please answer question number 9 Let I be the set of equivalence classes as defined as above. Define O as follows: if A and B areequivalence classes in T, A O B is the equivalence class of the sum of an element from A and anelement from B. For example, using the A and B from #6: (2,5) + (4,2) = (6,7), so, A OB is theequivalence class of (6,7): {(0,1), (1,2), (2,3), ...} Đis well defined because as we showed in #6, theresult of the operation does not depend on which representatives of the equivalence classes we choose.7. Show that {T,O}is a group.8. Match the elements in T with the integers and show that {T,O} has the same structure as theintegers with addition.Multiplication on the number pairs is defined as follows: (a, b) · (c, d) = (ad + bc, ac + bd).O in T isdefined in a similar way to O above from multiplication of number pairs.9. Show that the definition of 8 makes sense.10. Perhaps surprisingly, {T,8} has the same structure as the integers with multiplication. Checkthat on some examples. Why does this work?

Name: please answer question number 9 Let I be the set of equivalence classe
Uploaded: 2024-03-20T07:07:37.000Z
Channel: Bartleby
Description: please answer question number 9 Let I be the set of equivalence classes as defined as above. Define O as follows: if A and B areequivalence classes in T, A O B is the equivalence class of the sum of an element from A and anelement from B. For example

Multiplication on the number pairs is defined as : a, b . c, d =ad+bc , ac+bd ⊗ in T is defined as follows: If A and B are equivalence classes in T, A⊗B is the equivalence class of the multiplication of number pairs. i.e., If a,b∈A and c,d∈B, then the multiplication of an element from A and an element from B is defined as a, b . c, d =ad+bc , ac+bd It is required to prove that the definition of ⊗ makes sense. That is to prove that the operation is well-defined ( independent of the representatives chosen from each equivalence class)Let a, b, a', b' ∈A and c, d, c', d' ∈B Assume that a, b=a', b' that is a+b'=b+a' .............................(1) c, d=c', d' that is c+d'=d+c' ..............................(2) It is required to show that a, b . c, d =a', b' . c', d', that is ad+bc , ac+bd=a'd'+b'c' , a'c'+b'd' , that is ad+bc+a'c'+b'd'=ac+bd+a'd'+b'c' ...................(3)Consider ca+b'+da'+b+a'c+d'+b'c'+d ......................(4) Since from (1) and (2) , a+b'=b+a' and c+d'=d+c' Then (4) becomes ca+b'+da'+b+a'c+d'+b'c'+d = cb+a'+da+b'+a'd+c'+b'c+d' Multiplying out both sides of the above equation gives ca+cb'+da'+db+a'c+a'd'+b'c'+b'd = cb+ca'+da+db'+a'd+a'c'+b'c+b'd' Then using associative property in the above equation will imply ca+db+a'd'+b'c'+cb'+da'+a'c+b'd =da+cb+a'c'+b'd'+ b'c+a'd+ca'+b'd Now using commutative property gives ac+bd+a'd'+b'c'+cb'+da'+a'c+b'd =ad+bc+a'c'+b'd'+cb'+da'+a'c+b'd Using the cancellation law for addition will give ac+bd+a'd'+b'c'=ad+bc+a'c'+b'd' ⇒ ad+bc+a'c'+b'd'=ac+bd+a'd'+b'c' which is (3) Therefore it is proved that ad+bc+a'c'+b'd'=ac+bd+a'd'+b'c' That is, If a, b=a', b' and c, d=c', d' then a, b . c, d =a', b' . c', d' This implies that the operation is independent of the representatives chosen from each equivalence class A and B. That is the operation ⊗ defined is well-defined. Hence that the definition of ⊗ makes sense.