A function F from a set A to a set B is a relation with domain A and co-domain B that satisfies the following two properties: 1. For every element x in A, there is an element y in B such that (x, y) E F.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Please help me with this? How does number 1 differ from a relation? Please do not do it like the proof that was provide because it is wrong I just need a couple of sentences that answer the questions No proof please.
**Understanding Functions and Domains**

For the given condition 1, we must prove that the domain of \( f \) is \( A \). This means that for each element of \( A \), there exists some image in \( B \). In other words, each element of domain \( A \) must relate to a unique element in \( B \).

**Example Explanation:**

- Let \( A = \{1, 2, 3\} \) and \( B = \{a, b\} \).

- \( f_1 = \{(1, b), (2, a), (3, b)\} \) is a function.
  - Each element in \( A \) relates to one unique element in \( B \).

- \( f_2 = \{(1, b), (2, b), (1, a)\} \) is not a function.
  - Since 1 in \( A \) is related to both \( a, b \in B \), it doesn't meet the criteria for a function.

- \( f_3 = \{(1, b), (2, a)\} \) is not a function.
  - Since 3 in \( A \) has no image in \( B \), it does not fulfill the requirement for all elements in \( A \) to be associated with an element in \( B \).
Transcribed Image Text:**Understanding Functions and Domains** For the given condition 1, we must prove that the domain of \( f \) is \( A \). This means that for each element of \( A \), there exists some image in \( B \). In other words, each element of domain \( A \) must relate to a unique element in \( B \). **Example Explanation:** - Let \( A = \{1, 2, 3\} \) and \( B = \{a, b\} \). - \( f_1 = \{(1, b), (2, a), (3, b)\} \) is a function. - Each element in \( A \) relates to one unique element in \( B \). - \( f_2 = \{(1, b), (2, b), (1, a)\} \) is not a function. - Since 1 in \( A \) is related to both \( a, b \in B \), it doesn't meet the criteria for a function. - \( f_3 = \{(1, b), (2, a)\} \) is not a function. - Since 3 in \( A \) has no image in \( B \), it does not fulfill the requirement for all elements in \( A \) to be associated with an element in \( B \).
**Functions**

In Section 1.2, we showed that ordered pairs can be defined in terms of sets, and we defined Cartesian products in terms of ordered pairs. In this section, we introduced relations as subsets of Cartesian products. Thus, we can now define functions in a way that depends only on the concept of set. Although this definition is not obviously related to the way we usually work with functions in mathematics, it is satisfying from a theoretical point of view, and computer scientists like it because it is particularly well suited for operating with functions on a computer.

**Definition**

A *function* \( F \) from a set \( A \) to a set \( B \) is a relation with domain \( A \) and co-domain \( B \) that satisfies the following two properties:

1. For every element \( x \) in \( A \), there is an element \( y \) in \( B \) such that \( (x, y) \in F \).
Transcribed Image Text:**Functions** In Section 1.2, we showed that ordered pairs can be defined in terms of sets, and we defined Cartesian products in terms of ordered pairs. In this section, we introduced relations as subsets of Cartesian products. Thus, we can now define functions in a way that depends only on the concept of set. Although this definition is not obviously related to the way we usually work with functions in mathematics, it is satisfying from a theoretical point of view, and computer scientists like it because it is particularly well suited for operating with functions on a computer. **Definition** A *function* \( F \) from a set \( A \) to a set \( B \) is a relation with domain \( A \) and co-domain \( B \) that satisfies the following two properties: 1. For every element \( x \) in \( A \), there is an element \( y \) in \( B \) such that \( (x, y) \in F \).
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