Proving that the sum of fractions has an odd numerator and even denominator.I'm struggling to...

ScommaMaruj

ScommaMaruj

Answered

2022-07-01

Proving that the sum of fractions has an odd numerator and even denominator.
I'm struggling to show that, for all n > 1
1 + 1 2 + 1 3 + + 1 n = k m
where k is an odd number and m is an even number.
Proof: The proof is by induction on n
Base Case: 1 + 1 2 = 3 2
Assume the theorem is true for n and consider n + 1
1 + 1 2 + 1 3 + + 1 n + 1 n + 1
We know by the induction hypothesis that the first n terms take the form k m , where k is odd and m is even.
k m + 1 n + 1
To combine the terms, we find the least common multiple of m and n + 1. Since m is even, the least common multiple must be even.
Now What? I'm not sure how to show that the numerator remains odd. I don't think there's enough information to do just a case analysis on whether n + 1 is odd or even.
This question is from Manber's Introduction to Algorithms: A Creative Approach, which I'm using for personal development. He marked this question as "substantially more difficult."

Answer & Explanation

jugf5

jugf5

Expert

2022-07-02Added 18 answers

We actually need to prove by strong induction.
Suppose the result holds for all 2 , 3 , 4 , . . . , n 1. Now let's look at the case of n
We first look at the following sequences:
a c = 1 + 1 3 + . . . + 1 p where p is the largest odd number not exceeding n.
b d = 1 2 + . . . + 1 q where q is the largest even number not exceeding n
First we can conclude b is odd and d is even as
b d = 1 2 + . . . + 1 q = 1 2 ( 1 + 1 2 + 1 3 . . . + 1 ( q 2 ) )
By induction hypothesis we know for q > 2, ( 1 + 1 2 + 1 3 . . . + 1 ( q 2 ) ) has odd numerator and even denominator and hence d is even and b is odd.
For the case where q = 2, b = 1 and d = 2 so still d is even and b is odd.
Then let's look at c, we know c must be odd because 1 , 3 , 5 , . . . , p are all odd.
Now 1 + 1 2 + 1 3 + + 1 n = a c + b d = a d + b c c d . The numerator is odd and the denominator is even and we are good.
Bruno Pittman

Bruno Pittman

Expert

2022-07-03Added 4 answers

To complete your proof, first observe that
k m + 1 n + 1 = k ( n + 1 ) + m m ( n + 1 )
now m is even so let m = 2 α a,where α is the biggest power of 2 in m, so a is odd.
if n + 1 is odd obviously k ( n + 1 ) + m is odd and m ( n + 1 ) is even, so we are done.
So consider the case when n + 1 is even. Then n + 1 = 2 β b, where β is the biggest power of 2 in n + 1, so b is odd. so k ( n + 1 ) + m m ( n + 1 ) = k ( 2 β b ) + 2 α a ( 2 β b ) ( 2 α a ) = k ( 2 β b ) + 2 α a 2 α + β a b
now look at whether α or β is smaller. Say α is smaller, then
k ( 2 β b ) + 2 α a 2 α + β a b = k ( 2 β α b ) + a 2 β a b
now we have an even denominator, and on the numerator k ( 2 β α b ) is even and a is odd, so the numerator is odd.
Now if β is smaller, then
k ( 2 β b ) + 2 α a 2 α + β a b = k b + 2 α β a 2 α a b
now k b is odd so the numerator is odd, and our denominator is obviously even.
By induction, you have proved statement as required

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