case:**
For \( n = 1 \),
\[ n^3 - n = 1^3 - 1 = 0, \]
which is divisible by 3.
**Inductive Step:**
Assume the statement is true for some natural number \( k \), i.e.,
\[ k^3 - k \]
is divisible by 3.
**Inductive Hypothesis:**
Suppose \( k^3 - k = 3m \) for some integer \( m \).
**To Prove for \( n = k + 1 \):**
Consider \( (k+1)^3 - (k+1) \).
Expanding \( (k+1)^3 \):
\[ (k+1)^3 = k^3 + 3k^2 + 3k + 1. \]
So,
\[ (k+1)^3 - (k+1) = (k^3 + 3k^2 + 3k + 1) - (k + 1) \]
\[ = k^3 + 3k^2 + 3k + 1 - k - 1 \]
\[ = k^3 - k + 3k^2 + 3k. \]
By the inductive hypothesis, \( k^3 - k \) is divisible by 3.
Also, \( 3k^2 + 3k \) is clearly divisible by 3.
Thus, \( (k+1)^3 - (k+1) \) is divisible by 3, completing the inductive step.
**Conclusion:**
By mathematical induction, \( n^3 - n \) is divisible by 3 for all natural numbers \( n \).
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
