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

We have the set A = {1, 2, 3, 4, 5, 6}

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x A, xRx.

R is symmetric if for all x, y A, if xRy, then yRx.

R is transitive if for all x, y, z A, if xRy and yRz, then xRz.

We have

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

Every a, b A.

And A = {1, 2, 3, 4, 5, 6}

The relation R on set A can be defined as:

Put a = 1

b = a + 1

b = 1 + 1

b = 2

(a, b) ≡ (1, 2)

Put a = 2

b = 2 + 1

b = 3

(a, b) ≡ (2, 3)

Put a = 3

b = 3 + 1

b = 4

(a, b) ≡ (3, 4)

Put a = 4

b = 4 + 1

b = 5

(a, b) ≡ (4, 5)

Put a = 5

b = 5 + 1

b = 6

(a, b) ≡ (5, 6)

Put a = 6

b = 6 + 1

b = 7

(a, b) ≠ (6, 7) [ 7 A]

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

Check for Reflexivity:

For 1, 2, …, 6 A [ A = {1, 2, 3, 4, 5, 6}]

(1, 1) R

(2, 2) R

(6, 6) R

So, a A, then (a, a) R.

R is not reflexive.

R is not reflexive.

Check for Symmetry:

1, 2 A [ A = {1, 2, 3, 4, 5, 6}]

If (1, 2) R

Then, (2, 1) R

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

So, if (a, b) R, then (b, a) R

a, b A

R is not symmetric.

R is not symmetric.

Check for Transitivity:

1, 2, 3 A

If (1, 2) R and (2, 3) R

Then, (1, 3) R

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

So, if (a, b) R and (b, c) R, then (a, c) R.

a, b, c A

R is not transitive.

R is not transitive.

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