Answer
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Hint: In mathematics, a homogeneous relation R over a set $\mathrm{X}$ is transitive if for all elements $\mathrm{a}, \mathrm{b}, \mathrm{c}$ in $\mathrm{X},$ whenever $\mathrm{R}$ relates a to $\mathrm{b}$ and $\mathrm{b}$ to $\mathrm{c},$ then $\mathrm{R}$ also relates a to Each partial order as well as each equivalence relation needs to be transitive. In mathematics, the transitive property states that: If $a=b$ and $b=c,$ then $a=c .$ In other words, if a is related to $\mathrm{b}$ by some property, and $\mathrm{b}$ is related to $\mathrm{c}$ by the same property, then a is related to c by that property. The transitive property meme comes from the transitive property of equality in mathematics. In math, if $\mathrm{A}=\mathrm{B}$ and $\mathrm{B}=\mathrm{C},$ then $\mathrm{A}=\mathrm{C} .$ So, if $\mathrm{A}=5$ for example, then $\mathrm{B}$ and must both also be 5 by the transitive property. For example, humans eat cows and cows eat grass, so by the transitive property, humans eat grass.
Complete step-by-step answer:
Integer, Whole-valued positive or negative number or 0. The integers are generated from the set of counting numbers $1,2,3, \ldots$ and the operation of subtraction.
RULE 1: The product of a positive integer and a negative integer is negative.
RULE 2: The product of two positive integers is positive.
RULE 3: The product of two negative integers is positive.
RULE 4: The quotient of a positive integer and a negative integer is negative.
$\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x}-\mathrm{y}$ is an integer $\}$
Now, for every $\mathrm{x} \in \mathrm{Z},(\mathrm{x}, \mathrm{x}) \in \mathrm{R}$ as $\mathrm{x}-\mathrm{x}=0$ is an integer.
$\therefore \mathrm{R}$ is reflexive.
Now, for every $\mathrm{x}, \mathrm{y} \in \mathrm{Z}$ if $(\mathrm{x}, \mathrm{y}) \in \mathrm{R},$ then $\mathrm{x}-\mathrm{y}$ is an integer.
$\Rightarrow-(\mathrm{x}-\mathrm{y})$ is also an integer.
$\Rightarrow(\mathrm{y}-\mathrm{x})$ is an integer.
$\therefore(\mathrm{y}, \mathrm{x}) \in \mathrm{R}$
$\Rightarrow \mathrm{R}$ is symmetric.
Now,
Let $(\mathrm{x}, \mathrm{y})$ and $(\mathrm{y}, \mathrm{z}) \in \mathrm{R},$ where $\mathrm{x}, \mathrm{y}, \mathrm{z} \in \mathrm{Z}$
$\Rightarrow(\mathrm{x}-\mathrm{y})$ and $(\mathrm{y}-\mathrm{z})$ are integers.
$\Rightarrow \mathrm{x}-\mathrm{z}=(\mathrm{x}-\mathrm{y})+(\mathrm{y}-\mathrm{z})$ is an integer.
$\therefore(\mathrm{x}, \mathrm{z}) \in \mathrm{R}$
$\therefore \mathrm{R}$ is transitive.
Hence, $\mathrm{R}$ is reflexive, symmetric, and transitive.
Hence option 4 is correct answer
Note: Basic Properties of Reflections: (Reflection 1 ) A reflection maps a line to a line, a ray to a ray, a segment to a segment, and an angle to an angle. (Reflection 2) A reflection preserves lengths of segments. (Reflection 3) A reflection preserves measures of angles. The symmetric property of equality tells us that both sides of an equal sign are equal no matter which side of the equal sign they are on. Remember it states that if $x=y,$ then $y=x$.
The Symmetric Property states that for all real numbers $x$ and y, if $x=y,$ then $y=x .$ Transitive Property. The Transitive Property states that for all real numbers $\mathrm{x}, \mathrm{y},$ and z, if $x=y$ and $y=z,$ then $x=z$ . The symmetric property of equality is a simple property that says we can interchange the sides of an equation without changing the truth-value of the equation. That is, if a $=\mathrm{b},$ then $\mathrm{b}=\mathrm{a}$. Each side of the equation can be thought of as the mirror image of the other side.
Complete step-by-step answer:
Integer, Whole-valued positive or negative number or 0. The integers are generated from the set of counting numbers $1,2,3, \ldots$ and the operation of subtraction.
RULE 1: The product of a positive integer and a negative integer is negative.
RULE 2: The product of two positive integers is positive.
RULE 3: The product of two negative integers is positive.
RULE 4: The quotient of a positive integer and a negative integer is negative.
$\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x}-\mathrm{y}$ is an integer $\}$
Now, for every $\mathrm{x} \in \mathrm{Z},(\mathrm{x}, \mathrm{x}) \in \mathrm{R}$ as $\mathrm{x}-\mathrm{x}=0$ is an integer.
$\therefore \mathrm{R}$ is reflexive.
Now, for every $\mathrm{x}, \mathrm{y} \in \mathrm{Z}$ if $(\mathrm{x}, \mathrm{y}) \in \mathrm{R},$ then $\mathrm{x}-\mathrm{y}$ is an integer.
$\Rightarrow-(\mathrm{x}-\mathrm{y})$ is also an integer.
$\Rightarrow(\mathrm{y}-\mathrm{x})$ is an integer.
$\therefore(\mathrm{y}, \mathrm{x}) \in \mathrm{R}$
$\Rightarrow \mathrm{R}$ is symmetric.
Now,
Let $(\mathrm{x}, \mathrm{y})$ and $(\mathrm{y}, \mathrm{z}) \in \mathrm{R},$ where $\mathrm{x}, \mathrm{y}, \mathrm{z} \in \mathrm{Z}$
$\Rightarrow(\mathrm{x}-\mathrm{y})$ and $(\mathrm{y}-\mathrm{z})$ are integers.
$\Rightarrow \mathrm{x}-\mathrm{z}=(\mathrm{x}-\mathrm{y})+(\mathrm{y}-\mathrm{z})$ is an integer.
$\therefore(\mathrm{x}, \mathrm{z}) \in \mathrm{R}$
$\therefore \mathrm{R}$ is transitive.
Hence, $\mathrm{R}$ is reflexive, symmetric, and transitive.
Hence option 4 is correct answer
Note: Basic Properties of Reflections: (Reflection 1 ) A reflection maps a line to a line, a ray to a ray, a segment to a segment, and an angle to an angle. (Reflection 2) A reflection preserves lengths of segments. (Reflection 3) A reflection preserves measures of angles. The symmetric property of equality tells us that both sides of an equal sign are equal no matter which side of the equal sign they are on. Remember it states that if $x=y,$ then $y=x$.
The Symmetric Property states that for all real numbers $x$ and y, if $x=y,$ then $y=x .$ Transitive Property. The Transitive Property states that for all real numbers $\mathrm{x}, \mathrm{y},$ and z, if $x=y$ and $y=z,$ then $x=z$ . The symmetric property of equality is a simple property that says we can interchange the sides of an equation without changing the truth-value of the equation. That is, if a $=\mathrm{b},$ then $\mathrm{b}=\mathrm{a}$. Each side of the equation can be thought of as the mirror image of the other side.
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