Arithmetic+ 02: UNIZOR.COM - Math+ & Problems - Arithmetic

Notes to a video lecture on http://www.unizor.com

Arithmetic 02


Problem

Prove that
S_N = Σ_n∈[2,N]1/n
is not an integer number for any integer N.

Solution 1

Among fractional numbers in the sum S_N there are those of type 1/(2^k).
Let's choose among them the one with the maximum power k. This number is less than any other number of this type in a series and the closest to the end of a series among them. Let's call it the least binary fraction.

For example, in the sum
S₁₀ = Σ_n∈[2,10]1/n
we choose 1/8=1/(2³).

Now let's bring all the numbers in a series to the least common denominator (LCD).
For an example of S₁₀ the LCD=2520 and values of our fractions with this LCD are
1/2 = 1260/2520
1/3 = 840/2520
1/4 = 630/2520
1/5 = 504/2520
1/6 = 420/2520
1/7 = 360/2520
1/8 = 315/2520
1/9 = 280/2520
1/10 = 252/2520

Notice that the numerator of 1/8=315/2520 (a fraction of a type 1/(2^k) with maximum power k) equals 315, which is odd, while all other numerators are even.
That means, the sum of numerators with this common denominator is odd, while the least common denominator 2520 is even, which means that the sum S₁₀ cannot be an integer number.

Actually, this situation will be similar with any number of members in a series, that is for any N.
Here is why (and this is the required proof).

Let's analyze the process of determining the least common denominator (LCD).

Any number in the denominators of our series can be uniquely represented as a product of prime numbers.
Then LCD is formed from all the prime numbers in representation of all denominators with the quantity of each prime number being equal to the largest number of these primes among all denominators.

In our example of S₁₀ we have the following representation of denominators:
2 = 2
3 = 3
4 = 2·2
5 = 5
6 = 2·3
7 = 7
8 = 2·2·2
9 = 3·3
10 = 2·5

As we see, the prime numbers we have to use are 2,3,5,7 and the number of these primes in the LCD should be:
prime 2 should be taken 3 times because of denominator 8.
prime 3 should be taken 2 times because of denominator 9.
prime 5 should be taken once.
prime 7 should be taken once.
Therefore, the LCD is
2³·3²·5·7=2520

In general case, in the representation of LCD the number of 2's is the largest in a member 1/(2^k) with the largest k that we called above the least binary fraction.
Therefore, in the representation of the LCD as a product of prime numbers there are exactly k 2's and, obviously. other prime numbers, which are all odd, of course.

When we transform all the fractions in a series S_N to common denominator, we should multiply numerators (which are all 1's) by all primes in the LCD that are not in a representation of its denominator:

In our example of S₁₀
the numerator of a fraction 1/2 should be multiplied by 2³·3²·5·7/2 =
= 2²·3²·5·7 = 1260
the numerator of a fraction 1/3 should be multiplied by 2³·3²·5·7/3 =
= 2³·3·5·7 = 840
the numerator of a fraction 1/4 should be multiplied by 2³·3²·5·7/2² =
= 2·3²·5·7 = 630
the numerator of a fraction 1/5 should be multiplied by 2³·3²·5·7/5 =
= 2³·3²·7 = 504
the numerator of a fraction 1/6 should be multiplied by 2³·3²·5·7/(2·3) =
= 2²·3·5·7 = 420
the numerator of a fraction 1/7 should be multiplied by 2³·3²·5·7/7 =
= 2³·3²·5 = 360
the numerator of a fraction 1/8 should be multiplied by 2³·3²·5·7/2³ =
= 3²·5·7 = 315
the numerator of a fraction 1/9 should be multiplied by 2³·3²·5·7/3² =
= 2³·5·7 = 280
the numerator of a fraction 1/10 should be multiplied by 2³·3²·5·7/(2·5) =
= 2²·3²·7 = 252

As a result, the only numerator without any 2's in its representation as a product of primes will be the one in the one we called the least binary fraction.
That's why this numerator will be odd, while all others (with one or more 2's in their prime representation) will be even.
Therefore, the sum of all these numerators with the least common denominator will be odd and not divisible by the LCD, which is even.


Solution 2

Among the fractions of our series there are those with prime denominators, which we will call prime fractions.
Let's choose among all prime fractions the one with the largest prime denominator P_max (hence, the smallest in value among all prime fractions) and call it the least prime fraction.
So, P_max is the largest prime number not exceeding N.

At this point it's very important to state that all subsequent members of the series after 1/P_max up to the last one 1/N have denominators not multiple of P_max.
The reason why it is the case is based on so called Bertrand's postulate that states (and can be proven, but not at this moment) that for any integer M there exists a prime number p greater than M and less than 2·M−2.

If there is another fraction with denominator being a multiple of P_max than its denominator is, obviously, not less than 2·P_max and, according to the Bertrand's postulate, there must be another prime number between P_max and 2·P_max, so P_max is not the largest prime number not exceeding N.

In our example with S₁₀ P_max=7 and the least prime fraction is 1/7.

Now let's bring all the numbers in a series to the least common denominator (LCD) in order to sum up all numerators.
Since this least prime fraction has the largest prime denominator, this prime denominator will be represented only once in the LCD.

In our example with S₁₀ LCD is 2³·3²·5·7=2520 and, as we see, number 7 is presented only once.

The next step to calculate S_N is to bring all fractions to the least common denominator by multiplying each numerator (all numerators are 1's) by all prime numbers in the LCD that are not in a prime representation of a corresponding denominator.

Because the largest prime denominator in a series occurs only once in the LCD and does not occur in prime representation of any denominator other than that of the least prime fraction, this largest prime denominator will be represented in all new numerators of all fractions with common denominator except in the numerator of the least prime fraction.

Consequently, all new numerators, except the one of the least prime fraction will be divisible by P_max.

In our example with S₁₀ and P_max=7 the prime number 7 occurs exectly once in every new numerator except for 1/7.
Therefore, the sum of these new numerators will not be divisible by P_max and the result of the summation cannot be an integer number.