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Please explain the solution. I do not understand it. why and how do we set 1，2… 1000 in this case. Could u give an example to follow each step. Also what does the sum notation mean here. Why do we need do the computation at last and whats the meaning of setting B=B1\（B1 C1）.

Please explain the solution. I do not understand it. why and how do we set 1，2… 1000 in this case. Could u give an example to follow each step. Also what does the sum notation mean here. Why do we need do the computation at last and whats the meaning of setting B=B1\（B1 C1）.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please explain the solution. I do not understand it. why and how do we set 1，2… 1000 in this case. Could u give an example to follow each step. Also what does the sum notation mean here. Why do we need do the computation at last and whats the meaning of setting B=B1\（B1 C1）. thanks
09:10 Homework6Solutio... Note that IS well-denneu since the summ of any two (not necessarily distunet) elements of A is at most 2· 12 = 24 and at least 2 1 = 2. Since |A × A| = |A||A > 25, we know that f is not an injection by the pigeonhole principle. Hence, there exist (a, b), (c, d) E A × A such that (a, b) + (c, d) and f(a, b) = f(c, d), i.e. a+b= c+d. 3. Suppose that A C {1,2,..., 100} with JA| > 10. Show that there exist nonempty subsets B,C C A such that B, C are disjoint and the sum of the elements of B is equal to the sum of the elements of C. Solution: First, assume that |A| = 10. Define a function f : P(A) → {1,2,..., 1000} S(B) = 1 where Erenr denotes the sum of the elements of B. Note that f is well-defined since the sum of any 10 elements of {1, 2, ..., 100} is at least 10 -1 = 10 and at most 10 - 100 = 1000. Since |P(A)| = 24| = 210 = 1024 > 1000 = |{1,2, ..., 1000}|, Math 109 Homework 6 Due: November 13, 2021 at 11:59 PM we know that f is not an injection by the pigeonhole principle. Hence, there exist B1, C, € P(A) such that B, + C, and f(B) = f(C). Set B = B\ (B, nC) and C = C\ (B, nc). Then B and C are disjoint (do you see why?) and SI= zEB Σ) = f(B) – \rEBiNC = f(C) – Σ - (E)-(E.) Finally, since B and C are disjoint, at least one of them is nonempty, and since Ezent = Erec r, this implies the other is nonempty as well. Thus, B and C have the desired properties. Finally, if A > 10, then we may choose a subset A'CA with |A'| = 10. By the work above, there exist disjoint nonempty subsets B and C of A' (which are therefore also subsets of A) such that rERr =Erccr. 4. Fix n.k e Z. with k < n, and let A, = {1.2.....k}. A, = {k + 1.k + 2.....n}. 137 Dashboard Calendar To Do Notifications Inbox
Transcribed Image Text:09:10 Homework6Solutio... Note that IS well-denneu since the summ of any two (not necessarily distunet) elements of A is at most 2· 12 = 24 and at least 2 1 = 2. Since |A × A| = |A||A > 25, we know that f is not an injection by the pigeonhole principle. Hence, there exist (a, b), (c, d) E A × A such that (a, b) + (c, d) and f(a, b) = f(c, d), i.e. a+b= c+d. 3. Suppose that A C {1,2,..., 100} with JA| > 10. Show that there exist nonempty subsets B,C C A such that B, C are disjoint and the sum of the elements of B is equal to the sum of the elements of C. Solution: First, assume that |A| = 10. Define a function f : P(A) → {1,2,..., 1000} S(B) = 1 where Erenr denotes the sum of the elements of B. Note that f is well-defined since the sum of any 10 elements of {1, 2, ..., 100} is at least 10 -1 = 10 and at most 10 - 100 = 1000. Since |P(A)| = 24| = 210 = 1024 > 1000 = |{1,2, ..., 1000}|, Math 109 Homework 6 Due: November 13, 2021 at 11:59 PM we know that f is not an injection by the pigeonhole principle. Hence, there exist B1, C, € P(A) such that B, + C, and f(B) = f(C). Set B = B\ (B, nC) and C = C\ (B, nc). Then B and C are disjoint (do you see why?) and SI= zEB Σ) = f(B) – \rEBiNC = f(C) – Σ - (E)-(E.) Finally, since B and C are disjoint, at least one of them is nonempty, and since Ezent = Erec r, this implies the other is nonempty as well. Thus, B and C have the desired properties. Finally, if A > 10, then we may choose a subset A'CA with |A'| = 10. By the work above, there exist disjoint nonempty subsets B and C of A' (which are therefore also subsets of A) such that rERr =Erccr. 4. Fix n.k e Z. with k < n, and let A, = {1.2.....k}. A, = {k + 1.k + 2.....n}. 137 Dashboard Calendar To Do Notifications Inbox
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