Consider the ring Z/nZ, where n is a natural number. The elements of the ring are the residues of integers modulo n. For any a,b of Z/nZ, we define the addition and multiplication operations as follows: (a + b) mod n is the result of adding a and b modulo n (a*b) mod n is the result of multiplying a and b modulo n. Prove that Z/nZ is a commutative ring with unity.
Consider the ring Z/nZ, where n is a natural number. The elements of the ring are the residues of integers modulo n. For any a,b of Z/nZ, we define the addition and multiplication operations as follows: (a + b) mod n is the result of adding a and b modulo n (a*b) mod n is the result of multiplying a and b modulo n. Prove that Z/nZ is a commutative ring with unity.
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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Check if the solution is correct. Correct the mistakes.(sorry my english is bad)
(can you fix the marker on the image?)
![(II) Associativity of multiplication: for any a,b,c of Z/nZ, performed: ((a *b) * c) mod n = (a * (b
*c)) mod n.
by the definition of the multiplication operation and the associativity property of multiplication
of integers.
(III) The existence of a neutral element with respect to multiplication: for any a of Z/nZ, element
1 is neutral with respect to multiplication. Indeed: (a* 1) mod n = a mod n.
(IV) Multiplication commutativity: for any a and b of Z/nZ, (a * b) mod n = (b* a) mod n
is performed by the commutativity property of the multiplication operation in the ring Z/nZ:
(a*b) mod n = [(a mod n) * (b mod n)] mod n
since multiplication in integers is commutative, it is possible to swap a and b:
(b* a)mod n = [(b mod n)* (a mod n)]mod n.
From the property of the remainder when dividing by n:
(a * b)mod n = [(a mod n)* (b mod n)]mod n= = [(b mod n) * (a mod n)] mod n = (b * a) mod n
(V) Distributive law: for any a,b and c from Z/nZ executed:
(a* (b + c)) mod n = ((a*b) + (a * c)) mod n and
((a + b) * c) mod n = ((a* c) + (b* c)) mod n
To prove the first distributivity law using the remainder property when dividing by n and opening
the brackets: (a *(b + c)) mod n = (a* b + a * c) mod n
by the remainder property when dividing by n, for any integers x and y:
(x + y)mod n = ((x mod n)+ (y mod n))mod n.
Using this property, we get: (a * b + a* c) mod n = ((a * b) mod n + (a* c) mod n) mod n
Thus, the first distributive law is proved. The proof of the second law is carried out similarly. We
proved that Z/nZ is an abelian group and is also a commutative ring with unity.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19616a01-3b9f-4ded-9626-43dcd5e68cc7%2F6c671566-c10a-4fc1-98fc-966e00ea910d%2F0cbxfzj_processed.png&w=3840&q=75)
Transcribed Image Text:(II) Associativity of multiplication: for any a,b,c of Z/nZ, performed: ((a *b) * c) mod n = (a * (b
*c)) mod n.
by the definition of the multiplication operation and the associativity property of multiplication
of integers.
(III) The existence of a neutral element with respect to multiplication: for any a of Z/nZ, element
1 is neutral with respect to multiplication. Indeed: (a* 1) mod n = a mod n.
(IV) Multiplication commutativity: for any a and b of Z/nZ, (a * b) mod n = (b* a) mod n
is performed by the commutativity property of the multiplication operation in the ring Z/nZ:
(a*b) mod n = [(a mod n) * (b mod n)] mod n
since multiplication in integers is commutative, it is possible to swap a and b:
(b* a)mod n = [(b mod n)* (a mod n)]mod n.
From the property of the remainder when dividing by n:
(a * b)mod n = [(a mod n)* (b mod n)]mod n= = [(b mod n) * (a mod n)] mod n = (b * a) mod n
(V) Distributive law: for any a,b and c from Z/nZ executed:
(a* (b + c)) mod n = ((a*b) + (a * c)) mod n and
((a + b) * c) mod n = ((a* c) + (b* c)) mod n
To prove the first distributivity law using the remainder property when dividing by n and opening
the brackets: (a *(b + c)) mod n = (a* b + a * c) mod n
by the remainder property when dividing by n, for any integers x and y:
(x + y)mod n = ((x mod n)+ (y mod n))mod n.
Using this property, we get: (a * b + a* c) mod n = ((a * b) mod n + (a* c) mod n) mod n
Thus, the first distributive law is proved. The proof of the second law is carried out similarly. We
proved that Z/nZ is an abelian group and is also a commutative ring with unity.
Transcribed Image Text:Consider the ring Z/nZ, where n is a natural number. The elements of the ring are the residues of
integers modulo n. For any a,b of Z/nZ, we define the addition and multiplication operations as
follows:
(a + b) mod n is the result of adding a and b modulo n
(a*b) mod n is the result of multiplying a and b modulo n.
Prove that Z/nZ is a commutative ring with unity.
Decision. To prove that Z/nZ is a commutative ring with unity, first we prove that it is an abelian
group, that is, we check the fulfillment of the axioms:
(1) Closure with respect to addition: for any a,b of Z/nZ, the result of their addition will also be
the element Z/nZ: (a + b) mod n belongs to Z/nZ
Since a and b belong to the ring of integers Z, the result of their sum a + b will also be an integer
(II) Associativity of addition: for any a,b,c of Z/nZ, it is done: ((a + b) + c) mod n = (a + (b + c)) mod
n. by the definition of the addition operation and the associativity property of the addition of
integers.
(III) The existence of a neutral element with respect to addition: for any a of Z/nZ, element 0 is
neutral with respect to addition. Indeed: (a + 0) mod n = a mod n.
(IV) The existence of an inverse element with respect to addition: for any a of Z/nZ, there exists
an inverse element equal to (-a) such that: (a + (-a)) mod n = 0.
(V) Commutativity of addition: for any a,b of Z/nZ, it is fulfilled: (a + b) mod n = (b + a) mod n.
Similarly, it follows from the commutativity property of the addition operation in integers.
We have shown that (1)-(V) are true, hence (S,+) is an Abelian group.
Next we need to check that (S,+,*) is a commutative ring with unity.
(1) Closure with respect to multiplication: for any a,b from Z/nZ, the result of their multiplication
will also be an element from Z/nZ: (a*b) mod n belongs to Z/nZ.
since a and b belong to the ring of integers Z, their product a*b will also be an integer.
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