The student is familiar with the phrase ordered pair. In the ordered pair \((2,3),( - 2,4)\) and \((a,b)\text{;}\)\(2,\ - 2\) and \(a\) are the first coordinates while \(3,\ 4\) and \(b\) are the second coordinates.
Given sets \(A = \left\{ 3,\ 4 \right\}\) and \(B = \left\{ 2,\ 4,\ 5 \right\}\text{.}\) Then, the set \(\left\{ (3,2),(3,4),(3,5),(4,2),(4,4),(4,5) \right\}\) is the Cartesian product of \(A\) and \(B\text{,}\) and it is denoted by \(A \times B\text{.}\)
Suppose \(A\) and \(B\) are sets. The Cartesian product of \(A\) and \(B\text{,}\) denoted by \(A \times B\text{,}\) is the set which contains every ordered pair whose first coordinate is an element of \(A\) and second coordinate is an element of \(B\text{,}\) i.e.
From example 3.1, we can see that \(A \times B\) and \(B \times A\) are not equal. Recall that two sets are equal if one is a subset of the other and vice versa. To check equality of Cartesian products we need to define equality of ordered pairs.
Suppose \(R\) is a relation from a set \(A\) to a set \(B\text{.}\) Then, \(R\)⊆ \(A\)×\(B\) and hence for each \((a,b) \in A \times B\text{,}\) we have either \((a,b) \in R\) or \((a,b) \notin R\text{.}\) If \((a,b) \in R\text{,}\) we say “\(a\) is \(R-\)related (or simply related) to \(b\)”, and write \(\text{aRb}\text{.}\) If \((a,b) \notin R\text{,}\) we say that “\(a\) is not related to \(b\)”.
Let \(A = \left\{ 1,3,5,7 \right\}\) and \(B = \left\{ 6,8 \right\}\text{.}\) Let \(R\) be the relation “less than” from \(A\) to \(B\text{.}\) Then, \(R = \left\{ (1,6),(1,8),((3,6),(3,8),(5,6),(5,8),(7,8) \right\}\text{.}\)
If \(R\) is a relation from the set \(A\) to the set \(B\text{,}\) then the set \(B\) is called the codomain of the relation \(R\text{.}\) The range of relation is always a subset of the codomain.
The set \(R = \left \{ (4,7),(5,8),(6,\text{10}) \right \}\) is a relation from set \(A = \left\{ 1,2,3,4,5,6 \right\}\) to set \(B = \left\{ 6,7,8,9,\text{10} \right\}\text{.}\) The domain of \(R\) is \(\left\{ 4,5,6 \right\}\text{,}\) the range of \(R\) is \(\left\{ 7,8,\text{10} \right\}\) and the codomain of \(R\) is \(\left\{ 6,7,8,9,\text{10} \right\}\text{.}\)
The set of ordered pairs \(R = \left \{ (8,2),(6, - 3),(5,7),(5, - 3) \right \}\) is a relation between the sets \(\left \{ 5,6,8 \right \}\) and \(\left \{ 2, - 3,7 \right \}\text{,}\) where \(\left\{ 5,6,7 \right\}\) is the domain and \(\left\{ 2, - 3,7 \right\}\) is the range.
If \((a,b) \in R\) for a relation \(R\text{,}\) we say \(a\) is related to (or paired with) \(b\text{.}\) Note that \(a\) may also be paired with an element different from \(b\text{.}\) In any case, \(b\) is called the image of \(a\) while \(a\) is called the pre-image of \(b\)under \(R\text{.}\)
If the domain and/or range of a relation is infinite, we cannot list each element assignment, so instead we use set builder notation to describe the relation. The situation we will encounter most frequently is that of a relation defined by an equation or formula. For example,
is a relation for which the range value is 3 less than twice the domain value. Hence, \((0,\ - 3),(0\text{.}5,\ - 2)\)and \(( - 2, - 7)\) are examples of ordered pairs that are of the assignment.
Let \(R\) be the relation on \(A\) defined by \(R =\left\{(a,b):a,b \in A,a\,\text{is a factor of}\,b \right\}\text{.}\) Find the domain and range of \(R\text{.}\)
If \(R\) is a relation from \(A\) into \(B\text{,}\) then the inverse relation of \(R\text{,}\) denoted by \(R^{- 1}\text{,}\) is a relation from \(B\) to \(A\) and is given by:
Observe that \(Dom(R) = Range(R^{- 1})\) and \(Range(R) = Dom(R^{- 1})\text{.}\) For instance, if \(R = \left\{ (1,4),(9,\text{15}),(\text{10},2) \right\}\) is a relation on a set \(A = \left\{ 1,2,3,\cdots,\text{20} \right\}\text{,}\) then \(R^{- 1} = \left\{ (4,1),(\text{15},9),(2,\text{10}) \right\}\)
Mathematically, it is important for us to distinguish among the relations that assign a unique range element to each domain element and those that do not.
Each element in the domain, \(\left\{ 2,3,6 \right\}\text{,}\) is assigned no more than one value in the range, \(2\) is assigned only \(4\text{,}\)\(3\) is assigned only \(4\text{,}\) and \(6\) is assigned only \(-4\text{.}\) Therefore, it is a function.
Map or mapping, transformation and correspondence are synonyms for the word function. If \(f\) is a function and \((x,y) \in f\text{,}\) we say \(x\) is mapped to \(y\) by \(f\text{.}\)
If to the element \(x\) of \(A\) corresponds \(y\ ( \in B)\) under the function \(f\text{,}\) then we write \(f(x) = y\) and \(y\) is called the image of x under \(f\) and x is called a preimage of \(y\) under \(f\text{.}\)
In order to show that a relation \(f\) from \(A\) into \(B\) is a function, we first show that the domain of \(f\) is \(A\) and next we show that \(f\) well defined or single-valued, i.e. if \(x = y\) in \(A\text{,}\) then \(f(x) = f(y)\) in \(B\) for all \(x,y \in A\text{.}\)
Let \(A = \left\{ 1,2,3,4 \right\}\) and \(B = \left\{ 1,6,8,\text{11},\text{15} \right\}\text{.}\) Which of the following are functions from A to \(B\text{.}\)
\(f\) is a function because to each element of \(A\) there corresponds exactly one element of \(B\text{.}\) In the given function, the images of all element of \(A\) are the same.
As with relations, we can describe a function with an equation. For example, \(y=2x+1\) is a function, since each \(x\) will produce only one \(y\text{.}\)
More generally any real number \(x\) is mapped to its square. As the square of a number is unique, \(f\) maps every real number to a unique number. Thus, \(f\) is a function from \(\Re\) into \(\Re\text{.}\)
We will find it useful to use the following vocabulary: The independent variable refers to the variable representing possible values in the domain, and the dependent variable refers to the variable representing possible values in the range. Thus, in our usual ordered pair notation \((x,y)\text{,}\)\(x\) is the independent variable and \(y\) is the dependent variable.
Let \(f\) be the subset of \(Q \times Z\)defined by \(f = \left\{ \left( \frac{p}{q},p \right):p,q \in Z,\ q \neq 0 \right\}\text{.}\) Is \(f\) a function?
First we note that \(\text{Dom}(f) = Q\text{.}\) Then, \(f\) satisfies condition (i) in the definition of a function. Now, \(\left( \frac{2}{3},2 \right) \in f\text{,}\)\(\left( \frac{4}{6},4 \right) \in f\) and \(\frac{2}{3} = \frac{4}{6}\) but \(f\left( \frac{2}{3} \right) = 2 \neq 4 = f\left( \frac{4}{6} \right)\text{.}\) Thus \(f\) is not well defined. Hence, \(f\) is not a function from \(Q\) to \(Z\text{.}\)
First we show that \(f\) satisfies condition (i) in the definition. Let \(x\) be any element of \(Z\text{.}\) Then, \(x = x \cdot 1\text{.}\) Hence, \((x,x + 1) = (x \cdot 1,x + 1) \in f\text{.}\) This implies that \(x \in Dom(f)\text{.}\) Thus, \(Z \subseteq Dom(f)\text{.}\) However, \(Dom(f) \subseteq Z\) and so \(Dom(f) = Z\text{.}\) Now, \(\) and \(4 = 4 \cdot 1 = 2 \cdot 2\text{.}\) Thus, \((4 \cdot 1,4 + 1)\) and \((2 \cdot 2,2 + 2)\) are in \(f\text{.}\) Hence we find that \(4 \cdot 1 = 2 \cdot 2\) and \(f(4 \cdot 1) = 5 \neq 4 = f(2 \cdot 2)\text{.}\) This implies that \(f\) is not well defined, i.e, \(f\) does not satisfy condition (ii). Hence, \(f\) is not a function from \(Z\) to \(Z\text{.}\)
Let \(A = \left\{ 1,2,3 \right\}\) and \(B = \left\{ 1,2,3,\cdots,\text{10} \right\}\text{.}\) Let \(f:A \rightarrow B\) be the correspondence which assigns to each element in \(A\text{,}\) its square. Thus, we have \(f(1) = 1,\ f(2) = 4,\ f(3) = 9\text{.}\) Therefore, \(f\) is a function and \(\text{Dom}(f) = \left\{ 1,2,3 \right\}\text{,}\)\(\text{Range}(f) = \left\{ 1,4,9 \right\}\) and codomain of \(f\) is \(\left\{ 1,2,3,\cdots,\text{10} \right\}\text{.}\)
Let \(A = \left\{ 2,4,6,7,9 \right\},\ B = \text{IN}\text{.}\) Let \(x\) and \(y\) represent the elements in the sets \(A\) and \(B\text{,}\) respectively. Let \(f:A \rightarrow B\) be a function defined by \(f(x) = \text{15}x + \text{17},\ x \in A\text{.}\)
This implies that \(\text{Dom}(f) = \left\{ 2,4,6,7,9 \right\},\ \text{Range}(f) = \left\{ \text{47},\text{77},\text{107},\text{122},\text{152} \right\}\) and codomain of \(f\)is IN.
To determine whether \(y = - 3x + 5\) gives \(y\) as a function of \(x\text{,}\) we need to know whether each x-value uniquely determines a y-value. Looking at the equation \(y = - 3x + 5\text{,}\) we can see that once \(x\) is chosen we multiply it by – 3 and then add 5. Thus, for each x there is a unique \(y\text{.}\) Therefore, \(y = - 3x + 5\) is a function. It domain is the set of all real numbers.
Looking at the equation \(y = \frac{2x}{3x - 5}\) carefully, we can see that each \(x-value\) uniquely determines a \(y-value\) (one \(x-value\) can not produce two different \(y-values\)). Therefore, \(y = \frac{2x}{3x - 5}\) is a function.
As for its domain, we ask ourselves. Are there any values of \(x\) that must be excluded? Since \(y = \frac{2x}{3x - 5}\) is a fractional expression, we must exclude any value of \(x\) that makes the denominator equal to zero. We must have
Therefore, the domain consists of all real numbers except \(\frac{5}{3}\text{.}\) Thus, \(\text{Dom}(f) =\)\(\left\{ x:x \neq \frac{5}{3} \right\}\text{.}\)
For the equation \(y^{2} = x\text{,}\) if we choose \(x = 9\) we get \(y^{2} = 9\text{,}\) which gives \(y = \pm 3\text{.}\) In other words, there are two \(y -\)values associated with \(x = 9\text{.}\) Therefore, \(y^{2} = x\) is not a function.
Since we want \(3x - x^{2} = x(3 - x)\) to be non-negative, the sign analysis shows us that the domain is \(\left\{ x:0 \leq x \leq 3 \right\}\) or \(\lbrack 0,3\rbrack\text{.}\)
Let \(A = \left\{ 1,2,3,4,5,6 \right\}\text{.}\) Define a relation on \(A\) by \(R = \left\{ (x,y):y = x + 1 \right\}\text{.}\) Write down the domain, codomain and range of \(R\text{.}\) Find \(R^{- 1}\text{.}\)
