5. Let A be a set and let 3 be a binary operation on A. We say that a subset BCA is closed under 3 if for all a₁, a2 € B, a1 Ba2 € B. (a) Prove that, if subsets B and C of A are closed under 3, then the intersection, BNC, is also closed under 3. (b) For a € Z, define the set aZ = {ak k € Z}. Prove that aZ is closed under the addition operation on Z. (You can take for granted that the sum and product of two integers is an integer.)
5. Let A be a set and let 3 be a binary operation on A. We say that a subset BCA is closed under 3 if for all a₁, a2 € B, a1 Ba2 € B. (a) Prove that, if subsets B and C of A are closed under 3, then the intersection, BNC, is also closed under 3. (b) For a € Z, define the set aZ = {ak k € Z}. Prove that aZ is closed under the addition operation on Z. (You can take for granted that the sum and product of two integers is an integer.)
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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![5. Let A be a set and let 3 be a binary operation on A. We say that a subset BCA is
closed under 3 if for all a₁, a2 € B, a₁ßa2 € B.
(a) Prove that, if subsets B and C of A are closed under 3, then the intersection, BC,
is also closed under 3.
(b) For a € Z, define the set aZ = {ak : k € Z}. Prove that aZ is closed under the
addition operation on Z. (You can take for granted that the sum and product of two
integers is an integer.)
Let R be a ring and A a set. We denote by Rª the set of all functions from A to R. We can
use the ring structure on R to endow Rª with its own fabulous ring structure via pointwise
addition and multiplication: for f, g € RA, we define f+g: A → R and fg: A → R by the
formulas
(f+g) (a) = f(a) + g(a)
(fg)(a) = f(a)g(a)
for a € A. (Note that, in each of the displayed formulas, the arithmetic taking place on the
righthand side is that of the ring R.)
and
Proposition. With the operations of addition and multiplication just defined, RA is a ring.
The proof of the proposition entails verifying properties (R1)-(R6). We will prove associa-
tivity of addition on RA (half of (R1)), and you will prove associativity of multiplication (the
other half of (R1)) in the next problem (our proof for addition should serve as a model).
Proof. Let f, g, h € RA. [We need to prove that (f + g) + h = f + (g+h).] The functions
(f+g) +h and f+(g+h) have the same domain and codomain [(A and R, respectively)] by
definition, so it remains to show that, for each a € A, the value of (f+g) + h at a coincides
with the value of f+ (g+h) at a. Let's compute: for any a € A, we have
((f+g) + h)(a) = (f+g)(a) +h(a)
= (f(a) + g(a)) + h(a)
= f(a) + (g(a) + h(a))
= f(a) + (g+h)(a)
= (f+(g+h))(a).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Febb913e1-4986-4d74-b6ce-ad576ddf43d3%2F326c7e7f-2804-4672-aa09-203583511ea8%2Firpfrfe_processed.png&w=3840&q=75)
Transcribed Image Text:5. Let A be a set and let 3 be a binary operation on A. We say that a subset BCA is
closed under 3 if for all a₁, a2 € B, a₁ßa2 € B.
(a) Prove that, if subsets B and C of A are closed under 3, then the intersection, BC,
is also closed under 3.
(b) For a € Z, define the set aZ = {ak : k € Z}. Prove that aZ is closed under the
addition operation on Z. (You can take for granted that the sum and product of two
integers is an integer.)
Let R be a ring and A a set. We denote by Rª the set of all functions from A to R. We can
use the ring structure on R to endow Rª with its own fabulous ring structure via pointwise
addition and multiplication: for f, g € RA, we define f+g: A → R and fg: A → R by the
formulas
(f+g) (a) = f(a) + g(a)
(fg)(a) = f(a)g(a)
for a € A. (Note that, in each of the displayed formulas, the arithmetic taking place on the
righthand side is that of the ring R.)
and
Proposition. With the operations of addition and multiplication just defined, RA is a ring.
The proof of the proposition entails verifying properties (R1)-(R6). We will prove associa-
tivity of addition on RA (half of (R1)), and you will prove associativity of multiplication (the
other half of (R1)) in the next problem (our proof for addition should serve as a model).
Proof. Let f, g, h € RA. [We need to prove that (f + g) + h = f + (g+h).] The functions
(f+g) +h and f+(g+h) have the same domain and codomain [(A and R, respectively)] by
definition, so it remains to show that, for each a € A, the value of (f+g) + h at a coincides
with the value of f+ (g+h) at a. Let's compute: for any a € A, we have
((f+g) + h)(a) = (f+g)(a) +h(a)
= (f(a) + g(a)) + h(a)
= f(a) + (g(a) + h(a))
= f(a) + (g+h)(a)
= (f+(g+h))(a).
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