Mathenatucak Induction Problems in discrete math

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Homework Help Overview

The discussion revolves around mathematical induction problems in discrete mathematics, specifically focusing on proving divisibility properties involving polynomial expressions and Fibonacci numbers.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the base and inductive steps of mathematical induction, with attempts to clarify the correct formulation of the inductive hypothesis and the expansion of expressions. There are questions about the validity of certain algebraic manipulations and the need for clear statements to prove in the second problem.

Discussion Status

The discussion is ongoing, with some participants seeking clarification on the steps involved in the proofs. There is a mix of confusion and attempts to guide each other through the reasoning process, particularly regarding the first problem's inductive step and the requirements for the second problem.

Contextual Notes

Participants note the importance of having a clear statement to prove for the second problem, indicating a potential lack of information or clarity in the problem setup.

GoGoDancer12
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Homework Statement



Prove that 3 divides n3 + 2n whenever n is a positive integer.

Homework Equations


The Attempt at a Solution



Basis Step :
P(1) : [13 + 2(1) ] /3
[1+2] /3
[3]/3
1
Since 3/3 =1, P(1) is true

Inductive Step:
[ k3 + 2k ] /3 = [(k +1)3 + 2(k+1) ] /3

[k(k2 + 2k)] /3 = [(k)3 + (3k)2 + 5k +3] /3

Homework Statement



Prove that f1 2 + f2 2 +...+ fn 2 when n is a positive integer.

Homework Equations


The Fibonacci numbers f0, f1, f2..., are defined by the equations f0 = 0, f1 = 1 and fn = fn-1 + fn-2 for n = 2,3,4...

The Attempt at a Solution



Basic Step:
P(1) : f1 2 = f1 * f2

12 = (1)*(1)
1= 1
Since f1 2 = f1 * f2 , P(1) is true.

Inductive Step:
fk 2 = fk * fk+1

fk 2 + fk+1 2
= f1+fk+1 + (fk+1)
 
Last edited:
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For the first problem, I'm having some trouble understanding what you're doing.
You have established the base case, P(1). IOW, when n = 1, n^3 + 2n = 1 + 2 = 3, so the proposition is true for n = 1.

Assume that the proposition is true when n = k. IOW, assume that k^3 + 2k is divisible by 3. Another way to say this is that k^3 + 2k = 3m for some integer m.

Now, when n = k + 1, you want to show/prove that (k + 1)^3 + 2(k + 1) is divisible by 3. You do NOT want to show that k^3 + 2k = (k + 1)^3 + 2(k + 1), which is essentially what you wrote in your post.

To show that (k + 1)^3 + 2(k + 1) is divisible by 3, expand the two terms. You will need to use the assumption that k^3 + 2k = 3m.
 
I'm lost about showing that (k + 1)^3 + 2(k + 1) is divisible by 3. I expanded the right side of the equation [k^3 + 2k)] /3 = [(k)^3 + (3k)^2 + 5k +3] /3. So, do I replace on the left side [k(k^2 + 2k)] /3 with 3m??
 
Last edited:
GoGoDancer12 said:
I'm lost of showing that (k + 1)^3 + 2(k + 1) is divisible by 3. I expanded the right side of the equation [k^3 + 2k)] /3 = [(k)^3 + (3k)^2 + 5k +3] /3. So, do I replace on the left side [k(k^2 + 2k)] /3 with 3m??
This equation is bogus - [k^3 + 2k)] /3 = [(k)^3 + (3k)^2 + 5k +3] /3 - get rid of it, and also get rid of those /3 things that you have.

Take another look at what I said at the end of my previous post.
 
Ok, what about the second problem??
 
What is the second problem?
GoGoDancer12 said:
Prove that f1 2 + f2 2 +...+ fn 2 when n is a positive integer.
There's nothing there to prove. You have to have a statement of some kind in order to prove it.
 

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