  # 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.

### Comments for Group theory - binary operation on G

 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 = 3Thanks for writing. Staff www.solving-math-problems.com