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 09:10Homework6Solutio...Note that IS well-denneu since the summ of any two (not necessarily distunet) elementsof 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 nonemptysubsets B,C C A such that B, C are disjoint and the sum of the elements of B is equalto the sum of the elements of C.Solution: First, assume that |A| = 10. Define a functionf : P(A) → {1,2,..., 1000}S(B) = 1where Erenr denotes the sum of the elements of B. Note that f is well-definedsince the sum of any 10 elements of {1, 2, ..., 100} is at least 10 -1 = 10 and at most10 - 100 = 1000. Since|P(A)| = 24| = 210 = 1024 > 1000 = |{1,2, ..., 1000}|,Math 109Homework 6 Due: November 13, 2021 at 11:59 PMwe know that f is not an injection by the pigeonhole principle. Hence, there existB1, C, € P(A) such that B, + C, and f(B) = f(C). Set B = B\ (B, nC) andC = C\ (B, nc). Then B and C are disjoint (do you see why?) andSI=zEBΣ)= f(B) –\rEBiNC= f(C) –Σ- (E)-(E.)Finally, since B and C are disjoint, at least one of them is nonempty, and sinceEzent = Erec r, this implies the other is nonempty as well. Thus, B and C havethe desired properties.Finally, if A > 10, then we may choose a subset A'CA with |A'| = 10. By the workabove, there exist disjoint nonempty subsets B and C of A' (which are therefore alsosubsets 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}.137DashboardCalendarTo DoNotificationsInbox

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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

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

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
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