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Group theory - binary operation on G

by Rafael
(Philippines)











































Prove that * is a binary operation on G

      • ℚ = set of all RATIONAL NUMBERS

      • Set G = ℚ - {1}, set G includes all rational numbers EXCEPT for the number 1.

      • Define the operation * as: a*b = a+b-ab


   a. Prove that * is a binary operation on G, that is, if a and b are elements of G, then a*b is unique and a*b is an element of G

   b. Find the identity element of G?

   c. If a is an element of G, find the inverse of a

   d. Can we say that (G,*) is a group?

   e. If (G,*) defines a group, is it abelian?

   f. In G, find x if 3*x*2 =5 and then show that this value of x indeed satisfies this equation.


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Aug 17, 2012
Binary Operation on G
by: Staff


Answer:

Part I


   a. Prove that * is a binary operation on G, that is, if a and b are elements of G, then a*b is unique and a*b is an element of G

      Set G = ℚ - {1} means all rational numbers EXCEPT for the number 1

      A binary operation * on set G is an operation (or rule) which takes two elements of the set (a, b ∈ G) and produces a third element within the same set (c ∈ G): a * b = c.


      If the operation * is a binary operation on set G, G must be closed under the operation *.

      To prove * is a binary operation on set G you must prove that it is true for all arbitrary values of a and b (all a, b ∈ G)

      a*b = a + b - ab

      When a and b are rational numbers, it is obvious that a + b – ab will always produce another rational number.

     However,

      Set G = ℚ - {1} means all rational numbers EXCEPT for the number 1

      If a ≠ 1 and b ≠ 1, can a*b be made to = 1?

      If a*b can be calculated to = 1 under these conditions (when a ≠ 1 and b ≠ 1), then the operation * is not a binary operation for set G.

      a + b - ab = 1

      solve for “a” and “b”

      solution 1: b = 1, a ≠ 1

      solution 2: b∈ℚ, a = 1

      solution 3: a = 1, b ≠ 1

      solution 4: a∈ℚ, b = 1


      Since set G does not contain the number 1, then a ≠ 1 and b ≠ 1.

      As long as a ≠ 1 and b ≠ 1, a*b = a + b – ab will never equal the number 1.

      >>> The answer, then, is that the operation * is a valid binary operation on set G (set G includes all rational numbers EXCEPT for the number 1).

----------------------------

Aug 17, 2012
Binary Operation on G
by: Staff

----------------------------

Part II


   b. Find the identity element of G?

     Identity: Does there exist an identity element e∈G such that e*a=a and a*e=a for all a∈G?

      The identity element is 0 (zero)

      a*0 = a + 0 – a(0) = a

      0 * a = 0 + a – (0) a = a

      >>> The identity element is 0 (zero)


   c. If a is an element of G, find the inverse of a



     Inverses : If an inverse element exists (a⁻¹∈S) for each a∈S, then a*a⁻¹=e and a⁻¹*a=e.


      a*a⁻¹ = a + a⁻¹ – a(a⁻¹) ≠ 0, the identity element

      >>> The inverse element does not exist for the operation *.



   d. Can we say that (G,*) is a group?

      >>> (G,*) is not a group because it does not satisfy all four conditions (closure, associativity, identity, and inverses).


   e. If (G,*) defines a group, is it abelian?

      >>> (G,*) is not a group because it does not satisfy all four conditions (closure, associativity, identity, and inverses).



   f. In G, find x if 3*x*2 =5 and then show that this value of x indeed satisfies this equation.

      a*b = a + b - ab

      3*x = (3 + x - 3x)

      (3 + x - 3x)*2 = ((3 + x - 3x) + 2 - (3 + x - 3x)2)

      = (3 + x - 3x + 2 - 6 - 2x + 6x)

      = 2x – 1


      If 3*x*2 =5

      2x – 1 = 5

      x = 3


      >>> x = 3






Thanks for writing.

Staff
www.solving-math-problems.com



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