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Chapter 16: Problem 22

Let \(\mathrm{R}=\\{(1,3),(4,2),(2,4),(2,3),(3,1)\\}\) be a relation on the set \(A=\\{1,2,3,4\\} \ldots\) The relation \(\mathrm{R}\) is (a) reflexive (b) transitive (c) not symmetric (d) a function

Short Answer

Expert verified
\(\mathrm{R}\) is not reflexive, not transitive, not symmetric, and not a function.

Step by step solution

01

Understanding Reflexivity

For a relation to be reflexive on a set, every element in the set must be related to itself. For set \(A = \{1, 2, 3, 4\}\), the pairs \((1,1), (2,2), (3,3), (4,4)\) must be in \(\mathrm{R}\). Since none of these pairs are in \(\mathrm{R}\), the relation is not reflexive.
02

Examining Symmetry

A relation is symmetric if for every pair \((a, b)\) in the relation, the pair \((b, a)\) is also in the relation. Checking the pairs, \((1,3)\) is in \(\mathrm{R}\) but \((3,1)\) is also in \(\mathrm{R}\). However, the pair \((4,2)\) has no corresponding \((2,4)\) in \(\mathrm{R}\). Therefore, \(\mathrm{R}\) is not symmetric.
03

Checking Transitivity

A relation is transitive if whenever \((a, b)\) and \((b, c)\) are in the relation, then \((a, c)\) must also be in the relation. In \(\mathrm{R}\), we have \((1,3)\) and \((3,1)\), but \((1,1)\) is not in \(\mathrm{R}\). Similarly, \((2,4)\) and \((4,2)\) would require \((2,2)\) to be in \(\mathrm{R}\), which it is not. Hence, the relation is not transitive.
04

Determining if R is a Function

A relation is a function if every element of the first component in the pairs is related to exactly one element of the second component. The element '2' from set \(A\) appears in pairs \((2,4)\) and \((2,3)\), mapping to two different elements. Thus, \(\mathrm{R}\) is not a function.

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Key Concepts

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

Reflexive Relation
In mathematics, a relation is said to be reflexive if, for every element in the set, it is related to itself. This means that for a set \(A = \{1, 2, 3, 4\}\), the relation should include pairs such as \((1,1), (2,2), (3,3),\) and \((4,4)\).
However, when examining the relation \(\mathrm{R}=\{(1,3), (4,2), (2,4), (2,3), (3,1)\}\), none of these pairs are present. Therefore,
  • None of the elements in the set relate to themselves.
  • Thus, the relation \(\mathrm{R}\) is not reflexive.
Symmetric Relation
A symmetric relation means if \((a, b)\) is in the relation, then \((b, a)\) should also be in it. It's like a two-way street in mathematics, where the relationship between elements goes both directions.
When we look at our relation \(\mathrm{R}\):
  • We can see \((1,3)\) is present, and interestingly, so is \((3,1)\).
  • However, for the pair \((4,2)\), the corresponding \((2,4)\) is missing.
Therefore, because not all pairs have their reverse in \(\mathrm{R}\), this relation is not symmetric.
Transitive Relation
For a relation to be transitive, whenever there are two pairs such that the second element of the first pair is the same as the first element of the second pair, then there should also be a direct pair connecting the first element of the first pair to the second element of the second pair.
In simpler terms, if \((a, b)\) and \((b, c)\) are present, \((a, c)\) must exist for the relation to be transitive.
Looking at \(\mathrm{R}\) again:
  • We see \((1,3)\) and \((3,1)\), and thus \((1,1)\) should be there, but it's not.
  • Also, \((2,4)\) and \((4,2)\) should imply \((2,2)\), but this is missing as well.
Thus, \(\mathrm{R}\) fails the test for transitivity.
Function in Relation
A relation can also be scrutinized to check if it behaves as a function. In mathematical terms, a function means each input (or element from the set) corresponds to exactly one output. Looking at our set \(A=\{1,2,3,4\}\), each element should map to one distinct element in the relation \(\mathrm{R}\).
  • Inspecting \(\mathrm{R}\), the element '2' is connected to both '4' and '3', forming pairs \((2,4)\) and \((2,3)\).
  • This implies '2' has more than one output.
Therefore, this contradictory behavior means \(\mathrm{R}\) does not qualify as a function, since in a true function, each element must pair with only one unique element.

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Most popular questions from this chapter

Let \(\mathrm{R}=\left\\{(x, y): x, y \in N\right.\) and \(\left.x^{2}-4 x y+3 y^{2}=0\right\\}\), where \(N\) is the set of all natural numbers. Then the relation \(R\) is : [Online April 23, 2013] (a) reflexive but neither symmetric nor transitive. (b) symmetric and transitive. (c) reflexive and symmetric, (d) reflexive and transitive.

If \(A=\left\\{x \in z^{+}: x<10\right.\) and \(x\) is a multiple of 3 or 4\(\\}\), where \(z^{+}\)is the set of positive integers, then the total number of symmetric relations on \(A\) is \(\quad\) [Online May 12, 2012] (a) \(2^{5}\) (b) \(2^{15}\) (c) \(2^{10}\) (d) \(2^{20}\)

Consider the following relations: \(\mathrm{R}=\\{(x, y) \mid x, y\) are real numbers and \(x=w y\) for some rational number \(w\\} ; S=\left\\{\left(\frac{m}{n}, \frac{p}{q}\right) \mid m, n, p\right.\) and \(q\) are integers such that \(n, q \neq 0\) and \(q m=p n\\}\). Then (a) Neither \(\mathrm{R}\) nor \(\mathrm{S}\) is an equivalence relation (b) \(\mathrm{S}\) is an equivalence relation but \(\mathrm{R}\) is not an equivalence relation (c) \(\mathrm{R}\) and \(\mathrm{S}\) both are equivalence relations (d) \(\mathrm{R}\) is an equivalence relation but \(\mathrm{S}\) is not an equivalence relation

Let \(R\) be the real line. Consider the following subsets of the plane \(R \times R\) : \(S=\\{(x, y): y=x+1\) and \(0

Let \(\mathrm{N}\) denote the set of all natural numbers. Define two binary relations on \(N\) as \(R_{1}=\\{(x, y) \in N \times N: 2 x+y=10\\}\) and \(R_{2}=\\{(x, y) \in N \times N: x+2 y=10\\}\). Then [Online April 16, 2018] (a) Both \(R_{1}\) and \(R_{2}\) are transitive relations (b) Both \(R_{1}\) and \(R_{2}\) are symmetric relations (c) Range of \(R_{2}\) is \(\\{1,2,3,4\\}\) (d) Range of \(R_{1}\) is \(\\{2,4,8\\}\)
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