Class 8: Relations and Mapping

$\displaystyle \textbf{Ordered Pair}$

$\displaystyle \text{Two elements }a\text{ and }b\text{ listed in a specific order form an ordered pair,}$
$\displaystyle \text{denoted by }(a,b).$

$\displaystyle \text{In the ordered pair }(a,b),\ a\text{ is called the first component and }b$
$\displaystyle \text{is called the second component.}$

$\displaystyle \text{The order of the components is important. In general,}$

$\displaystyle (a,b)\ne(b,a).$

$\displaystyle \text{For example, }(5,4)\ne(4,5).$

$\displaystyle \text{Two ordered pairs are equal if and only if their corresponding components}$
$\displaystyle \text{are equal. Thus,}$

$\displaystyle (a,b)=(c,d)\iff a=c\text{ and }b=d.$

$\displaystyle \text{For example, if }(a,b)=(2,7),\text{ then }a=2\text{ and }b=7.$

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$\displaystyle \textbf{Cartesian Product of Two Sets}$

$\displaystyle \text{Let }A\text{ and }B\text{ be two non-empty sets. The Cartesian product}$
$\displaystyle A\times B\text{ is the set of all ordered pairs }(a,b)\text{ such that}$
$\displaystyle a\in A\text{ and }b\in B.$

$\displaystyle A\times B=\{(a,b):a\in A\text{ and }b\in B\}.$

$\displaystyle \textbf{Example 1:}$

$\displaystyle \text{Let }A=\{a,b\}\text{ and }B=\{1,2,3\}.$

$\displaystyle \text{Then}$

$\displaystyle A\times B=\{(a,1),(a,2),(a,3),(b,1),(b,2),(b,3)\}.$

$\displaystyle \text{Thus, every element of }A\text{ is paired with every element of }B.$

$\displaystyle \text{Also, if }n(A)=m\text{ and }n(B)=n,\text{ then}$

$\displaystyle n(A\times B)=mn.$

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$\displaystyle \textbf{Relation}$

$\displaystyle \text{Let }A\text{ and }B\text{ be two non-empty sets. A relation }R\text{ from }A$
$\displaystyle \text{to }B\text{ is any subset of the Cartesian product }A\times B.$

$\displaystyle R\subseteq A\times B.$

$\displaystyle \text{Therefore, every element of a relation }R\text{ is an ordered pair }(a,b),$
$\displaystyle \text{where }a\in A\text{ and }b\in B.$

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$\displaystyle \textbf{Representation of a Relation}$

$\displaystyle \text{A relation can be represented in different ways, such as roster form,}$
$\displaystyle \text{an arrow diagram and set-builder form.}$

$\displaystyle \textbf{1. Roster Form}$

$\displaystyle \text{When a relation is written by listing all its ordered pairs, it is said to be}$
$\displaystyle \text{represented in roster form.}$

$\displaystyle \textbf{Example 2:}$

$\displaystyle \text{Let }A=\{1,9,16,25\}\text{ and }B=\{1,2,3,4,5\}.$

$\displaystyle \text{Let }R\text{ be the relation from }A\text{ to }B\text{ defined by}$
$\displaystyle \text{``the first component is the square of the second component.''}$

$\displaystyle \text{Then, in roster form,}$

$\displaystyle R=\{(1,1),(9,3),(16,4),(25,5)\}.$

$\displaystyle \text{Notice that in every ordered pair }(a,b)\in R,\text{ we have }a=b^2.$

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$\displaystyle \textbf{2. Arrow Diagram}$

$\displaystyle \text{Let }R\text{ be a relation from }A\text{ to }B.\text{ Write the elements of }A$
$\displaystyle \text{and }B\text{ in two separate sets and draw an arrow from }a\in A\text{ to }b\in B$
$\displaystyle \text{whenever }(a,b)\in R.$

$\displaystyle \text{Thus, the relation in Example 2 is represented by the arrows}$

$\displaystyle 1\rightarrow1,\qquad9\rightarrow3,\qquad 16\rightarrow4,\qquad25\rightarrow5.$

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$\displaystyle \textbf{3. Set-Builder Form}$

$\displaystyle \text{A relation }R\text{ from }A\text{ to }B\text{ can be represented in set-builder}$
$\displaystyle \text{form by stating the rule connecting the two components.}$

$\displaystyle R=\{(a,b):a\in A,\ b\in B\text{ and }a\text{ is related to }b\}.$

$\displaystyle \text{For Example 2,}$

$\displaystyle R=\{(a,b):a\in A,\ b\in B\text{ and }a=b^2\}.$

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$\displaystyle \textbf{Domain and Range of a Relation}$

$\displaystyle \text{Let }R\text{ be a relation from }A\text{ to }B.$

$\displaystyle \text{Domain}(R)=\text{the set of all first components of the ordered pairs in }R.$

$\displaystyle \text{Range}(R)=\text{the set of all second components of the ordered pairs in }R.$

$\displaystyle \text{The domain is a subset of }A\text{ and the range is a subset of }B.$

$\displaystyle \text{Thus,}$

$\displaystyle \text{Domain}(R)\subseteq A,\qquad\text{Range}(R)\subseteq B.$

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$\displaystyle \textbf{Rule of a Relation}$

$\displaystyle \text{The property that describes how the first component is related to the second}$
$\displaystyle \text{component of each ordered pair is called the rule of the relation.}$

$\displaystyle \textbf{Example 3:}$

$\displaystyle \text{Let }A=\{2,3,4\}\text{ and }B=\{1,3,5\}.$

$\displaystyle \text{Let }R\text{ be the relation ``is less than'' from }A\text{ to }B.$
$\displaystyle \text{Find }R\text{ and write its domain and range.}$

$\displaystyle \text{Solution:}$

$\displaystyle R=\{(a,b):a\in A,\ b\in B\text{ and }a<b\}.$

$\displaystyle \text{Checking the elements of }A\times B\text{ for }a<b,\text{ we get}$

$\displaystyle R=\{(2,3),(2,5),(3,5),(4,5)\}.$

$\displaystyle \therefore \text{Domain}(R)=\{2,3,4\}.$

$\displaystyle \therefore \text{Range}(R)=\{3,5\}.$

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$\displaystyle \textbf{Function or Mapping}$

$\displaystyle \text{Let }A\text{ and }B\text{ be two non-empty sets. A function or mapping }f$
$\displaystyle \text{from }A\text{ to }B\text{ assigns each element }x\in A\text{ to exactly one}$
$\displaystyle \text{element }f(x)\in B.$

$\displaystyle \text{We write}$

$\displaystyle f:A\rightarrow B.$

$\displaystyle \text{The element }f(x)\text{ is called the image of }x\text{ under }f.$

$\displaystyle \text{For }f:A\rightarrow B\text{ to be a function:}$

$\displaystyle \text{(i) Every element of }A\text{ must have an image in }B.$

$\displaystyle \text{(ii) Every element of }A\text{ must have exactly one image in }B.$

$\displaystyle \text{Different elements of }A\text{ may, however, have the same image in }B.$

$\displaystyle \textbf{Example 1:}$

$\displaystyle \text{Let }A=\{1,2,3\}\text{ and }B=\{2,4,6,8\}.$

$\displaystyle \text{Consider the rule }f(x)=2x.$

$\displaystyle f(1)=2,\qquad f(2)=4,\qquad f(3)=6.$

$\displaystyle \text{Each element of }A\text{ has exactly one image in }B.$

$\displaystyle \therefore f:A\rightarrow B\text{ is a function.}$

$\displaystyle \text{Notice that }8\in B\text{ need not be the image of an element of }A.$

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$\displaystyle \textbf{Representation of a Function}$

$\displaystyle \text{A function can be represented in different ways. Three common methods are}$
$\displaystyle \text{an arrow diagram, roster form and equation form.}$

$\displaystyle \textbf{1. Arrow Diagram}$

$\displaystyle \text{In an arrow diagram, an arrow is drawn from every element of the domain to}$
$\displaystyle \text{its unique image in the co-domain.}$

$\displaystyle \text{For the function }f(x)=2x\text{ in Example 1, the arrows are}$

$\displaystyle 1\rightarrow2,\qquad2\rightarrow4,\qquad3\rightarrow6.$

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$\displaystyle \textbf{2. Roster Form}$

$\displaystyle \text{A function can be represented as the set of ordered pairs }(x,f(x)),$
$\displaystyle \text{where }x\text{ belongs to the domain.}$

$\displaystyle f=\{(1,2),(2,4),(3,6)\}.$

$\displaystyle \\$

$\displaystyle \textbf{3. Equation Form}$

$\displaystyle \text{If a function is defined by a formula or rule, it can be represented by an}$
$\displaystyle \text{equation relating the input and output.}$

$\displaystyle \text{For example, if }f(x)=2x\text{ and }y=f(x),\text{ then}$

$\displaystyle y=2x.$

$\displaystyle \\$

$\displaystyle \textbf{Example 2:}$

$\displaystyle \text{Let }A=\{1,2,3,4,5\}\text{ and }B=\{1,4,9,16,20,25\}.$

$\displaystyle \text{Define }f:A\rightarrow B\text{ by }f(x)=x^2.$

$\displaystyle \text{Represent the function by an arrow diagram, roster form and equation form.}$

$\displaystyle \text{Solution:}$

$\displaystyle f(1)=1,\qquad f(2)=4,\qquad f(3)=9,\qquad f(4)=16,\qquad f(5)=25.$

$\displaystyle \text{Arrow Diagram:}$

$\displaystyle 1\rightarrow1,\qquad2\rightarrow4,\qquad3\rightarrow9,\qquad 4\rightarrow16,\qquad5\rightarrow25.$

$\displaystyle \text{Roster Form:}$

$\displaystyle f=\{(1,1),(2,4),(3,9),(4,16),(5,25)\}.$

$\displaystyle \text{Equation Form:}$

$\displaystyle y=x^2.$

$\displaystyle \text{Notice that }20\in B\text{ has no pre-image in }A.\text{ This does not prevent}$
$\displaystyle f\text{ from being a function from }A\text{ to }B.$

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$\displaystyle \textbf{Domain, Co-Domain and Range of a Function}$

$\displaystyle \text{Let }f:A\rightarrow B\text{ be a function. Then}$

$\displaystyle \text{Domain}(f)=A.$

$\displaystyle \text{Co-domain}(f)=B.$

$\displaystyle \text{Range}(f)=\{f(x):x\in A\}.$

$\displaystyle \text{Therefore, the range of a function is a subset of its co-domain:}$

$\displaystyle \text{Range}(f)\subseteq B.$

$\displaystyle \text{The range need not be equal to the co-domain.}$

$\displaystyle \text{For Example 2,}$

$\displaystyle \text{Domain}(f)=\{1,2,3,4,5\},$

$\displaystyle \text{Co-domain}(f)=\{1,4,9,16,20,25\},$

$\displaystyle \text{Range}(f)=\{1,4,9,16,25\}.$

$\displaystyle \\$

$\displaystyle \textbf{Function as a Relation}$

$\displaystyle \text{Let }A\text{ and }B\text{ be two non-empty sets and let }R\text{ be a relation}$
$\displaystyle \text{from }A\text{ to }B.$

$\displaystyle R\text{ is a function from }A\text{ to }B\text{ if and only if:}$

$\displaystyle \text{(i) Domain}(R)=A,\text{ and}$

$\displaystyle \text{(ii) every element of }A\text{ occurs as the first component of exactly one}$
$\displaystyle \text{ordered pair in }R.$

$\displaystyle \text{Equivalently, no two distinct ordered pairs in }R\text{ can have the same}$
$\displaystyle \text{first component with different second components.}$

$\displaystyle \textbf{Example 3:}$

$\displaystyle \text{Let }A=\{1,2,3,4\}\text{ and }B=\{3,4,5,6,7\}.$

$\displaystyle R_1=\{(1,3),(2,4),(3,5)\},$

$\displaystyle R_2=\{(1,3),(1,7),(2,4),(3,5),(4,6)\},$

$\displaystyle R_3=\{(1,3),(2,4),(3,5),(4,6)\}.$

$\displaystyle \text{Determine which of the above relations is a function from }A\text{ to }B.$

$\displaystyle \text{Solution:}$

$\displaystyle \text{Domain}(R_1)=\{1,2,3\}\ne A.$

$\displaystyle \therefore R_1\text{ is not a function from }A\text{ to }B.$

$\displaystyle \text{In }R_2,\text{ the element }1\in A\text{ is associated with both }3\text{ and }7.$

$\displaystyle \text{Thus, }1\text{ has two different images.}$

$\displaystyle \therefore R_2\text{ is not a function from }A\text{ to }B.$

$\displaystyle \text{Domain}(R_3)=\{1,2,3,4\}=A,\text{ and each element of }A$
$\displaystyle \text{has exactly one image in }B.$

$\displaystyle \therefore R_3\text{ is a function from }A\text{ to }B.$

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$\displaystyle \textbf{Real-Valued Functions}$

$\displaystyle \text{A function whose values are real numbers is called a real-valued function.}$

$\displaystyle \text{In particular, if both the domain and co-domain are subsets of }\mathbb{R},$
$\displaystyle \text{then }f\text{ is a real function of a real variable.}$