Question #62050

Which of the following are binary operations. Justify your answer.
i) The operation · defined on Q by a·b = a(b− 1).
ii) The operation · defined on [0,π] by x·y = cosxy.
Also, for those operations which are binary operations, check whether they are
associative and commutative.

Expert's answer

Answer on Question #62050 – Math – Abstract Algebra

Question

Which of the following are binary operations. Justify your answer.

i) The operation \cdot defined on QQ by ab=a(b1)a \cdot b = a(b - 1).

ii) The operation \cdot defined on [0,π][0, \pi] by xy=cosxyx \cdot y = \cos xy.

Also, for those operations which are binary operations, check whether they are associative and commutative.

Solution

A binary operation f(x,y)f(x, y) is an operation that applies to two quantities or expressions xx and yy.

A binary operation on a nonempty set AA is a map f:A×AAf: A \times A \to A such that

1. ff is defined for every pair of elements in AA, and

2. ff uniquely associates each pair of elements in AA to some element of AA.

i) The operation \cdot defined on QQ by ab=a(b1)a \cdot b = a(b - 1) is binary

The operation is defined for all a,bQa, b \in Q. For any a,bQa, b \in Q, a(b1)Qa(b - 1) \in Q.

If a=a1a = a_1 and b=b1b = b_1, hence a(b1)=a1(b11)a(b - 1) = a_1(b_1 - 1).

There exist aa and bb in QQ such that a(b1)(a1)ba(b - 1) \neq (a - 1)b, hence the operation is not commutative. For example: 4(351)35(41)4(\frac{3}{5} - 1) \neq \frac{3}{5}(4 - 1).

There exist aa, bb and cc in QQ such that a(b(c1)1)(a(b1))(c1)a(b(c - 1) - 1) \neq (a(b - 1))(c - 1), hence the operation is not associative. For example: 35(2(31)1)35(21)(31)\frac{3}{5}(2(3 - 1) - 1) \neq \frac{3}{5}(2 - 1)(3 - 1).

Answer: The operation \cdot defined on QQ by ab=a(b1)a \cdot b = a(b - 1) is binary, not associative and not commutative.

ii) The operation \cdot defined on [0,π][0, \pi] by xy=cosxyx \cdot y = \cos xy is not binary. Because there exist aa and bb in [0,π][0, \pi] such that cosab[0,π]\cos ab \notin [0, \pi].

For example, cosπ1=1[0,π]\cos \pi \cdot 1 = -1 \notin [0, \pi].

Answer: The operation \cdot defined on [0,π][0, \pi] by xy=cosxyx \cdot y = \cos xy is not binary.

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