First note that R* x R is a non-empty set. Let us verify the properties first. Closure. Let (a, b), (c,d) E R* x R. Then we have (a, b) o (c,d) = (ac, bc+d) e R* x R. This is because a, c#0 therefore ac #0 and bc + de R. ● ● ● Associative. Note that we have ● ((a,b) o (c,d)) o (e. f) = (ac, bc + d) o (e. f) = (ace, bce +de + f) (a,b) o ((c,d) o (e, ƒ)) = (a,b) o (ce, de + f) = (ace, bce +de + f) (2) (3) Equations (2) and (3) prove that o is associative. Identity. We have to show that R* x R has an identity element. This means that there should exists an element (c, d) such that (a, b)o(c,d) = (a, b) for all (a, b) e R* x R. This gives that (ac, bc + d) = (a, b). This gives us that ac = a ⇒ ac-a=0⇒ a(c-1) = 0. Since a 0 as it belongs to R* we deduce that c = 1. Again we get bc + d = b. We already know that c = 1 and so we get b + d = b and this gives us d = 0 which means (1, 0) is the required identity element. Inverse. We have to show that inverse exist. This for every (a, b) € R* x R there exist (c,d) such that (a, b) o (c,d) = (1,0). Solving this we get (ac, bc + d) = (1,0). We get that ac = 1 ⇒ c = 1. Also we get bc + d = 0. Using the fact that c = 1/a we get + d = 0 ⇒ d = -b/a. Hence the inverse is (1/a, -b/a). All the axioms have been verified and this proves that R* x R is a group.
First note that R* x R is a non-empty set. Let us verify the properties first. Closure. Let (a, b), (c,d) E R* x R. Then we have (a, b) o (c,d) = (ac, bc+d) e R* x R. This is because a, c#0 therefore ac #0 and bc + de R. ● ● ● Associative. Note that we have ● ((a,b) o (c,d)) o (e. f) = (ac, bc + d) o (e. f) = (ace, bce +de + f) (a,b) o ((c,d) o (e, ƒ)) = (a,b) o (ce, de + f) = (ace, bce +de + f) (2) (3) Equations (2) and (3) prove that o is associative. Identity. We have to show that R* x R has an identity element. This means that there should exists an element (c, d) such that (a, b)o(c,d) = (a, b) for all (a, b) e R* x R. This gives that (ac, bc + d) = (a, b). This gives us that ac = a ⇒ ac-a=0⇒ a(c-1) = 0. Since a 0 as it belongs to R* we deduce that c = 1. Again we get bc + d = b. We already know that c = 1 and so we get b + d = b and this gives us d = 0 which means (1, 0) is the required identity element. Inverse. We have to show that inverse exist. This for every (a, b) € R* x R there exist (c,d) such that (a, b) o (c,d) = (1,0). Solving this we get (ac, bc + d) = (1,0). We get that ac = 1 ⇒ c = 1. Also we get bc + d = 0. Using the fact that c = 1/a we get + d = 0 ⇒ d = -b/a. Hence the inverse is (1/a, -b/a). All the axioms have been verified and this proves that R* x R is a group.
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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ABSTRACT ALGEBRA
UPVOTE WILL BE GIVEN. PLEASE WRITE THE COMPLETE SOLUTIONS LEGIBLY. NO LONG EXPLANATION NEEDED.
Transcribed Image Text:First note that R* x R is a non-empty set. Let us verify the properties first.
Closure. Let (a, b), (c,d) E R* x R. Then we have (a, b) o (c,d) =
(ac, bc+d) e R* x R. This is because a, c#0 therefore ac #0 and
bc + de R.
●
●
●
Associative. Note that we have
●
((a,b) o (c,d)) o (e. f) = (ac, bc+d) o (e, f)
= (ace, bce +de + f)
(a,b) o ((c,d) o (e, ƒ)) = (a,b) o (ce, de + f)
= (ace, bce +de + f)
(2)
(3)
Equations (2) and (3) prove that o is associative.
Identity. We have to show that R* x R has an identity element. This
means that there should exists an element (c, d) such that (a, b)o(c,d) =
(a, b) for all (a, b) e R* x R. This gives that (ac, bc + d) = (a, b).
This gives us that ac = a ⇒ ac-a=0⇒ a(c-1) = 0. Since a 0
as it belongs to R* we deduce that c = 1. Again we get bc + d = b.
We already know that c = 1 and so we get b + d = b and this gives
us d = 0 which means (1, 0) is the required identity element.
Inverse. We have to show that inverse exist. This for every (a, b) €
R* x R there exist (c,d) such that (a, b) o (c,d) = (1,0). Solving this
we get (ac, bc + d) = (1,0). We get that ac = 1 ⇒ c = 1. Also we
get bc + d = 0. Using the fact that c = 1/a we get + d = 0 ⇒
d = -b/a. Hence the inverse is (1/a, -b/a).
a
All the axioms have been verified and this proves that R* x R is a group.
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