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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.

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.1: Sets And Geometry
Problem 11E: For the sets gives in Exercise 9, is there a distributive relationship for union with respect with...
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ABSTRACT ALGEBRA UPVOTE WILL BE GIVEN. PLEASE WRITE THE COMPLETE SOLUTIONS LEGIBLY. NO LONG EXPLANATION NEEDED.
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.
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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