Question

Which of the following are binary operations on S = {x ∈ R | x > 0}? Justify your
answer.
i) The operation Δ defined by xΔy = x(y− 2).
ii) The operation ∇ defined by x∇y = e^x+y.
Also, for those operations which are binary operations, check whether they are
associative and commutative.

Answer 1

Answer on Question #64892 – Math – Linear Algebra

Question

Which of the following are binary operations on $S = \{x \in \mathbb{R} \mid x > 0\}$? Justify your answer.

i) The operation $\Delta$ defined by $x\Delta y = x(y - 2)$.

ii) The operation $\nabla$ defined by $x\nabla y = e^{\wedge}x + y$.

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

Solution

$$\mathbb{S} = \{x \in \mathbb{R} \mid x > 0\}.$$

i. $x\Delta y = x(y - 2)$

Binary operation: $S \times S \to S$, so for any $x \in S, y \in S$, $x(y - 2)$ must be an element of $S$.

So $x(y - 2)$ must be positive for any positive $x, y$. If $0 < y < 2$, then $x(y - 2) < 0 \Rightarrow x\Delta y \notin S$.

So $x\Delta y = x(y - 2)$ is not a binary operation on $\mathbb{S} = \{x \in \mathbb{R} \mid x > 0\}$;

ii. $x\nabla y = e^{x} + y$

Binary operation: $S \times S \to S$, so for any $x \in S, y \in S$, $e^{x} + y$ must be an element of $S$.

That is true, because for any positive $x, y - e^{x} + y$ is also positive $\Rightarrow x\Delta y \in S$.

So $x\nabla y = e^{x} + y$ is a binary operation on $\mathbb{S} = \{x \in \mathbb{R} \mid x > 0\}$;

Associativity:

$$\begin{array}{l} \left(\left(x \nabla y\right) \nabla z = x \nabla \left(y \nabla z\right)\right) \\ \left(x \nabla y\right) \nabla z = \left(e^{x} + y\right) \nabla z = e^{e^{x} + y} + z = e^{e^{x}} * e^{y} + z; \\ x \nabla (y \nabla z) = x \nabla (e^{y} + z) = e^{x} + e^{y} + z; \\ e^{e^{x}} * e^{y} + z \neq e^{x} + e^{y} + z \Rightarrow (x \nabla y) \nabla z \neq x \nabla (y \nabla z) \\ \end{array}$$

Operation is not associative.

Commutativity:

$$\begin{array}{l} \left(x \nabla y = y \nabla x\right) \\ \left(x \nabla y\right) = \left(e^{x} + y\right); \\ \left(y \nabla x\right) = \left(e^{y} + x\right); \\ e^{x} + y \neq e^{y} + x \Rightarrow (x \nabla y) \neq (y \nabla x) \\ \end{array}$$

Operation is not commutative.

Answer: ii) is a binary relation, not associative, not commutative.

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