mbt uk Entrance Examination 2008, the highlight of -

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Entrance Examination 2008, the highlight of scan

ab, No. EP (divisor 6 ≠ 0), called P is a number field. For example, Q is the rational number field set; number set F = {port + b, 72Ia,[link widoczny dla zalogowanych], bEQ} is the number of domains. The following proposition: ① the set of integers is a number field; ② If the rational set of QM, the number set number M will be the domain; ③ an unlimited number of domains will be set; ④ there are infinitely many number fields. * Reference Mathematical Review Newsletter (2o08 on 2l of) Proposition 43, which number is correct --. (You think that the right number of propositions are filled) to analyze whether the problem can be individually tested to meet the \1 + i suddenly appears M, so it is not a number field; ③ ④ is clearly established. Comments to higher mathematics of the title \The key is solving the \Example 9 (2008 Entrance Beijing volumes (Science) 20 questions) for each are positive integers of the series A: Port l, day 2, ..., define transformations T1,[link widoczny dla zalogowanych], Tl the series A transformation into a series of Tl (A): n, al-1. 2-1, ..., the mouth of a 1. For all non-negative integer for each series B: bl, b2, ... ... b, the definition of transformation, the series B of the descending order, and then remove all the zero entries are series T2 (B); and define S (B) = 2 (b1 +2 bz + ... + \(chi = 0,[link widoczny dla zalogowanych],1,2, ...). (I) If the sequence A0 for the 5,3,2 and write series A1, Az; (II) are positive integers for each of the finite number of column A, prove that S ( Tl (A)) = S (A); (Ⅲ) Proof: For any given positive integers each are finite number of column A., there exists a positive integer K, when the time is ≥ K, S (A +1 ): S (Ak). (I) solution A0: 5,3,2; Tl (A0): 3,4,2,1; Al: T2 (Tl (Ao)): 4,3,2,1; 1 (A1): 4,3,2,1,0; A2 = T2 (Tl (A1)); 4,3,2,1. (Ⅱ) proved that each set of positive integers are finite number of column A is al , a2, ..., a,[link widoczny dla zalogowanych], then TI (A) is,. 1-1, a2-1, ..., port-1 to S (T1 (A)) = 2 [+2 (al-1) +3 ( a2-1) + ... + (7z +1) (a-1)] +2 + (port l of a 1) + (a2-1) 2 + ... + (day-1). and S (A) = 2 (al +2 az + ... + write) + port} + a; + ... + population, so S (T1 (A)) a S (A) = 2 [a 2-3 one ... one (+1)] +2 ( mouth l + a2 + ... + a) +7 z2-2 (al + a2 + ... + port) +: A (+1) +2 + n: 0, so S (T1 (A)): S (A). (Ⅲ ) to prove that each set A is non-negative integer numbers are bad U port l, 2, ..., a. When there is 1 ≤ f <≤ n, so ai ≤ mouth, the exchange of series A, No. J i-items entries received series B, then S (B) a S (A) = 2 (+ ja-ia a) = 2 (i-) (aj-ai) ≤ 0. When there is l ≤ m <, make a, JI + 1: a2 = ...: a = 0, if the count out port l, port 2, ..., a; to C, then S (C) = S (A). Therefore S (D 2 (A)) ≤ S ( A). which for any given series of A0, the A + l = T2 (T1 (A)) (VII: 0,1,2, ...), we know that S (A +1) ≤ S (T1 (A) .) again by (II) shows that S (T1 (A)) = S (Ak), so S (+1) ≤ S (). that is, for a ∈ N, either there are S (Ak +1) = S (Ak ), or a S (A +1) ≤ S (A) a 1. because the S (A) is an integer greater than 2, so after a finite steps, there must be S (Ak) = S (Ak +1) = S (A +2) = ..., that there is a positive integer K, the time when Yu ≥ K, S (ten 1): S (Ak). Reviews of the question as to the number of carriers, the use of advanced mathematics used in the term \words and the symbols, not only to examine the series of such knowledge, and examine the logical thinking ability, problem solving is a breakthrough, \ More articles related to topics：

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