Recent content by cheiney

Thanks for the help! It clarified the step I was missing. From there, I would get (a^2+b^2)/2>=((a+b)^2)/4 and then take the square root to get RMS>=AM.

Homework Statement I am asked to discover and prove the inequality relationship between root mean square, arithmetic mean, geometric mean, and harmonic mean. Homework Equations Let a,b, be non-negative integers. (a-b)2 ≥ 0 and (√a-√b)2 ≥ 0 The Attempt at a Solution Using (a-b)2 ≥...

Okay I understand what you are saying. What if I said: If you sum 1^a+2^a+3^a+...+n^a (where n is a prime number) and divide this sum by n, you end up with a remainder of 0?

Ah, I didn't think of that. What if I specify n consecutive integers starting from 1?

Thanks everyone for your posts. I figured out the answer to both parts. Part a) , the remainder is 0. Part b) , If you sum n integers, where n is prime, and divide the sum by n, you will always get a remainder of 0. P.S. Phinds and Mark44, I wrote it incorrectly. The remainder should be 1...

Homework Statement (a) Find the remainder when 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divided by 5. (b) Generalize this resultHomework Equations Congruence Modulo a\equivb mod n also a=n*q+b where q is some integer.The Attempt at a Solution The remainder for 1^99 would be 1. The remainder for...

Yes. The only 2 approaches I could really think of is that, since they didn't specify that n is greater than or equal to 2, so if all of q_j are negative and if j is an even number, then it would hold up. The other approach would be to describe n as a member of a set with "primes" in the sense...

Homework Statement Assume n = p_1*p_2*p_3*...*p_r = q_1*q_2*q_3*...*q_s, where the p's and q's are primes. We can assume that none of the p's are equal to any of the q's. Why? Homework Equations The Attempt at a Solution I am completely stuck on this. My understanding of the...

Another example could be where a_2 or b_2 was a negative integer. When a_1 is negative, the expression would result in fractions that are not contained in set Z of integers. Is this what you were possibly hinting at?

Prior to this question, we were prompted to test the theorem 10 times, and in each case the theorem was shown to be true. So, I did *attempt* to disprove it during those 10 attempts, but in each case it seemed to hold up. I know that a-b has to be a multiple of n by the definition of congruence...

Homework Statement I am required to prove/disprove the theorem: If a_1 is congruent to b_1 (mod n) and a_2 is congruent to b_2 (mod n), then (a_1)^(a_2) is congruent to (b_1)^(b_2) (mod n). Homework Equations a_1 is congruent to b_1(mod n) can also be expressed as b_1=a_1+q*n...