91Ó°ÊÓ

Exercises \(34-38\) deal with these relations on the set of real numbers: \(\begin{aligned} R_{1}=&\left\\{(a, b) \in \mathbf{R}^{2} | a>b\right\\}, \text { the greater than relation, } \\ R_{2}=&\left\\{(a, b) \in \mathbf{R}^{2} | a \geq b\right\\}, \text { the greater than or equal to relation, } \end{aligned}\) \(\begin{aligned} R_{3}=\left\\{(a, b) \in \mathbf{R}^{2} | a < b\right\\}, \text { the less than relation, } \\ R_{4}= \left\\{(a, b) \in \mathbf{R}^{2} | a \leq b\right\\}, \text { the less than or equal to relation, } \end{aligned}\) \(R_{5}=\left\\{(a, b) \in \mathbf{R}^{2} | a=b\right\\},\) the equal to relation, \(R_{6}=\left\\{(a, b) \in \mathbf{R}^{2} | a \neq b\right\\},\) the unequal to relation. Find $$ \begin{array}{llll}{\text { a) } R_{2} \circ R_{1}} & {\text { b) } R_{2} \circ R_{2}} \\ {\text { c) } R_{3} \circ R_{5}} & {\text { d) } R_{4} \circ R_{1}} \\ {\text { e) } R_{5} \circ R_{3}} & {\text { f) } R_{3} \circ R_{6}} \\\ {\text { g) } R_{4} \circ R_{6}} & {\text { h) } R_{6} \circ R_{6}}\end{array} $$

Step by step solution

01

Understand Relation Composition

To find the composition of two relations, say \(R_i \circ R_j\), one needs to identify pairs \(a, b\) and \(b, c\) such that \(a R_i b\) and \(b R_j c\) are true. This leads to \(a R_{i+j} c\).

02

Find \(R_{2} \circ R_{1}\)

For \( R_{2} \circ R_{1} \), identify pairs \((a, b) \in R_1\) and \((b, c) \in R_2\). This means \( a > b \) and \( b \geq c \). Combining these, we get \( a > c \). Therefore, \(R_{2} \circ R_{1} = R_{1}\).

03

Find \(R_{2} \circ R_{2}\)

For \( R_{2} \circ R_{2} \), identify pairs \((a, b) \in R_2\) and \((b, c) \in R_2\). This means \( a \geq b \) and \( b \geq c \). Combining these, we get \( a \geq c \). Therefore, \(R_{2} \circ R_{2} = R_{2}\).

04

Find \(R_{3} \circ R_{5}\)

For \( R_{3} \circ R_{5} \), identify pairs \((a, b) \in R_5\) and \((b, c) \in R_3\). This means \( a = b \) and \( b < c \). Combining these, we get \( a < c \). Therefore, \(R_{3} \circ R_{5} = R_{3}\).

05

Find \(R_{4} \circ R_{1}\)

For \( R_{4} \circ R_{1} \), identify pairs \((a, b) \in R_1\) and \((b, c) \in R_4\). This means \( a > b \) and \( b \leq c \). Combining these, we get \( a > c \). Therefore, \(R_{4} \circ R_{1} = R_{1}\).

06

Find \(R_{5} \circ R_{3}\)

For \( R_{5} \circ R_{3} \), identify pairs \((a, b) \in R_3\) and \((b, c) \in R_5\). This means \( a < b \) and \( b = c \). Combining these, we get \( a < c \). Therefore, \(R_{5} \circ R_{3} = R_{3}\).

07

Find \(R_{3} \circ R_{6}\)

For \( R_{3} \circ R_{6} \), identify pairs \((a, b) \in R_6\) and \((b, c) \in R_3\). This means \( a eq b \) and \( b < c \). Combining these, we get \( a < c \). Therefore, \(R_{3} \circ R_{6} = R_{3}\).

08

Find \(R_{4} \circ R_{6}\)

For \( R_{4} \circ R_{6} \), identify pairs \((a, b) \in R_6\) and \((b, c) \in R_4\). This means \( a eq b \) and \( b \leq c \). This composition does not lead to any specific pattern, so \(R_{4} \circ R_{6} = R \backslash R_{5}\), the relation that includes all pairs except for the equality.

09

Find \(R_{6} \circ R_{6}\)

For \( R_{6} \circ R_{6} \), identify pairs \((a, b) \in R_6\) and \((b, c) \in R_6\). This means \( a eq b \) and \( b eq c \). Since \( a \) and \( c \) can still be different, we get \( a eq c \). Therefore, \(R_{6} \circ R_{6} = R_{6}\).

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Key Concepts

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

Greater Than Relation

The greater than relation, denoted as \( R_1 \), involves pairs \((a, b)\) in real numbers such that \( a > b \). This relation is straightforward: it simply means that the first number is larger than the second.

For example:

• In the pair \((5, 3)\), we say 5 is greater than 3, so \( (5, 3) \in R_1 \).
• However, in the pair \((2, 4)\), since 2 is not greater than 4, \( (2, 4) otin R_1 \).

Composition of relations applies here too. If you have two conditions, \( a > b \) and \( b > c \), you can infer \( a > c \). This is useful in solving problems where you need to find a combined relation.

Less Than Relation

The less than relation, represented as \( R_3 \), consists of pairs \((a, b)\) where \( a < b \). This is essentially the opposite of the greater than relation.

Consider these examples:

• In the pair \((2, 5)\), we see 2 is less than 5, so \( (2, 5) \in R_3 \).
• For the pair \((7, 1)\), 7 is not less than 1, hence \( (7, 1) otin R_3 \).

When dealing with the composition of this relation with another, you follow similar logic. If you know \( a < b \) and \( b < c \), then it is clear that \( a < c \).

Equal To Relation

The equal to relation, denoted as \( R_5 \), includes pairs \((a, b)\) where \( a = b \). This relation is all about equality, meaning the first number is exactly the same as the second.

Here are a few illustrations:

• In the pair \((4, 4)\), 4 is equal to 4, so \( (4, 4) \in R_5 \).
• Conversely, in the pair \((3, 7)\), 3 is not equal to 7, thus \( (3, 7) otin R_5 \).

In terms of relation composition, if we combine \( a = b \) and \( b = c \), we get \( a = c \) since they maintain equality through the chain. This kind of relation is fully transitive.