Chapter 1: Relations and Functions

Ans: The given relation is: \[\text{R = }\left\{ \left( \text{1, 3} \right)\text{, }\left( \text{2, 6} \right)\text{, }\left( \text{3, 9} \right)\text{, }\left( \text{4, 12} \right) \right\}\]

Since \[\left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{ }...\] and \[\left( \text{14, 14} \right)\notin R\].

We conclude that \[\text{R}\] is not reflexive.

Since \[\left( \text{1, 3} \right)\in \text{R}\], but \[\left( \text{3, 1} \right)\notin \text{R}\]. since \[\text{3}\left( \text{3} \right)\text{-1}\ne \text{0}\]

We conclude that \[\text{R}\] is not symmetric.

Since \[\left( \text{1, 3} \right)\] and \[\left( \text{3, 9} \right)\in \text{R}\], but\[\left( \text{1, 9} \right)\notin \text{R}\text{. }\left[ \text{3}\left( \text{1} \right)\text{-9}\ne \text{0} \right]\].

We conclude that \[\text{R}\] is not transitive.

Therefore, the relation \[\text{R}\] is not reflexive, symmetric or transitive.

1. Relation $\text{R}$ in the set \[\text{N}\] of natural numbers defined as \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: y = x + 5}\] and \[\text{x<4 }\!\!\}\!\!\text{ }\]

Ans: The given relation is: \[\text{R = }\left\{ \left( \text{1, 6} \right)\text{, }\left( \text{2, 7} \right)\text{, }\left( \text{3, 8} \right) \right\}\].

Since \[\left( \text{1, 1} \right)\notin \text{R}\].

We conclude that \[\text{R}\] is not reflexive.

Since \[\left( \text{1, 6} \right)\in \text{R}\] but \[\left( \text{6, 1} \right)\notin \text{R}\].

We conclude that \[\text{R}\] is not symmetric.

In the given relation \[\text{R}\] there is not any ordered pair such that \[\left( \text{x, y} \right)\] and \[\left( \text{y, z} \right)\] both \[\in \text{R}\], therefore we can say that \[\left( \text{x, z} \right)\] cannot belong to \[\text{R}\].

Therefore \[\text{R}\] is not transitive.

Hence, the given relation \[\text{R}\] is not reflexive, symmetric or transitive.

1. Relation $\text{R}$in the set \[\text{A =  }\!\!\{\!\!\text{ 1, 2, 3, 4, 5, 6 }\!\!\}\!\!\text{ }\] as \[\text{R=  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: y}\]is divisible by \[\text{x }\!\!\}\!\!\text{ }\]

Ans: The given relation is \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{ : y}\] is divisible by \[\text{x }\!\!\}\!\!\text{ }\]

As we know that any number except \[\text{0}\] is divisible by itself, therefore \[\left( \text{x, x} \right)\in \text{R}\].

We conclude that \[\text{R}\] is reflexive.

Since \[\left( \text{2, 4} \right)\in \text{R}\] (because 4 is divisible by 2, but \[\left( \text{4, 2} \right)\notin \text{R}\] (since 2 is not divisible by 4.

We conclude that \[\text{R}\] is not symmetric.

Assuming that \[\left( \text{x, y} \right)\] and \[\left( \text{y, z} \right)\in \text{R}\], \[\text{y}\] is divisible by \[\text{x}\] and \[\text{z}\] is divisible by \[\text{y}\]. Hence \[\text{z}\] is divisible by \[\text{x}\Rightarrow \left( \text{x, z} \right)\in \text{R}\].

We conclude that \[\text{R}\] is transitive.

Hence, the given relation \[\text{R}\] is reflexive and transitive but it is not symmetric.

1. Relation \[\text{R}\] in the set \[\text{Z}\] of all integers defined as \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x - y }\!\!\}\!\!\text{ }\] is as integer

Ans: The given relation is \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x - y}\] is an integer\[\text{ }\!\!\}\!\!\text{ }\]

If \[\text{x}\in \text{Z, }\left( \text{x, x} \right)\in \text{R}\] because \[\text{x-x = 0}\] is an integer.

Hence, we conclude that \[\text{R}\] is reflexive.

For \[\text{x, y}\in \text{Z}\], if \[\left( \text{x, y} \right)\in \text{R}\], then \[\text{x - y}\] is an integer and therefore \[\left( \text{y-x} \right)\] is also an integer.

Therefore, we conclude that \[\left( \text{y, x} \right)\in \text{R}\]and hence \[\text{R}\] is symmetric.

Assuming that \[\left( \text{x, y} \right)\] and \[\left( \text{y, z} \right)\in \text{R}\], where \[\text{x, y, z}\in \text{Z}\].

We can say that \[\left( \text{x-y} \right)\] and \[\left( \text{y-z} \right)\] are integers.

so, \[\left( \text{x, z} \right)\in \text{R}\] 

Hence, we conclude that \[\text{R}\] is transitive.

Therefore the given relation \[\text{R}\] is reflexive, symmetric, and transitive.

1. Relation \[\text{R}\] in the set \[\text{A}\] of human beings in a town at a particular time given by

1. The relation is: \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x and y}\] work at the same place\[\}\]

Ans: The given relation is: \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}\] and \[\text{y}\] work at the same place\[\text{ }\!\!\}\!\!\text{ }\]

This implies that \[\left( \text{x, x} \right)\in \text{R}\].

Hence, we conclude that \[\text{R}\] is reflexive.

Now, \[\left( \text{x, y} \right)\in \text{R}\], then \[\text{x}\] and \[\text{y}\] work at the same place, which means \[\text{y}\] and \[\text{x}\] also work at the same place. Therefore, \[\left( \text{y, x} \right)\in \text{R}\].

Hence, we conclude that \[\text{R}\] is symmetric.

Let us assume that \[\left( \text{x, y} \right)\text{, }\left( \text{y, z} \right)\in \text{R}\]. 

Then, we can say that \[\text{x}\] and \[\text{y}\] work at the same place and \[\text{y}\] and \[\text{z}\] work at the same place. Which means that \[\text{x}\] and \[\text{z}\] also work at the same place.

Therefore, \[\left( \text{x, z} \right)\in \text{R}\].

Hence, we conclude that \[\text{R}\] is transitive.

Therefore, the given relation \[\text{R}\] is reflexive, symmetric and transitive.

1. The relation is: \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x and y}\] live in the same locality\[\text{ }\!\!\}\!\!\text{ }\]

Ans: The given relation is \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x and y}\] live in the same locality\[\text{ }\!\!\}\!\!\text{ }\]

Since, \[\left( \text{x, x} \right)\in \text{R}\]. 

Therefore, we conclude that \[\text{R}\] is reflexive.

Since \[\left( \text{x, y} \right)\in \text{R}\], \[\text{x}\] and \[\text{y}\] live in the same locality. Therefore, \[\text{y}\] and \[\text{x}\] also live in the same locality, so, \[\left( \text{y, x} \right)\in \text{R}\].

Hence, \[\text{R}\] is symmetric.

Let \[\left( \text{x, y} \right)\in \text{R}\] and \[\left( \text{y, z} \right)\in \text{R}\]. Hence \[\text{x}\] and \[\text{y}\] live in the same locality and \[\text{y}\] and \[\text{z}\] also live in the same locality. Which means that \[\text{x}\] and \[\text{z}\] also live in the same locality.

Therefore, \[\left( \text{x, z} \right)\in \text{R}\].

Hence, we conclude that \[\text{R}\] is transitive.

Therefore, the given relation \[\text{R}\] is reflexive, symmetric and transitive.

1. \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}\] is exactly \[\text{7}\]cm taller than \[\text{y }\!\!\}\!\!\text{ }\]

Ans: The given relation is: \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}\] is exactly \[\text{7}\] cm taller than \[\text{y }\!\!\}\!\!\text{ }\]

Since, \[\left( \text{x, x} \right)\notin \text{R}\]. 

Therefore, we conclude that \[\text{R}\] is not reflexive.

Let \[\left( \text{x, y} \right)\in \text{R}\] , Since \[\text{x}\] is exactly \[\text{7}\] cm taller than \[\text{y}\], therefore \[\text{y}\] is obviously not taller than \[\text{x}\], so, \[\left( \text{y, x} \right)\notin \text{R}\].

Hence, \[\text{R}\] is not symmetric.

Assuming that \[\left( \text{x, y} \right)\text{, }\left( \text{y, z} \right)\in \text{R}\], we can say that \[\text{x}\] is exactly \[\text{7}\] cm taller than \[\text{y}\]and \[\text{y}\] is exactly \[\text{7}\] cm taller than \[\text{z}\]. Which means that \[\text{x}\] is exactly \[\text{14}\] cm taller than \[\text{z}\]. So, \[\left( \text{x, z} \right)\notin \text{R}\]. 

Hence, \[\text{R}\] is not transitive.

Therefore, the given relation \[\text{R}\] is not reflexive, symmetric or transitive.

1. \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}\] is wife of \[\text{y }\!\!\}\!\!\text{ }\]

Ans: The given relation is: \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}\] is the wife of \[\text{y }\!\!\}\!\!\text{ }\].

Since, \[\left( \text{x, x} \right)\notin \text{R}\] 

Therefore, we conclude that \[\text{R}\] is not reflexive.

Let \[\left( \text{x, y} \right)\in \text{R}\] , Since \[\text{x}\] is the wife of \[\text{y}\], therefore \[\text{y}\] is obviously not the wife of \[\text{x}\], so, \[\left( \text{y, x} \right)\notin \text{R}\].

Hence, \[\text{R}\] is not symmetric.

Assuming that \[\left( \text{x, y} \right)\text{, }\left( \text{y, z} \right)\in \text{R}\], we can say that \[\text{x}\] is the wife of \[\text{y}\]and \[\text{y}\] is the wife of \[\text{z}\], which is not possible. So, \[\left( \text{x, z} \right)\notin \text{R}\]. 

Hence, \[\text{R}\] is not transitive.

Therefore the given relation \[\text{R}\]is not reflexive, symmetric or transitive.

1. \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}\] is father of \[\text{y }\!\!\}\!\!\text{ }\] 

Ans: The given relation is: \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}\] is the father of \[\text{y }\!\!\}\!\!\text{ }\]

Since, \[\left( \text{x, x} \right)\notin \text{R}\] 

Therefore, we conclude that \[\text{R}\] is not reflexive.

Let \[\left( \text{x, y} \right)\in \text{R}\] , Since \[\text{x}\] is the father of \[\text{y}\], therefore \[\text{y}\] is obviously not the father of \[\text{x}\], so, \[\left( \text{y, x} \right)\notin \text{R}\].

Hence, \[\text{R}\] is not symmetric.

Assuming that \[\left( \text{x, y} \right)\text{, }\left( \text{y, z} \right)\in \text{R}\], we can say that \[\text{x}\] is the father of \[\text{y}\]and \[\text{y}\] is the father of \[\text{z}\], then \[\text{x}\] is not the father of  \[\text{z}\]. So, \[\left( \text{x, z} \right)\notin \text{R}\]. 

Hence, \[\text{R}\] is not transitive.

Therefore the given relation \[\text{R}\]is not reflexive, symmetric or transitive.

Ans: The given relation is: \[\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le {{\text{b}}^{\text{2}}} \right\}\]

Since \[\left( \dfrac{\text{1}}{\text{2}}\text{, }\dfrac{\text{1}}{\text{2}} \right)\notin \text{R}\]. (Since $\frac{1}{2}$ is not less than $\frac{1}{4}$)

Therefore, \[\text{R}\] is not reflexive.

Since \[\left( \text{1, 4} \right)\in \text{R}\] as \[\text{1}{{\text{4}}^{\text{2}}}\], but \[\left( \text{4, 1} \right)\notin \text{R}\] as \[{{\text{4}}^{2}}\] is not less than \[{{\text{1}}^{2}}\].

Therefore \[\text{R}\] is not symmetric.

Assuming that \[\left( \text{3, 2} \right)\text{, }\left( \text{2, 1}\text{.5} \right)\in \text{R}\], so, \[\text{3}{{\text{2}}^{\text{2}}}\text{=4}\] and \[\text{2}{{\left( \text{1}\text{.5} \right)}^{\text{2}}}\text{=2}\text{.25}\] but \[\text{3}{{\left( \text{1}\text{.5} \right)}^{\text{2}}}\text{=2}\text{.25}\].

Hence, \[\text{R}\] is not transitive.

Therefore, the given relation \[\text{R}\] is neither reflexive, nor symmetric, nor transitive.

Ans: The given relation is \[\text{R = }\left\{ \left( \text{a, b} \right)\text{: b=a+1} \right\}\] defined in the set \[\text{A=}\left\{ \text{1, 2, 3, 4, 5, 6} \right\}\].

So, \[\text{R=}\left\{ \left( \text{1, 2} \right)\text{, }\left( \text{2, 3} \right)\text{, }\left( \text{3, 4} \right)\text{, }\left( \text{4, 5} \right)\text{, }\left( \text{5, 6} \right) \right\}\]

Since, \[\left( \text{a, a} \right)\notin \text{R,a}\in \text{A}\].

\[\left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{4, 4} \right)\text{, }\left( \text{5, 5} \right)\text{, }\left( \text{6, 6} \right)\notin \text{R}\]

Therefore, \[\text{R}\] is not reflexive

Since, \[\left( \text{1, 2} \right)\in \text{R}\], but \[\left( \text{2, 1} \right)\notin \text{R}\].

Therefore \[\text{R}\] is not symmetric.

Since \[\left( \text{1, 2} \right)\text{, }\left( \text{2, 3} \right)\in \text{R}\], but \[\left( \text{1, 3} \right)\notin \text{R}\].

Hence, \[\text{R}\] is not transitive.

Therefore, the given relation \[\text{R}\] is neither reflexive, nor symmetric, nor transitive

Ans: The given relation is \[\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le \text{b} \right\}\].

Since, \[\left( \text{a, a} \right)\in R\].

Therefore, \[\text{R}\] is reflexive.

Since, \[\left( \text{2, 4} \right)\in R\] (as \[\text{2,4}\]), but \[\left( \text{4, 2} \right)\notin R\](as \[\text{4,2}\]).

Therefore \[\text{R}\] is not symmetric.

Assuming that \[\left( \text{a, b} \right)\text{, }\left( \text{b, c} \right)\in R\], \[\text{a}\le \text{b}\] and \[b\le c\], therefore, \[a\le c\].

Hence, \[\text{R}\] is transitive.

Therefore, the given relation \[\text{R}\] is reflexive and transitive but not symmetric.

Ans: The given relation is: \[\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le {{\text{b}}^{\text{3}}} \right\}\]

Since \[\left( \dfrac{\text{1}}{\text{2}}\text{, }\dfrac{\text{1}}{8} \right)\notin \text{R}\].  (Since $\frac{1}{2}$ is not less than $\frac{1}{8}$)

Therefore, \[\text{R}\] is not reflexive.

Since \[\left( \text{1, 4} \right)\in \text{R}\] as \[\text{1}{{\text{4}}^{\text{3}}}\], but \[\left( \text{4, 1} \right)\notin \text{R}\] as \[\text{4}\] is not less than \[{{\text{1}}^{\text{3}}}\].

Therefore \[\text{R}\] is not symmetric.

Assuming that \[\left( \text{3, }\dfrac{\text{3}}{\text{2}} \right)\text{, }\left( \dfrac{\text{3}}{\text{2}}\text{, }\dfrac{\text{6}}{\text{5}} \right)\in \text{R}\], so, \[\text{3}{{\left( \dfrac{\text{3}}{\text{2}} \right)}^{\text{3}}}\] and \[\dfrac{\text{3}}{\text{2}}\text{}{{\left( \dfrac{\text{6}}{\text{5}} \right)}^{\text{3}}}\] but \[\text{3}{{\left( \dfrac{\text{6}}{\text{5}} \right)}^{\text{3}}}\notin R\].

Hence, \[\text{R}\] is not transitive.

Therefore, the given relation \[\text{R}\] is neither reflexive, nor symmetric, nor transitive.

Ans: The given relation is \[\text{R = }\left\{ \left( \text{1, 2} \right)\text{, }\left( \text{2, 1} \right) \right\}\] on the set \[\text{A=}\left\{ \text{1, 2, 3} \right\}\].

Since \[\left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\notin \text{R}\]

Therefore, \[\text{R}\] is not reflexive.

Since, \[\left( \text{1, 2} \right)\in R\] and \[\left( \text{2, 1} \right)\in R\].

Therefore \[\text{R}\] is symmetric.

Since, \[\left( \text{1, 2} \right)\in R\] and \[\left( \text{2, 1} \right)\in R\], but \[\left( \text{1, 1} \right)\notin \text{R}\].

Hence, \[\text{R}\] is not transitive.

Therefore, the given relation \[\text{R}\] is symmetric but neither reflexive nor transitive.

Ans: The given relation is: \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{x,}\,\,\text{y} \right)\text{: x}\] and \[\text{y}\] have the same number of pages\[\text{ }\!\!\}\!\!\text{ }\]

Since  \[\left( \text{x, x} \right)\in \text{R}\] as \[\text{x}\] and \[\text{x}\] have the same number of pages.

Therefore, \[\text{R}\] is reflexive.

Let \[\left( \text{x, y} \right)\in R\], so \[\text{x}\] and \[\text{y}\] have the same number of pages, \[\left( \text{y, x} \right)\in \text{R}\] therefore \[\text{y}\] and \[\text{x}\] will also have the same number of pages.

Therefore \[\text{R}\] is symmetric.

Assuming \[\left( \text{x, y} \right)\in R\] and \[\left( \text{y, z} \right)\in R\]. \[\text{x}\] and \[\text{y}\] have the same number of pages and \[\text{y}\] and \[\text{z}\] also have the same number of pages. Therefore, \[\text{x}\] and \[\text{z}\] will also have the same number of pages. So, \[\left( \text{x, z} \right)\in R\].

Hence, \[\text{R}\] is transitive.

Therefore, the given relation \[\text{R}\] is an equivalence relation.

Ans: Let \[\text{a}\in \text{A}\],

So, \[\left| \text{a-a} \right|\text{ = 0}\]  (which is an even number).

Therefore, \[\text{R}\] is reflexive.

Let \[\left( \text{a, b} \right)\in \text{R}\],

Now, \[\left| \text{a-b} \right|\] is even,

Hence \[\left| \text{a-b} \right|\] and \[\left| \text{b-a} \right|\] are both even

Therefore, \[\left( \text{b, a} \right)\in \text{R}\] 

Therefore \[\text{R}\] is symmetric

Let \[\left( \text{a, b} \right)\in \text{R}\]and \[\left( \text{b, c} \right)\in \text{R}\],

\[\Rightarrow \left| \text{a-b} \right|\] is even and \[\left| \text{b-c} \right|\] is even

\[\Rightarrow \left| \text{a-c} \right|\] is even.

\[\Rightarrow \left( \text{a, c} \right)\in \text{R}\] 

Therefore, \[\text{R}\] is transitive.

Therefore, the given relation \[\text{R}\] is an equivalence relation.

All the elements of the set \[\left\{ \text{1, }\!\!~\!\!\text{ 3, 5} \right\}\] are all odd. Hence, the modulus of the difference of any two elements will be an even number. So, all the elements of this set are related to each other.

All elements of \[\left\{ \text{2, 4} \right\}\] are even while all the elements of \[\left\{ \text{1, }\!\!~\!\!\text{ 3, 5} \right\}\] are odd so no element of \[\left\{ \text{1, }\!\!~\!\!\text{ 3, 5} \right\}\] can be related to any element of\[\left\{ 2,\text{ }4 \right\}\].

Therefore, the absolute value of the difference between the two elements (from each of these two subsets) will not be an even value.

Ans: The given set \[\text{A =  }\!\!\{\!\!\text{ x}\in \text{Z: 0}\le \text{x}\le \text{12 }\!\!\}\!\!\text{  = }\left\{ \text{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} \right\}\] 

The given relation is: \[\text{R =  }\!\!\{\!\!\text{ }\left( \text{a, b} \right)\text{ : }\left| \text{a -- b} \right|\] is a multiple of \[\text{4 }\!\!\}\!\!\text{ }\].

Let \[a\in A\],

\[\left( \text{a, a} \right)\in R\] as \[\left| \text{a-a} \right|\text{=0}\] is a multiple of \[\text{4}\].

Therefore, \[\text{R}\] is reflexive.

Let, \[\left( \text{a, b} \right)\in \text{R}\Rightarrow \left| \text{a-b} \right|\] is a multiple of \[\text{4}\].

\[\Rightarrow \left| \text{-}\left( \text{a-b} \right) \right|\text{=}\left| \text{b-a} \right|\] is a multiple of \[\text{4}\].

\[\Rightarrow \left( \text{b, a} \right)\in \text{R}\]

Therefore \[\text{R}\] is symmetric.

\[\left( \text{a, b} \right)\text{, }\left( \text{b, c} \right)\in R\].

\[\Rightarrow \left| \text{a-b} \right|\] is a multiple of \[\text{4}\] and \[\left| \text{b-c} \right|\] is a multiple of \[\text{4}\].

\[\Rightarrow \left( \text{a-b} \right)\] is a multiple of \[\text{4}\] and \[\left( \text{b-c} \right)\] is a multiple of \[\text{4}\].

\[\Rightarrow \left( \text{a-c} \right)\text{=}\left( \text{a-b} \right)\text{+}\left( \text{b-c} \right)\] is a multiple of \[\text{4}\].

\[\Rightarrow \left| \text{a-c} \right|\] is a multiple of \[\text{4}\].

\[\Rightarrow \left( \text{a, c} \right)\in \text{R}\]

Therefore, \[\text{R}\] is transitive.

Therefore, the given relation \[\text{R}\] is an equivalence relation.

The set of elements related to 1 is \[\left\{ \text{1, }\!\!~\!\!\text{ 5, 9} \right\}\] 

1. \[\text{ }\!\!~\!\!\text{ R = }\left\{ \left( \text{a, b} \right)\text{ : a = b} \right\}\] 

Ans: The given relation is: \[\text{R = }\left\{ \left( \text{a, b} \right)\text{ : a = b} \right\}\].

\[a\in A,\left( \text{a, a} \right)\in R\], since \[\text{a = a}\].

Therefore, \[\text{R}\] is reflexive.

Let \[\left( \text{a, b} \right)\in \text{R}\Rightarrow \text{a=b}\].

\[\Rightarrow b\text{=a}\Rightarrow \left( \text{b, a} \right)\in R\]

Therefore \[\text{R}\] is symmetric.

\[\left( \text{a, b} \right)\text{, }\left( \text{b, c} \right)\in R\]

\[\Rightarrow a\text{=b}\] and \[b\text{=c}\]

\[\Rightarrow a\text{=c}\]

\[\Rightarrow \left( \text{a, c} \right)\in R\]

Therefore, \[\text{R}\] is transitive.

Therefore, the given relation \[\text{R}\] is an equivalence relation.

The set of elements related to \[\text{1}\] is \[\left\{ \text{1} \right\}\]. 

Ans: Let us assume the relation \[\text{R= }\left\{ \left( \text{5, 6} \right)\text{, }\left( \text{6, 5} \right) \right\}\] in set \[\text{A= }\left\{ \text{5, 6, 7} \right\}\].

So, the relation \[\text{R}\] is not reflexive as \[\left( \text{5, 5} \right)\text{, }\left( \text{6, 6} \right)\text{, }\left( \text{7, 7} \right)\notin \text{R}\].

The relation \[\text{R}\] is symmetric as \[\left( \text{5, 6} \right)\in R\] and \[\left( \text{6, 5} \right)\in R\].

The relation \[\text{R}\] is not transitive as \[\left( \text{5, 6} \right)\text{, }\left( \text{6, 5} \right)\in R\] , but \[\left( \text{5, 5} \right)\notin \text{R}\].

Therefore, the given relation \[\text{R}\] is symmetric but not reflexive or transitive.

1. Transitive but neither reflexive nor symmetric.

Ans: Let us assume the relation \[\text{R = }\left\{ \left( \text{a, b} \right)\text{ : a  b} \right\}\] 

So, the relation \[\text{R}\] is not reflexive because for \[a\in R\], \[\left( \text{a, a} \right)\notin \text{R}\] since a cannot be strictly less than itself.

Let \[\left( {1,2} \right) \in R\left( {as1 < 2} \right)\]

Since \[\text{2}\] is not less than \[\text{1}\], \[\left( \text{2, 1} \right)\notin \text{R}\]. 

Therefore \[\text{R}\] is not symmetric.

Let \[\left( \text{a, b} \right)\text{, }\left( \text{b, c} \right)\in R\].

\[\Rightarrow \left( \text{a, c} \right)\in R\]

Therefore, \[\text{R}\] is transitive.

So, the relation \[\text{R}\] is transitive but not reflexive and symmetric.

1. Reflexive and symmetric but not transitive.

Ans: Let us assume the relation \[R=\left\{ \left( \text{4, 4} \right)\text{, }\left( \text{6, 6} \right)\text{, }\left( \text{8, 8} \right),\text{ }\left( \text{4, 6} \right),\text{ }\left( \text{6, 4} \right),\text{ }\left( \text{6, 8} \right),\text{ }\left( \text{8, 6} \right) \right\}\] in set \[\text{A= }\left\{ \text{4, 6, 8} \right\}\]. 

The relation \[\text{R}\] is reflexive since for \[a\in R\], \[\left( \text{a, a} \right)\in R\]. 

The relation \[\text{R}\] is symmetric since \[\left( \text{a, b} \right)\in R\Rightarrow \left( \text{b, a} \right)\in R\] for \[a,b\in R\]. 

The relation \[\text{R}\] is not transitive since \[\left( \text{4, 6} \right)\text{, }\left( \text{6, 8} \right)\in R\], but \[\left( \text{4, 8} \right)\notin \text{R}\].

Therefore the relation \[\text{R}\] is reflexive and symmetric but not transitive.

1. Reflexive and transitive but not symmetric.

Ans: Let us take the relation \[\text{ }\!\!~\!\!\text{ R = }\left\{ \left( \text{a, b} \right)\text{ : }{{\text{a}}^{\text{3}}}\ge {{\text{b}}^{\text{3}}} \right\}\].

Since \[\left( \text{a, b} \right)\in \text{R}\].

Therefore \[\text{R}\] is reflexive.

Since \[\left( \text{2, 1} \right)\in \text{R}\], but \[\left( \text{1, 2} \right)\notin \text{R}\],

Therefore \[\text{R}\] is not symmetric.

Let \[\left( {a,b} \right),\left( {b,c} \right) \in R\]

\[\Rightarrow {{\text{a}}^{\text{3}}}\ge {{\text{b}}^{\text{3}}}\] and \[{{\text{b}}^{\text{3}}}\ge {{\text{c}}^{\text{3}}}\]

\[\Rightarrow {{\text{a}}^{\text{3}}}\ge {{\text{c}}^{\text{3}}}\]

\[\Rightarrow \left( \text{a, c} \right)\in \text{R}\]

Therefore \[\text{R}\] is transitive.

Therefore the relation \[\text{R}\] is reflexive and transitive but not symmetric.

1. Symmetric and transitive but not reflexive.

Ans: Let us take a relation \[\text{R=}\left\{ \left( \text{-5, -6} \right)\text{, }\left( \text{-6, -5} \right)\text{, }\left( \text{-5, -5} \right) \right\}\] in set \[\text{A=}\left\{ \text{-5, -6} \right\}\].

The relation \[\text{R}\] is not reflexive as \[\left( \text{-6, -6} \right)\notin \text{R}\].

Since \[\left( \text{-5, -6} \right)\in \text{R}\] and \[\left( \text{-6, -5} \right)\in \text{R}\].

Therefore \[\text{R}\] is symmetric.

Since \[\left( \text{-5, -6} \right)\text{, }\left( \text{-6, -5} \right)\in \text{R}\] and \[\left( \text{-5, -5} \right)\in \text{R}\].

Therefore \[\text{R}\] is transitive.

Therefore the relation \[\text{R}\] is symmetric and transitive but not reflexive.