Question

$1)$ Mentioned below defines a relation on $N:$

$\left(i\right)$ $x$ is greater than $y,x,y\in N$

Determine whether the relation is reflexive, symmetric and transitive.

$2)$ Mentioned below defines a relation on $N:$

$\left(ii\right)$ $x+y=10,$
$x,y\in N$

Determine whether the relation is reflexive, symmetric and transitive.

$3)$ Mentioned below defines a relation on $N:$

$\left(iii\right)$ $xy$ is square of an integer $x,y\in N$

Determine whether the relation is reflexive, symmetric and transitive.

$4)$ Mentioned below defines a relation on $N:$

$\left(iv\right)$ $x+4y=10,x,y\in N$

Determine whether the relation is reflexive, symmetric and transitive.

Answer 1

$1)$ Testing Reflexivity

Given relation on $N:$

$x$ is greater than $y,x,y\in N$

If $(x,x)ϵR,$ then $x>x,$ which is not true for any $x\in N$

$\therefore (x,x)\notin R$

So, $R$ is not reflexive.

Testing whether a relation is symmetric

Let $(x,y)\in R$

$\Rightarrow x>y$

$\Rightarrow y<x$ not possible

$\Rightarrow (y,x)\notin R$

Hence, $R$ is not symmetric.

Testing whether a relation is transitive

Let $(x,y)\in R$ and $(y,z)\in R$

$\Rightarrow x>y$ and $y>z$

$\Rightarrow x>z$

$\Rightarrow (x,z)\in R$

Hence, $R$ is transitive.

Final Answer:$R$ is only transitive.

$2)$ Testing whether relation is reflexive.

Given relation on $N:$

$x+y=10,x,y\in N$

If $(x,x)\in R,$ then $x+x=10,\forall x\in N$

Since, $1+1\ne 10$

Therefore, $(1,1)\notin R$

Hence, $R$ is not reflexive

Testing whether relation is symmetric

Let $(x,y)\in R$

$\Rightarrow x+y=10$

$\Rightarrow y+x=10$

$(\because a+b=b+a,a,b\in N)$

$\Rightarrow (y,x)\in R$

Therefore, $R$ is symmetric

Testing whether relation is transitive

Since, $x+y=10,x,y\in N$

$\because (1,9)\in R$ and $(9,1)\in R$

But $(1,1)\notin R,$ as $1+1\ne 10$

Therefore, $R$ is not transitive.

Hence, $R$ is only symmetric.

$3)$ Checking whether the relation is reflexive

Given:

$R=\left\{\right(x,y):xy\text{is a square of an integer}‘x,y\in N\}$

As, $x^2$ is square of an integer for any $x\in N$

Therefore, $(x,x)\in R,\forall x\in N$

Hence, $R$ is reflexive

Checking whether the relation is symmetric

Let $(x,y)\in R$

$\Rightarrow xy$ is a square of an integer

$\Rightarrow yx$ is a square of an integer

$(\because$ multiplication is commutative in $N)$

$\Rightarrow (y,x)\in R$

Hnece, $R$ is symmetric

Checking whether the relation is transitive

Let $(x,y)\in R$ and $(y,z)\in R$

$\Rightarrow xy$ is square of an integer and

$yz$ is square of an integer

Let $xy=m^2$ and $yz=n^2,$ for some $m,n\in Z$

Then$x=\frac{m^2}{y}$ and $z=\frac{n^2}{y}$

$\because xz=\frac{m^2{n}^2}{y^2}$

which is square of an integer

$\therefore (x,z)\in R$

Hence $R$ is transitive

Hence, $R$ is reflexive, symmetric and transitive.

$4)$ Writing the relation R is roster form

Given:

$R:\left\{\right(x,y):x+4y=10,x,y\in N\}$

In roster form: $R=\left\{\right(6,1),(2,2\left)\right\}$

Clearly, $(1,1)\notin R$

Therefore, $R$ is not reflexive

Now, Since, $(6,1)\in R,$ but $(1,6)\notin R$

Therefore, $R$ is not symmetric

Since, there are no elements of the form $(a_1,a_2),(a_2,a_3)$ in the given relation.

Hence, $R$ is transitive.

$\therefore R$ R is only transitive and it is neither reflexive nor symmetric.