Can anyone solve this mathematical induction problem?

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Discussion Overview

The discussion revolves around a mathematical induction problem concerning the sum of the harmonic series, specifically proving that for all n > 1, the sum can be expressed as a fraction k/m where k is an odd number and m is an even number. The participants explore the base case and the inductive step, examining the conditions under which the numerator and denominator maintain their parity.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the base case for n=2, asserting that P(2) = 3/2, which fits the form k/m with k odd and m even.
  • The inductive step is outlined, where it is assumed that P(n) holds true for some arbitrary n, leading to the need to prove P(n+1).
  • In the inductive step, the expression for P(n+1) is derived, and the participant divides the analysis into two cases based on the parity of n+1.
  • Case 1 discusses when n+1 is odd, concluding that the numerator remains odd and the denominator even, which is claimed to be true.
  • Case 2 addresses when n+1 is even, where the participant struggles to show that the numerator remains odd while the denominator is even.
  • Another participant suggests rewriting the denominator and multiplying by a positive integer to demonstrate the parity of the numerator and denominator.
  • A later reply challenges the assertion that k(n+1) is odd when (n+1) is even, arguing that the product of an odd number and an even number is always even.

Areas of Agreement / Disagreement

Participants express differing views on the parity of the numerator and denominator in the case where n+1 is even. There is no consensus on the correctness of the claims made regarding the oddness or evenness of the resulting expressions.

Contextual Notes

The discussion includes unresolved mathematical steps, particularly in the inductive proof, and relies on assumptions about the parity of k and m without fully establishing conditions for all cases.

ctothe
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Homework Statement


Prove that for all n>1,
P(n) :[itex]1 + 1/2 + 1/3 +...+1/n = k/m[/itex]

where k is an odd number an m is an even number.

Homework Equations


The Attempt at a Solution


1)Base Case n =2
P(2) = k/m
[itex]3/2 = k/m[/itex] which is true.

2) Inductive Step
Assume P(n) is true for some arbitrary n.

3) Prove P(n+1)
[itex]P(n+1) = k/m[/itex]
[itex]P(n) +1/(n+1) = k/m[/itex]
We know/assume that P(n) has an odd numerator and an even denominator. So,
[itex]k/m + 1/(n+1) = k/m[/itex]
[itex](k(n+1) +m)/ (m+mn)[/itex]

So i divided the problem into two cases:

Case 1: n+1 is odd
if n+1 is odd then k(n+1) + m is odd and m +mn is even which is true.

Case 2: n+1 is even
here's where the problem is
 
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ctothe said:
1. Case 2: n+1 is even
here's where the problem is


Let me rewrite your denominator mn + m = m(n+1).

Multiply m(n+1) by 2p.

Show that there exists a positive integer value for p such that the numerator must wind up odd whereas the denominator will always be even.
 
ctothe said:

Homework Statement


Prove that for all n>1,
P(n) :[itex]1 + 1/2 + 1/3 +...+1/n = k/m[/itex]

where k is an odd number an m is an even number.


Homework Equations





The Attempt at a Solution


1)Base Case n =2
P(2) = k/m
[itex]3/2 = k/m[/itex] which is true.

2) Inductive Step
Assume P(n) is true for some arbitrary n.

3) Prove P(n+1)
[itex]P(n+1) = k/m[/itex]
[itex]P(n) +1/(n+1) = k/m[/itex]
We know/assume that P(n) has an odd numerator and an even denominator. So,
[itex]k/m + 1/(n+1) = k/m[/itex]
[itex](k(n+1) +m)/ (m+mn)[/itex]

So i divided the problem into two cases:

Case 1: n+1 is odd
if n+1 is odd then k(n+1) + m is odd and m +mn is even which is true.

Case 2: n+1 is even
here's where the problem is

k is odd. So, k(n+1) is odd, no matter if (n+1) is odd or even. Since m is even, k(n+1) + m is odd.
 
CEL said:
k is odd. So, k(n+1) is odd, no matter if (n+1) is odd or even. Since m is even, k(n+1) + m is odd.

Huh?

If k is odd and (n+1) is even, k(n+1) is even!

Odd x even = even always!
 

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