Question

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as

R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

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Answer 1

Answer:

Let A = {1, 2, 3, 4, 5, 6}.

A relation R is defined on set A as:

R = {(a, b): b = a + 1}

∴R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

We can find (a, a) ∉ R, where a ∈ A.

For instance,

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R

∴R is not reflexive.

It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R.

∴R is not symmetric.

Now, (1, 2), (2, 3) ∈ R

But,

(1, 3) ∉ R

∴R is not transitive

Hence, R is neither reflexive, nor symmetric, nor transitive.

Answer 2

$\large{\underline{\rm{\green{\bf{Question:-}}}}}$

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

$\large{\underline{\rm{\green{\bf{Given:-}}}}}$

R = {(a, b): b = a + 1}

$\large{\underline{\rm{\green{\bf{To \: Find:-}}}}}$

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6}

$\large{\underline{\rm{\green{\bf{Solution:-}}}}}$

Given that, R = {(a, b): b = a + 1}

Now, for this relation we have to check whether it is reflexive, transitive and symmetric reflexivity:

Reflexive:

Let a be an arbitrary element of R.

Then, a = a + 1 cannot be true for all a ∈ A.

$\implies \sf (a, \: a) \: \notin R$

So, R is not reflexive on A.

Symmetry:

Let (a, b) ∈ R

$\implies \sf b=a+1$

$\implies \sf -a=-b+1$

$\implies \sf a=b-1$

Thus, (b, a) ∉ R

So, R is not symmetric on A.

Transitivity:

Let (1, 2) and (2, 3) ∈ R

$\implies \sf 2=1+1 \: and \: 3$

2 + 1  is true.

But 3 ≠ 1 + 1

$\implies \sf (1, \: 3) \: \notin R$

So, R is not transitive on A.