Question

R={(a,b):b=a+1} on the set A={1,2,3,4,5,6}
R={(x,y): x-2y-3=0} on the set of all real numbers R
show that these set are whether transitive or reflexive or symmetric


ASAP ...PLSSS.​

Answer 1

Answer:

set 1 st is not reflexive as if u put any value in a from set A it will not return the same value hence time is not reflexive

symmetric : this set is also not symmetric as if we put a= 1 we get b= 2 and this pair becomes (1,2)

now put a=2 we get b= 3 hence it's not symmetric

also it's not transitive

do the same processes to find set 2

Answer 2

Consider the relation $\sf{R:A\to A}$ on the set $\sf{A=\{1,\ 2,\ 3,\ 4,\ 5,\ 6\}}$ defined as,

$\longrightarrow\sf{R=\{(a,\ b):b=a+1\}}$

We see that,

$\longrightarrow\sf{a\neq a+1}$

$\Longrightarrow\sf{(a,\ a)\notin R}$

Therefore, R is not reflexive.

We see that,

$\longrightarrow\sf{b=a+1\quad\implies\quad a=b-1}$

$\Longrightarrow\sf{a\neq b+1}$

$\Longrightarrow\sf{(b,\ a)\notin R}$

Therefore, R is not symmetric.

Assume $\sf{(b,\ c)\in R,}$ such that,

$\longrightarrow\sf{c=b+1}$

$\longrightarrow\sf{c=a+1+1}$

$\longrightarrow\sf{c=a+2}$

$\Longrightarrow\sf{c\neq a+1}$

$\longrightarrow\sf{(a,\ c)\notin R}$

Therefore, R is not transitive.

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Consider the relation $\sf{R:\mathbb{R}\to\mathbb{R}}$ defined as,

$\longrightarrow\sf{R=\{(x,\ y):x-2y-3=0\}}$

We see that if we take $\sf{x=y}$ in such a way that $\sf{(x,\ x)\in R}$ as an assumption,

$\longrightarrow\sf{x-2x-3=0}$

$\longrightarrow\sf{-x-3=0}$

$\longrightarrow\sf{x=-3}$

This gives a unique solution and hence this condition is not applicable to all real numbers.

$\Longrightarrow\sf{(x,\ x)\notin R}$

Therefore, R is not reflexive.

By definition of our relation, we have,

$\longrightarrow\sf{y=\dfrac{x-3}{2}\quad\quad\dots(1)}$

Assume there exists a relation,

$\longrightarrow\sf{y-2x-3=0.}$

in such a way that $\sf{(y,\ x)\in R}$ as an assumption.

From (1),

$\longrightarrow\sf{\dfrac{x-3}{2}-2x-3=0.}$

$\longrightarrow\sf{x=-3}$

This also gives a unique solution and not applicable to all real numbers.

$\Longrightarrow\sf{(y,\ x)\notin R}$

Therefore, R is not symmetric.

Assume there exists a relation of $\sf{y}$ and $\sf{z}$ as,

$\longrightarrow\sf{y-2z-3=0}$

such that $\sf{(y,\ z)\in R}$ as an assumption.

From (1),

$\longrightarrow\sf{\dfrac{x-3}{2}-2z-3=0}$

$\longrightarrow\sf{x-4z-9=0}$

$\Longrightarrow\sf{x-2z-3\neq0}$

$\Longrightarrow\sf{(x,\ z)\notin R}$

Therefore, R is not transitive.