Continued Fractions

Show that each of the following holds: [1,1,1,1,1,1,....]^2 = [2,1,1,1,1,1,....] [1,2,2,2,2,2,....]^2 = [2] [1,1,1,1,1,1,....][0,1,1,1,1,1,....] = [1] In each case make a conjecture about a possible generalisation, and explore it (i.e. attempt to prove your conjectures true or false). Note:We can write down any ...continues

The Last Outpost is a tourist stop in a western resort community.

1. C-V-P Analysis The Last Outpost is a tourist stop in a western resort community. Kerry Yost, the owner of the shop, sells hand-woven blankets for an average price of $30 per blanket. Kerry buys the blankets from weavers at an average cost of $21. In addition, he has selling expenses of $3 per blanket. Kerry rents the buil ...continues

1. Mrs Bollo's second grade class of thirty student conducted a pet ownership survey . Results of the survey indicate that eight students own a cat, 15 students own a dog, and 5 students ...

1. Mrs Bollo's second grade class of thirty student conducted a pet ownership survey . Results of the survey indicate that eight students own a cat, 15 students own a dog, and 5 students own both a cat and a dog, How many of the students surveyed own only a cat? 2. Please use a venn diagram to answer the questions? At east ...continues

Real number proofs

-(-a)=a -(a)*-(b)=a*b if a doesn't equal 0 then 1/a doesn't equal 0 if a doesn't equal 0 then (-1)/a = 1/(-a) a(b-c)=ab-ac

Real number proofs

prove 1 a/b=(ac)/(bc),c can't equal 0 2 a-b < 0, "if and only if" a < b 3 if a < b then -b < -a 4 if 0 < a < b, then (1/b) < (1/a 5 a^2 greater than or equal to 0

Prime Problem

See the attached file. Show that if p is an odd prime then....

Polynomial

Let p be a prime. Show that if f(x) ≡ 0 (mod p) has a solution x=a (mod p) then there is a polynomial q(x) such that f(x) ≡ (x-a)q(x) (mod p). Note the ='s here denote congruences.

Congruence Problem

Please provide a detailed solution to the following problem: Let p be an odd prime and a such that (a, p) = 1. Show that the equation x^2 = a (mod p^i) has a solution for all i if it has a solution for i = 1. How many solutions are there?

Show that if n>1 then n does not divide ((2^n)-1).

Please provide a detailed proof for the following problem: Show that if n>1 then n does not divide ((2^n)-1). [Hint: If p is the smallest prime divisor of n then (n, p-1)=1

Divisibility and Congruences

Please see the attached file for the fully formatted problem.