Question

Which of the following are binary operations on S = {x ∈ R | x > 0}? Justify your answer. i) The operation defined by xy=|ln(xy)| where lnx is the natural logarithm. ii.)The operation ∆ defined by x∆y=x2+y3. Also, for those operations which are binary operations, check whether they are associative or commutative.

Answer 1

ANSWER on Question #81130 – 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 $x\Delta y = |\ln (xy)|$ where $\ln (x)$ is the natural logarithm.

ii) The operation $\Delta$ defined by $x\Delta y = x^2 + y^3$. Also, for those operations which are binary operations, check whether they are associative or commutative.

SOLUTION

First, let us recall the definition of a binary operation.

A binary operation on a set $S$ is a map which sends elements of the Cartesian product $S \times S$ to $S$:

$$\Delta : S \times S \rightarrow S$$

that is, a binary operation, it is an operation that takes any two elements $x$ and $y$ from the set $S$, does something with them (depends on the definition), and produces the third number $c$, which must be in the set $S$.

(More information: https://en.wikipedia.org/wiki/Binary_operation)

Next we recall the definition of associativity and commutativity.

A binary operation $\Delta$ on a set $S$ is called associative if it satisfies the associative law:

$$(x \Delta y) \Delta z = x \Delta (y \Delta z), \quad \forall x, y, z \in S$$

(More information: https://en.wikipedia.org/wiki/Associative_property)

A binary operation $\Delta$ on a set $S$ is called commutative if:

$$x \Delta y = y \Delta x, \quad \forall x, y \in S$$

(More information: https://en.wikipedia.org/wiki/Commutative_property)

Now we turn to the solution of the problem.

i)

$$x \Delta y = | \ln (x y) |$$

This is not a binary operation. If we choose $x = y = 1 > 0$, then,

$$1 \Delta 1 = | \ln (1 \cdot 1) | = | \ln (1) | = | 0 | = 0 \gg 0$$

Conclusion,

$$x \Delta y = | \ln (x y) | \text{ is not a binary operation for } S = \{x \in \mathbb{R} \mid x > 0 \}$$

ii)

$$x \Delta y = x ^ {2} + y ^ {3}$$

As we know

$$x, y \in S \to \left\{ \begin{array}{l} x > 0 \\ y > 0 \end{array} \right. \to \left\{ \begin{array}{l} x ^ {2} > 0 \\ y ^ {3} > 0 \end{array} \right. \to \boxed {x ^ {2} + y ^ {3} > 0}$$

Conclusion,

$$x \Delta y = x ^ {2} + y ^ {3} \text{ is a binary operation for } S = \{x \in \mathbb{R} \mid x > 0 \}$$

Associative:

$$\begin{array}{l} (x \Delta y) \Delta z = (x \Delta y) ^ {2} + z ^ {3} = (x ^ {2} + y ^ {3}) ^ {2} + z ^ {3} = \left[ \begin{array}{c} \text{We use the binomial formula} \\ (a + b) ^ {2} = a ^ {2} + 2 a b + b ^ {2} \end{array} \right] = \\ = ((x ^ {2}) ^ {2} + 2 \cdot x ^ {2} \cdot y ^ {3} + (y ^ {3}) ^ {2}) + z ^ {3} = x ^ {4} + 2 x ^ {2} y ^ {3} + y ^ {6} + z ^ {3} \\ \end{array}$$ $$\boxed {(x \Delta y) \Delta z = x ^ {4} + 2 x ^ {2} y ^ {3} + y ^ {6} + z ^ {3}}$$ $$\begin{array}{l} x \Delta (y \Delta z) = x ^ {2} + (y \Delta z) ^ {3} = x ^ {2} + (y ^ {2} + z ^ {3}) ^ {3} = \left[ \begin{array}{c} \text{We use the binomial formula} \\ (a + b) ^ {3} = a ^ {3} + 3 a ^ {2} b + 3 a b ^ {2} + b ^ {3} \end{array} \right] = \\ = x ^ {2} + ((y ^ {2}) ^ {3} + 3 \cdot (y ^ {2}) ^ {2} \cdot z ^ {3} + 3 \cdot y ^ {2} \cdot (z ^ {3}) ^ {2} + (z ^ {3}) ^ {3}) = x ^ {2} + y ^ {6} + 3 y ^ {4} z ^ {3} + 3 y ^ {2} z ^ {6} + z ^ {9} \\ \end{array}$$ $$\boxed {x \Delta (y \Delta z) = x ^ {2} + y ^ {6} + 3 y ^ {4} z ^ {3} + 3 y ^ {2} z ^ {6} + z ^ {9}}$$

Then,

$$\left\{ \begin{array}{c} (x \Delta y) \Delta z = x ^ {4} + 2 x ^ {2} y ^ {3} + y ^ {6} + z ^ {3} \\ x \Delta (y \Delta z) = x ^ {2} + y ^ {6} + 3 y ^ {4} z ^ {3} + 3 y ^ {2} z ^ {6} + z ^ {9} \end{array} \right. \to \boxed {(x \Delta y) \Delta z \neq x \Delta (y \Delta z)}$$

Conclusion,

$$x \Delta y = x ^ {2} + y ^ {3} \text{ is not an associative binary operation}$$

Commutative:

$$\left\{ \begin{array}{l} x \Delta y = x ^ {2} + y ^ {3} \\ y \Delta x = y ^ {2} + x ^ {3} \end{array} \right. \to x \Delta y \neq y \Delta z$$

Conclusion,

$$x \Delta y = x ^ {2} + y ^ {3} \text{ is not a commutative binary operation}$$

ANSWER

i)

$$x \Delta y = | \ln (x y) | \text{ is not a binary operation for } S = \{x \in \mathbb{R} \mid x > 0 \}$$

ii)

$$x \Delta y = x ^ {2} + y ^ {3} \text{ is a binary operation for } S = \{x \in \mathbb{R} \mid x > 0 \}$$ $$x \Delta y = x ^ {2} + y ^ {3} \text{ is not an associative binary operation}$$ $$x \Delta y = x ^ {2} + y ^ {3} \text{ is not a commutative binary operation}.$$

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