Question

Integers are closed under
$$\begin{array}{rl}& \text{A}\text{. Addition}\\ & \text{B}\text{. Subtraction}\\ & \text{C}\text{. Multiplication}\\ & \text{D}\text{. Division}\end{array}$$

Answer 1

Hint: We know that a set is closed under an operation if the performance of that operation on members of the set always produces a member of that set. Now we have to check whether integers are closed under addition, subtraction, multiplication, and division. Let us consider two integers. Now we will check that in each and every operation whether the integers are closed or not.

Complete step-by-step solution:
Before solving the question, we should know that a set is closed under an operation if the performance of that operation on members of the set always produces a member of that set.
Now we have to check whether integers are closed under addition, subtraction, multiplication, and division.
Now, let us consider two integers 4 and -7.
Let us assume
$$\begin{array}{rl}& \text{a=4}.....\text{(1)}\\ & \text{b=-7}....\text{(2)}\end{array}$$
So, it is clear that $$\text{a,b}\in I$$.
CLOSURE PROPERTY UNDER ADDITION:
Now we have to check $$\text{a+b}\in I$$ is satisfied or not.
Let us assume the value of $$\text{a+b}$$ is equal to C.
$$\Rightarrow c=a+b....(3)$$
Now let us substitute equation (1) and equation (2) in equation (3), then we get
$$\begin{array}{rl}& \Rightarrow c\text{=4-7}\\ & \Rightarrow c\text{=-3}....\text{(4)}\end{array}$$
From equation (4), it is clear that the sum of a and b is equal to -3.
We know that $$c\in I$$. So, we can say that $$\text{a+b}\in I$$ is satisfied.
Hence, we can say that $$\text{a+b}$$ is closed under addition.
CLOSURE PROPERTY UNDER SUBTRACTION:
Now we have to check $$\text{a-b}\in I$$ is satisfied or not.
Let us assume the value of $$\text{a-b}$$ is equal to d.
$$\Rightarrow d=a-b....(5)$$
Now let us substitute equation (1) and equation (2) in equation (3), then we get
$$\begin{array}{rl}& \Rightarrow d\text{=4-(-7)}\\ & \Rightarrow \text{d=4+7}\\ & \Rightarrow d\text{=11}...\text{(6)}\end{array}$$
From equation (6), it is clear that the difference of a and b is equal to 11.
We know that $$d\in I$$. So, we can say that $$\text{a-b}\in I$$ is satisfied.
Hence, we can say that $$\text{a-b}$$ is closed under subtraction.
CLOSURE PROPERTY UNDER MULTIPLICATION:
Now we have to check $$\text{a}\times \text{b}\in I$$ is satisfied or not.
Let us assume the value of $$\text{a}\times \text{b}$$ is equal to e.
$$\Rightarrow e=a\times b....(7)$$
Now let us substitute equation (1) and equation (2) in equation (3), then we get
$$\begin{array}{rl}& \Rightarrow e\text{=4}\times (-7)\\ & \Rightarrow e\text{=-28}...\text{(8)}\end{array}$$
From equation (8), it is clear that the product of a and b is equal to -28
We know that $$e\in I$$. So, we can say that $$\text{a}\times \text{b}\in I$$ is satisfied.
Hence, we can say that $$\text{a}\times \text{b}$$ is closed under multiplication.
CLOSURE PROPERTY UNDER DIVISION:
Now we have to check $$\text{a}÷\text{b}\in I$$ is satisfied or not.
Let us assume the value of $$\text{a}÷\text{b}$$ is equal to f.
$$\Rightarrow f=a÷b....(9)$$
Now let us substitute equation (1) and equation (2) in equation (3), then we get
$$\begin{array}{rl}& \Rightarrow f\text{=4}÷(-7)\\ & \Rightarrow f\text{=-}{\displaystyle \frac{4}{7}}\\ & \Rightarrow f=-0.5555.....(10)\end{array}$$
From equation (10), it is clear that the division of a and b is equal to -0.5555.
We know that $$f\notin I$$. So, we can say that $$\text{a}÷\text{b}\notin I$$ is satisfied.
Hence, we can say that $$\text{a}÷\text{b}$$ is not closed under division.
So, we can say that option A, option B, and Option C are correct.

Note: Students may have a misconception that a set is not closed under an operation if performance of that operation on members of the set always produces a member of that set. According to this definition, it is clear that integers are closed under division. But we know that integers are closed under addition, subtraction, and multiplication but not closed under division. So, this misconception should be avoided.