Notice that going the other way if is a fraction with positive top and bottom which satisfies , , and , then it is one of the lowest-terms fractions. For for some k, and then and the last fraction is one of the original fractions.
How many of the lowest-terms fractions have "d" on the bottom? Since the "a" on top is a positive number relatively prime to d, there are such fractions.
Summing over all d's which divide n gives. But since every lowest-terms fraction has some such "d" on the bottom, this sum accounts for all the fractions and there are n of them. For example, suppose. For instance, suppose , so. If , the result is immediate by convention. Each term is , where d is 1 the first term or a product of distinct primes. So the expression above is. I can run the sum over all divisors, because if d has a repeated prime factor. Now simply multiply by n:.
The formula in the theorem is useful for hand-computations of. For example, , so. Likewise, , so. An arithmetic function f is multiplicative if implies.
Since , the two products have no primes in common. Moreover, the primes that appear in either of the products are exactly the prime factors of. Note that if , then. That is, the prime powers in the prime factorization of n are relatively prime. Recall that if p is prime, then.
These observations combined with the fact that is multiplicative give. The second formula follows from the first by factoring out common factors from each term. If is the highest power of p which divides n, then the corresponding term in the second formula for is just and so, if you don't know the value of r in , the only term you can assume will appear in is.
While the formula in the earlier theorem provided a way of computing , the formula in the corollary is often useful in proving results about.
If , then is even. In fact, if n has k odd prime factors, then. This is even if. So suppose that n has k odd prime factors. Each odd prime power factor in the prime factorization of n gives a term in the product for in the corollary, and each term is even since it's a difference of odd numbers. Hence, is divisible by. For example, consider. There are 3 odd prime factors, so should be divisible by 8. And in fact,.
Find all positive integers n such that. I'll do this in steps. First, I'll show that no prime can divide n. At that point, with , I'll get bounds on a, b, and c. That will leave me with 16 cases, which I can check directly. The formula in the second corollary tells us that there is at least a term in the product for. Book Name —rd sharma class 11 maths volume I and volume II.
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