Can anyone solve this mathematical induction problem?

[ctothe]

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

 

[rude man]

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 2^p.

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.

 

[CEL]

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.

 

[rude man]

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!

 

[rezeew]

. i couldn't figure out how to prove that this case is also true.I would first commend the person for attempting to solve the problem using mathematical induction. It shows a good understanding of the concept and a willingness to think critically and logically.

I would then offer some suggestions for their attempted solution. In the inductive step, instead of assuming that P(n) is true, I would suggest using the inductive hypothesis, which states that P(n) is true for some arbitrary n. This allows for a stronger and more general proof.

For the second case, where n+1 is even, I would suggest breaking it down further into two subcases: n is even and n is odd. This is because when n is even, n+1 is odd and when n is odd, n+1 is even. This will help in showing that the numerator and denominator in the expression (k(n+1) + m)/(m+mn) are both odd and even, respectively, regardless of whether n+1 is even or odd.

I would also suggest considering the fact that the sum of two odd numbers is even, and the sum of an even number and an odd number is odd. This can help in simplifying the expression and showing that the numerator and denominator are both odd and even, respectively.

Overall, I would encourage the person to keep exploring and trying to solve the problem. Mathematical induction can be tricky, but with practice and perseverance, they will be able to master it.