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Problem 39 (Binary Operation Introduction)

In which of the following scenarios do we have a binary operation on a set $G$. Justify your answers.

  1. Addition $a+b$ of integers.
  2. Multiplication $A\cdot B$ of 3 by 3 matrices.
  3. The dot product $\vec u\cdot \vec v$ of two dimensional vectors.
  4. The cross product $\vec u\times \vec v$ of three dimensional vectors.
  5. The scalar product $c\cdot \vec v$.
  6. Composition $f\circ g$ of permutations on the same set $X$.
  7. Composition $f\circ g$ of functions from $X$ to $Y$.

Solution

Part 1.

Let $a,b \in \mathbb{Z}$. Let $c = a + b$. Clearly, $c \in \mathbb{Z}$. Thus, addition is a binary operation on $\mathbb{Z}$.

Part 2.

Let $A,B$ be $3 \times 3$ matrices. Let $C = AB$. From Linear Algebra, we know that multiplication of $3 \times 3$ matrices results in a $3 \times 3$ matrix. Thus, multiplication is a binary operation on the set of all $3\times3$ matrices.

Part 3.

I claim that the dot product is not a binary operation on the set of all two-dimensional vectors. Let $a = [0,0]$ and let $b = [1,0]$. Thus, $a\cdot b = 0$. Since $0$ is not in the set of all two-dimensional vectors, the dot product is not a binary operation on the set of all two-dimensional vectors.

Part 4.

Let $\overrightarrow{v},\overrightarrow{u}$ be defined as three-dimensional vectors. Let $\overrightarrow{w} = \overrightarrow{v} \times \overrightarrow{u}$. Since the cross product of two three-dimensional vectors yields a three dimensional vector, we know that $\overrightarrow{w}$ is a three dimensional vector. Thus, the cross product is a binary operation on the set of three-dimensional vectors.

Part 5.

Let $\sigma_1,\sigma_2$ be defined as permutations on a set $X$. Let $\sigma_3 = \sigma_1 \circ \sigma_3$. Because a permutation is a function from $X$ into $X$, we know that $\sigma_3 \in X$. Thus, the compositions of permutations is a binary operation on the set they are defined on.

Part 6.

I claim that function composition from a set $X$ to a set $Y$ is not a binary operation. Let $f$ be a function from ${[0,1]} \rightarrow {[2,3]}$ be defined by $f(x) = x+2$. Let $g$ be defined the same. Consider the input $g(0.5)$. Clearly, this is $2.5$. Since $f$ is only defined on the interval ${[0,1]}$, we know that $(f \circ g)(0.5)$ is not defined. Thus, function composition from an arbitrary set to another arbitrary set is not a binary operation.

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