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Let \(X=\\{0,1, \ldots, 6,7\\}\) and \(Y=\\{8,10,12, \ldots, 20,22\\} .\) Define \(F: X \rightarrow Y\) as \(F(x)=2 x+8\). List the ordered pairs of the relation that define this function.

Short Answer

Expert verified
The ordered pairs are \((0, 8), (1, 10), (2, 12), (3, 14), (4, 16), (5, 18), (6, 20), (7, 22)\).

Step by step solution

01

Understand the Function Transformation

The function is given by \( F: X \rightarrow Y \) with \( F(x) = 2x + 8 \). This means for each element \( x \) in the set \( X \), the function \( F \) transforms it into an element in \( Y \) using the formula \( F(x) = 2x + 8 \). Our task is to compute \( F(x) \) for each \( x \) in the set \( X \).
02

Evaluate the Function for Each Element of X

Compute \( F(x) \) for each element of \( X = \{0, 1, 2, 3, 4, 5, 6, 7\} \). - \( F(0) = 2(0) + 8 = 8 \) - \( F(1) = 2(1) + 8 = 10 \) - \( F(2) = 2(2) + 8 = 12 \) - \( F(3) = 2(3) + 8 = 14 \) - \( F(4) = 2(4) + 8 = 16 \) - \( F(5) = 2(5) + 8 = 18 \) - \( F(6) = 2(6) + 8 = 20 \) - \( F(7) = 2(7) + 8 = 22 \)
03

Formulate the Relation as Ordered Pairs

For each \( x \) in \( X \), \( F(x) \) is calculated and the function \( F \) can be written in terms of ordered pairs \((x, F(x))\). Therefore, the function is represented as a set of ordered pairs: \( \{(0, 8), (1, 10), (2, 12), (3, 14), (4, 16), (5, 18), (6, 20), (7, 22)\} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ordered Pairs
In mathematics, ordered pairs are a fundamental concept used to describe relationships between elements of two sets. An ordered pair is written in the form
  • \((x, y)\) - where \(x\) and \(y\) are elements belonging to two different sets.
The order is important here; \((x, y)\) is not the same as \((y, x)\) unless \(x = y\).
In the context of our function \(F: X \rightarrow Y\), ordered pairs are used to relate each element \(x\) in set \(X\) with a corresponding element \(F(x)\) in set \(Y\).
For instance, after calculating the function, we see the ordered pair \((0, 8)\), linking \(x = 0\) in set \(X\) with \(F(x) = 8\) in set \(Y\). Similarly, the ordered pair \((7, 22)\) indicates \(x = 7\) in set \(X\) corresponds to \(F(x) = 22\) in set \(Y\).
This pairing helps visually see the mapping of inputs to outputs, making function relations clear and organized.
Function Evaluation
Function evaluation is the process of determining the value of a function for specific inputs. When evaluating a function, like \(F(x) = 2x + 8\), we substitute each \(x\) from the set \(X\) into the expression.
  • For example, to evaluate \(F(x)\) for \(x = 3\), we calculate:\[ F(3) = 2(3) + 8 = 14 \]
This tells us that when \(x\) is 3, the result of the function, or \(F(x)\), is 14. Such evaluation is performed for every element of the input set.
Each result from this operation contributes to forming the ordered pairs of the function's relation.
In educational exercises, function evaluation enhances understanding of how functions operate and transform inputs to outputs—key for grasping basic and advanced mathematical concepts.
Set Mapping
Set mapping refers to connecting each element of one set to an element of another set using a defined rule or function. The function \(F: X \rightarrow Y\) defines such a mapping, where each element \(x\) from set \(X\) is linked through a function to an element in set \(Y\).
  • Mapping is significant because it determines how elements from one set pair with elements in another set based on the rule given.
In our example, the mapping rule is \(F(x) = 2x + 8\). When we input values from \(X = \{0,1,2,3,4,5,6,7\}\), each is transformed into a corresponding value in \(Y\) using this function.
Mapping helps to structure relationships in expressions of functions and has practical applications in analyzing connections between different data sets in various fields, such as computer science and economics.

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Most popular questions from this chapter

Let \(A\) and \(B\) be sets with \(A_{1}, A_{2} \subseteq A,\) and let \(F: A \rightarrow B\). Let \(F\left(A_{i}\right)\) denote \(\mid F(x)\) : \(\left.x \in A_{i}\right\\}\) for \(i=1,2\). Show that: (a) If \(A_{1} \subseteq A_{2},\) then \(F\left(A_{1}\right) \subseteq F\left(A_{2}\right)\). (b) \(F\left(A_{1} \cup A_{2}\right)=F\left(A_{1}\right) \cup F\left(A_{2}\right)\). (c) \(F\left(A_{1} \cap A_{2}\right) \subseteq F\left(A_{1}\right) \cap F\left(A_{2}\right)\). (d) \(F\left(A_{1}\right)-F\left(A_{2}\right) \subseteq F\left(A_{1}-A_{2}\right)\). (e) \(A_{1} \subseteq F^{-1}\left(F\left(A_{1}\right)\right)\). (f) Find an example in which \(A_{1} \subset A_{2}\) but \(F\left(A_{1}\right)=F\left(A_{2}\right)\). (g) Find an example in which \(A_{1} \neq F^{-1}\left(F\left(A_{1}\right)\right)\).

For sets \(X, Y,\) and \(Z\), let \(F: X \rightarrow Y\) and \(G: Y \rightarrow Z\) be \(I-I\) correspondences. Prove that \((G \circ F)^{-1}=F^{-1} \circ G^{-1}\)

(a) Let \(F: A \rightarrow B\) be a function. Prove that \(F\) is onto if and only if \(F^{-1}\left(B_{1}\right) \neq B\) for each nonempty subset \(B_{1}\) of \(B\). (b) Let \(F: A \rightarrow B\) be a function. Prove that \(F\) is onto if and only if \(F\left(F^{-1}\left(B_{1}\right)\right)=\) \(B_{1}\) for all \(B_{1} \subseteq B\).

Let \(S=\\{(0,8),(1,10),(2,12),(3,14),(4,16),(5,18),(6,20),(7,22)]\). Is \(S\) a function? Why, or why not? Find \(S^{-1},\) and list its elements. Is \(S^{-1}\) a function? Why, or why not? Identify the domain of \(S^{-1}\)

Let \(X\) be a set, and let \(\mathcal{F}_{X}\) be the set of all \(I-I\) functions from \(X\) onto \(X\). We have two operations on functions in \(\mathcal{F}_{X}: \circ\) and -1 . Prove the following statements called group axioms. (If the results are already proved in the book, note where to find the proofs.) (a) For all \(F, G \in \mathcal{F}_{X}, F \circ G \in \mathcal{F}\). (b) For all \(F, G, H \in \mathcal{F}_{X},(F \circ G) \circ H=F \circ(G \circ H)\) (Associative Law). (c) For all \(F \in \mathcal{F}_{X}, F \circ I d_{X}=I d_{X} \circ F=F\). (Identity Axiom). (d) For all \(F \in \mathcal{F}_{X}\), there exists an \(F^{-1}\) such that \(F \circ F^{-1}=F^{-1} \circ F=I d_{X}\) (Inverse Axiom).

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