By Arun-Kumar S.

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**Example text**

Hence (j − i)d mod q = 0. Therefore q | j − i or q | d and neither is possible. Therefore we have R = {a mod q, (a + d) mod q, . . (a + (q − 1)d) mod q} = {0, . . e. for every positive integer k, there exist k consecutive composite members. Proof: This can be easily seen as ∀ positive integers k we have (k + 1)! + 2, . . , (k + 1)! + k + 1. 1) j | (k + 1)! + j, ∀j ∈ 2, . . 1 pα || n means pα | n but pα+1 | n 45 46 CHAPTER 9. 7 If for prime p and n ≥ 1 pα || n! Clearly n = 0 and n = 1 are trivial cases.

Similarly, by symmetry, we have if x ∈ Q then x ∈ P . Hence P = Q, and therefore pi = qi . Next, we will show that ei = di for all 1 ≤ i ≤ k. Suppose ei = di for some 1 ≤ i ≤ k. Let ci = max(ei , di ). Once again, pci i | n is one representation and not in the other. That is impossible, therefore ei = di for all ✷ 1 ≤ i ≤ k. 4 There are an infinite number of prime numbers. Proof: We present a proof by contradiction. Assume that there are a finite number m of primes which are p1 , p2 , . , pm . Consider the natural number p = p1 p2 .

Recall that √ 1+ 5 = [1, 1, 1, . 2 A periodic continued fraction is a continued fraction [a0 , a1 , . . , an , . ] such that. an = an+h for a fixed positive integer h and all sufficiently large n. We call h the period of the continued fraction. 1 Consider the periodic continued fraction [1, 2, 1, 2, . ] = [1, 2]. 1 1) A periodic continued fraction represent a quadratic irrationals. 2) Any quadratic irrational has SPCF representation. 1 Every quadratic irrational has SPCF representation. Proof Outline : Let say that x is a quadratic irrational.