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It is given that

R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (3, 2)}

We know that

(1, 1), (2, 2), (3, 3), (4, 4) ∈ R is reflexive.

We know that

(1, 2) ∈ R but (2, 1) ∈ R

Hence, R is not symmetric.

The same way

(1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∈ R

Hence, R is transitive.

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.

Let there be a set A.

A = {1, 2, 3, 4}

We need to define a relation on A which is reflexive and symmetric but not transitive.

Let there be a set A.

A = {1, 2, 3, 4}

Reflexive relation:

R = {(1, 1), (2, 2), (3, 3), (4, 4)} …(1)

Symmetric relation:

R = {(3, 4), (4, 3)} …(2)

Combine results (1) and (2), we get

R = {(1, 1), (2, 2), (3, 3), (4, 4), (3, 4), (4, 3)}

Check for Transitivity:

If (3, 4) ∈ R and (4, 3) ∈ R

Then, (3, 3) ∈ R

∀ 3, 4 ∈ A [∵ A = {1, 2, 3, 4}]

So eliminate (3, 3) from R, we get

R = {(1, 1), (2, 2), (4, 4), (3, 4), (4, 3)}

Check for Transitivity:

If (4, 3) ∈ R and (3, 4) ∈ R

Then, (4, 4) ∈ R

∀ 3, 4 ∈ A

So, eliminate (4, 4) from R, we get

R = {(1, 1), (2, 2), (3, 4), (4, 3)}

Thus, the relation which is reflexive and symmetric but not transitive is:

R = {(1, 1), (2, 2), (3, 4), (4, 3)}

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.

Let there be a set A.

A = {1, 2, 3, 4}

We need to define a relation on A which is reflexive and transitive but not symmetric.

Let there be a set A.

A = {1, 2, 3, 4}

Reflexive relation:

R = {(1, 1), (2, 2), (3, 3), (4, 4)} …(1)

Transitive relation:

R = {(3, 4), (4, 1), (3, 1)} …(2)

Combine results (1) and (2), we get

R = {(1, 1), (2, 2), (3, 3), (4, 4), (3, 4), (4, 1), (3, 1)}

Check for Symmetry:

If (3, 4) ∈ R

Then, (4, 3) ∉ R

∀ 3, 4 ∈ A [∵ A = {1, 2, 3, 4}]

One example is enough to prove that R is not symmetric.

Thus, the relation which is reflexive and transitive but not symmetric is:

R = {(1, 1), (2, 2), (3, 3), (4, 4), (3, 4), (4, 1), (3, 1)}

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.

Let there be a set A.

A = {1, 2, 3, 4}

We need to define a relation on A which is symmetric and transitive but not reflexive.

It is not possible to define such relation which is symmetric and transitive but not reflexive. As every relation which is symmetric and transitive will use identity ordered pair of the form (x, x) to balance the relation (to make the relation symmetric and transitive). Without such identity pair both, symmetry and transitivity will not be possible.

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.

Let there be a set A.

A = {1, 2, 3, 4}

We need to define a relation which is symmetric but neither reflexive nor transitive.

Let there be a set A.

A = {1, 2, 3, 4}

Symmetric Relation:

{(1, 3), (3, 1)}

This is neither reflexive nor transitive.

∵ (1, 1) ∉ R

(3, 3) ∉ R

Hence, R is not reflexive.

∵ (1, 3) ∈ R and (3, 1) ∈ R

Then, (1, 1) ∉ R

Hence, R is not transitive.

Thus, the relation which is symmetric but neither nor transitive is:

R = {(1, 3), (3, 1)}

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.

Let there be a set A.

A = {1, 2, 3, 4}

We need to define a relation which is transitive but neither reflexive nor symmetric.

Let there be a set A.

A = {1, 2, 3}

Transitive Relation:

R = {(2, 4), (4, 1), (2, 1)}

This is neither reflexive nor symmetric.

∵ (1, 1) ∉ R

(2, 2) ∉ R

(4, 4) ∉ R

Hence, R is not reflexive.

∵ if (2, 4) ∈ R

Then, (4, 2) ∉ R

Hence, R is not symmetric.

Thus, the relation which is transitive but neither reflexive nor symmetric is:

R = {(2, 4), (4, 1), (2, 1)}

It is given that

R = {(a, b): a = b²} for all a, b ∈ N

Reflexivity-

By substituting 2 we get

2 ≠ 2² where 2 is not related to 2

It can be written as

(2, 2) ∉ R

Hence, R is not reflexive.

Symmetry-

By substituting 4 we get

4 = 2²

It can be written as

(4, 2) ∈ R and (2, 4) ∉ R

Hence, R is not symmetric.

Transitive-

Here, we know that (16, 4) ∈ R and (16, 2) ∉ R

Therefore, R is not transitive.

Given is:

R = {(1, 2), (2, 3)} on the set A.

A = {1, 2, 3}

Right now, we have

R = {(1, 2), (2, 3)}

Symmetric Relation:

We know (1, 2) ∈ R

Then, (2, 1) ∈ R

Also, (2, 3) ∈ R

Then, (3, 2) ∈ R

So, add (2, 1) and (3, 2) in R, so that we get

R’ = {(1, 2), (2, 1), (2, 3), (3, 2)}

Transitive Relation:

We need to make the relation R’ transitive.

So, we know (1, 2) ∈ R and (2, 1) ∈ R

Then, (1, 1) ∈ R

Also, (2, 3) ∈ R and (3, 2)

Then, (2, 2) ∈ R

Also, (2, 1) ∈ R and (1, 2) ∈ R

Then, (2, 2) ∈ R

Also, (3, 2) ∈ R and (2, 3) ∈ R

Then, (3, 3) ∈ R

Add (1, 1), (2, 2) and (3, 3) in R’, we get

R’’ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}

Thus, we have got a relation which is reflexive, symmetric and transitive.

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}

The ordered pair added are (1, 1), (2, 2), (3, 3), (3, 2).

We know that

R = {(a, b): a² + b² = 1}

Reflexive-

R is the set of real numbers

If a ∈ R then a² + a² ≠ 1 where a = 2, 3 ….

Therefore, R is not reflexive.

Symmetric-

Consider (a, b) ∈ R where a² + b² = 1

We get

(b, a) ∈ R and (a, b) ∈ R

Hence, R is symmetric.

Transitive-

Consider (a, b) ∈ R and (b, c) ∈ R

We know that

(cos 30^o, sin 30^o) ∈ R and (sin 30^o, cos 30^o) ∈ R

So (cos 30^o, cos 30^o) ∉ R

Therefore, R is not transitive.

Here, R is symmetric but neither reflexive nor transitive.

(i) Reflexive-

d(A, A) < 2 we get (A, A) ∈ R

(ii) Symmetric-

(A, B) ∈ R

We get

d (A, B) < 2 and d (B, A) < 2

So (B, A) ∈ R

(iii) Transitive-

If A (0, 0), B (1.5, 0) and C (3, 0) are the point s we get

d (A, B) = 1.5, d (B, C) = 1.5 and d (A, C) = 3

So we get

d (A, C) < 2 is not true

Therefore, R is reflexive and symmetric but not transitive.

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

x + y = 10, x, y ∈ N

This relation is defined on N (set of Natural Numbers)

The relation can also be defined as

R = {(x, y): x + y = 10} on N

Check for Reflexivity:

∀ x ∈ N

We should have, (x, x) ∈ R

⇒ x + x = 10, which is not true everytime.

Take x = 4.

x + x = 10

⇒ 4 + 4 = 10

⇒ 8 = 10, which is not true.

That is 8 ≠ 10.

So, ∀ x ∈ N, then (x, x) ∉ R

⇒ R is not reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ x + y = 10

Now, replace x by y and y by x. We get

y + x = 10, which is as same as x + y = 10.

⇒ y + x = 10

⇒ (y, x) ∈ R

So, if (x, y) ∈ R, and then (y, x) ∈ R ∀ x, y ∈ N

⇒ R is symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

⇒ x + y = 10 and y + z = 10

⇒ x + z = 10, may or may not be true.

Let us take x = 6, y = 4 and z = 6

x + y = 10

⇒ 6 + 4 = 10

⇒ 10 = 10, which is true.

y + z = 10

⇒ 4 + 6 = 10

⇒ 10 = 10, which is true.

x + z = 10

⇒ 6 + 6 = 10

⇒ 12 = 10, which is not true

That is, 12 ≠ 10

⇒ x + z ≠ 10

⇒ (x, z) ∉ R

So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∉ R

∀ x, y, z ∈ N

⇒ R is not transitive.

Hence, the relation is symmetric but neither reflexive nor transitive.

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

xy is the square of an integer. x, y ∈ N.

This relation is defined on N (set of Natural Numbers)

The relation can also be defined as

R = {(x, y): xy = a², a = √(xy), a ∈ N} on N

Check for Reflexivity:

∀ x ∈ N

We should have, (x, x) ∈ R

⇒ xx = a², where a = √(xx)

⇒ x² = a², where a = √(x²)

which is true every time.

Take x = 1 and y = 4

xy = a²

⇒ 1 × 4 = (√(1 × 4))² [∵ a = √(xy)]

⇒ 4 = (√4)²

⇒ 4 = (2)²

⇒ 4 = 4

So, ∀ x ∈ N, then (x, x) ∈ R

⇒ R is reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ xy = a², where a = √(xy)

Now, replace x by y and y by x. We get

yx = a², which is as same as xy = a²

where a = √(yx)

⇒ yx = a²

⇒ (y, x) ∈ R

So, if (x, y) ∈ R, and then (y, x) ∈ R ∀ x, y ∈ N

⇒ R is symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

⇒ xy = a² and yz = a²

⇒ xz = a², may or may not be true.

Let us take x = 8, y = 2 and z = 50

xy = a², where a = √(xy)

⇒ (8)(2) = (√(8 × 2))²

⇒ 16 = (4)²

⇒ 16 = 16, which is true.

yz = a²

⇒ (2)(50) = (√(2 × 50))²

⇒ 100 = (10)²

⇒ 100 = 100, which is true

xz = a²

⇒ (8)(50) = (√(8 × 50))²

⇒ 400 = (20)²

⇒ 400 = 400

We won’t be able to find a case to show a contradiction.

⇒ xz = a²

⇒ (x, z) ∈ R

So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∈ R

∀ x, y, z ∈ N

⇒ R is transitive.

Hence, the relation is symmetric and transitivity, but not reflexive.

First let us define what range and domain are.

Range: The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually) after we have substituted the domain. In plain English, the definition means: The range is the resulting y-values we get after substituting all the possible x-values.

Domain: The domain of definition of a function is the set of "input" or argument values for which the function is defined. That is, the function provides an "output" or value for each member of the domain.

We have been given that, R is a relation defined on N.

N = set of natural numbers

R = {(x, y): x, y ∈ N, 2x + y = 41}

We have the function as

2x + y = 41

⇒ y = 41 – 2x

As y ∈ N (Natural number)

⇒ 41 – 2x ≥ 1

⇒ –2x ≥ 1 – 41

⇒ –2x ≥ –40

⇒ 2x ≤ 40

⇒ x ≤ 20

So, the domain is first 20 natural numbers.

As, 2x + y = 41

⇒ 2x = 41 – y

⇒ 41 – y/ 2

As x ∈ N (Natural number)

⇒ 41 – y/2 ≥ 1

⇒ 41 – y ≥ 2

⇒ –y ≥ 2 – 41

⇒ –y ≥ –39

⇒ y ≤ 39

So, the range is first 39 natural numbers.

We have relation R defined on set N.

R = {(x, y): x, y ∈ N, 2x + y = 41}

Check for Reflexivity:

∀ x ∈ N

If (x, x) ∈ R

⇒ 2x + x = 41

⇒ 3x = 41

⇒ x = 41/3

⇒ x = 13.67

But, x ≠ 13.67 as x ∈ N.

⇒ (x, x) ∉ R

So, ∀ x ∈ N, then (x, x) ∉ R

⇒ R is not reflexive.

∴ R is not reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ 2x + y = 41 …(i)

Now, replace x by y and y by x, we get

⇒ 2y + x = 41 …(ii)

Take x = 20 and y = 1.

Equation (i) ⇒ 2(20) + 1 = 41

⇒ 40 + 1 = 41

⇒ 41 = 41, holds true.

Equation (ii) ⇒ 2(1) + 20 = 41

⇒ 2 + 20 = 41

⇒ 22 = 41, which is not true as 22 ≠ 41.

⇒ (y, x) ∉ R

So, if (x, y) ∈ R, then (y, x) ∉ R.

∀ x, y ∈ N

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

⇒ 2x – y = 41 and 2y – z = 41

⇒ 2x – z = 41, may be true or not.

Let us sole these to find out.

We have

2x – y = 41 …(iii)

2y – z = 41 …(iv)

Multiply 2 by equation (i), we get

4x – 2y = 82 …(v)

Adding equation (v) and (iv), we get

(4x – 2y) + (2y – z) = 82 + 41

⇒ 4x – z = 123

⇒ 2x + 2x – z = 123

⇒ 2x – z = 123 – 2x

Take x = 40 (as x ∈ N)

⇒ 2x – z = 123 – 2(40)

⇒ 2x – z = 123 – 80

⇒ 2x – z = 43 ≠ 41

⇒ (x, z) ∉ R

So, if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∉ R.

∀ x, y, z ∈ N

⇒ R is not transitive.

∴ R is not transitive.

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

x + 4y = 10, x, y ∈ N

This relation is defined on N (set of Natural Numbers)

The relation can also be defined as

R = {(x, y): 4x + y = 10} on N

Check for Reflexivity:

∀ x ∈ N

We should have, (x, x) ∈ R

⇒ 4x + x = 10, which is obviously not true everytime.

Take x = 4.

4x + x = 10

⇒ 16 + 4 = 10

⇒ 20 = 10, which is not true.

That is 20 ≠ 10.

So, ∀ x ∈ N, then (x, x) ∉ R

⇒ R is not reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ 4x + y = 10

Now, replace x by y and y by x. We get

4y + x = 10, which may or may not be true.

Take x = 1 and y = 6

4x + y = 10

⇒ 4(1) + 6 = 10

⇒ 4 + 6 = 10

⇒ 10 = 10

4y + x = 10

⇒ 4(6) + 1 = 10

⇒ 24 + 1 = 10

⇒ 25 = 10, which is not true.

⇒ 4y + x ≠ 10

⇒ (y, x) ∉ R

So, if (x, y) ∈ R, and then (y, x) ∉ R ∀ x, y ∈ N

⇒ R is not symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

Then, (x, z) ∈ R

We have

4x + y = 10

⇒ y = 10 – 4x

Where x, y ∈ N

So, put x = 1

⇒ y = 10 – 4(1)

⇒ y = 10 – 4

⇒ y = 6

Put x = 2

⇒ y = 10 – 4(2)

⇒ y = 10 – 8

⇒ y = 2

We can’t take y >2, because if we put y = 3

⇒ y = 10 – 4(3)

⇒ y = 10 – 12

⇒ y = –2

But, y ≠ –2 as y ∈ N

So, only ordered pairs possible are

R = {(1, 6), (2, 2)}

This relation R can never be transitive.

Because if (a, b) ∈ R, then (b, c) ∉ R

⇒ R is not reflexive.

Hence, the relation is neither reflexive nor symmetric nor transitive.

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

x > y, x, y ∈ N

This relation is defined on N (set of Natural Numbers)

The relation can also be defined as

R = {(x, y): x > y} on N

Check for Reflexivity:

∀ x ∈ N

We should have, (x, x) ∈ R

⇒ x > x, which is not true.

1 can’t be greater than 1.

2 can’t be greater than 2.

16 can’t be greater than 16.

Similarly, x can’t be greater than x.

So, ∀ x ∈ N, then (x, x) ∉ R

⇒ R is not reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ x > y

Now, replace x by y and y by x. We get

y > x, which may or not be true.

Let us take x = 5 and y = 2.

x > y

⇒ 5 > 2, which is true.

y > x

⇒ 2 > 5, which is not true.

⇒ y > x, is not true as x > y

⇒ (y, x) ∉ R

So, if (x, y) ∈ R, but (y, x) ∉ R ∀ x, y ∈ N

⇒ R is not symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

⇒ x > y and y > z

⇒ x > y > z

⇒ x > z

⇒ (x, z) ∈ R

So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∈ R

∀ x, y, z ∈ N

⇒ R is transitive.

Hence, the relation is transitive but neither reflexive nor symmetric.

Recall that,

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 relation R = {(a, a), (b, c), (a, b)} on A.

A = {a, b, c}

For Transitive:

If (a, b) ∈ R and (b, c) ∈ R

Then, (a, c) ∈ R

∀ a, b, c ∈ A

For Reflexive:

∀ a, b, c ∈ R

Then, (a, a) ∈ R

(b, b) ∈ R

(c, c) ∈ R

We need to add (b, b), (c, c) and (a, c) in R.

We get

R = {(a, a), (b, b), (c, c), (a, b), (b, c), (a, c)}

We have the relation R such that

R = {(1, 2), (1, 1), (2, 3)}

R is defined on set A.

A = {1, 2, 3}

Recall that,

A relation R defined on a set A is called transitive if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R, ∀ a, b, c ∈ A.

For transitive relation:

Note in R,

(1, 2) ∈ R and (2, 3) ∈ R

Then, (1, 3) ∈ R

So, add (1, 3) in R.

R = {(1, 2), (1, 1), (2, 3), (1, 3)}

Now, we can see that R is transitive.

Hence, the ordered pair to be added is (1, 3).

Given,

a ∈ { - 1, 2, 3, 4, 5} and b ∈ {0, 3, 6}, 

To find : the ordered pair (a, b) such that a + b = 5 

then the ordered pair (a, b) such that a + b = 5 are as follows 

(a, b)∈ {( - 1, 6), (2, 3), (5, 0)}

Given the ordered pairs (x,-1) and (5, y) belong to the set,

{(a, b) : b = 2a - 3} 

To find : values of x and y 

solving for first order pair 

⇒ (x,-1) = {(a, b): b = 2a - 3} 

⇒ x = a and b = -1 

If b = - 1 then 2a = - 1 + 3 = 2 

So, a = 1

⇒ x = 1 

Similarly, 

Solving for second order pair.

⇒ (5, y) = {(a, b): b = 2a - 3} 

⇒ a = 5 and y = b 

If a = 5 then b = 2×5 - 3 

So, b = 7 

⇒ y = 7

Given as

The ordered pairs (x, – 1) and (5, y) belong to the set {(a, b): b = 2a – 3}

By solving for first order pair

(x, – 1) = {(a, b): b = 2a – 3}

x = a and -1 = b

On taking b = 2a – 3

If b = – 1 then 2a = – 1 + 3

= 2

a = 2/2

= 1

Therefore, a = 1

Here x = a, x = 1

Similarly, on solving for second order pair

(5, y) = {(a, b): b = 2a – 3}

5 = a and y = b

On taking b = 2a – 3

If a = 5 then b = 2×5 – 3

= 10 – 3

= 7

Therefore, b = 7

Here, y = b, y = 7

Thus, values of x and y are, x = 1 and y = 7

Given: A × B = {(a, b): b = 3a – 2} 

and {(x, -5), (2, y)} Є A × B 

For (x, -5) Є A × B 

b = 3a – 2 

⇒ -5 = 3(x) – 2 

⇒ -5 + 2 = 3x 

⇒ -3 = 3x 

⇒ x = -1 

For (2, y) Є A × B 

b = 3a – 2 

⇒ y = 3(2) – 2

⇒ y = 6 – 2 

⇒ y = 4 

Hence, the value of x = -1 and y = 4

Given A = {2, 3, 4}, B = {1, 3, 7} and R = {(x,y) : x ϵ A, y ϵB and x < y}

According to the condition x < y

⇒ R= A × B

⇒ R={(2,3), (2,7), (3,7), (4,7)}

⇒ R^-1= B × A

⇒ R^-1= { (3,2), (7,2), (7,3), (7,4)}

Given A = {3, 5, 7}, B = {2, 6, 10} and R = {(x, y) : x and y are relatively prime }

According to the condition, x and y are prime numbers.

⇒ R= A × B

⇒ R= {(3,2), (5,2), (7,2)}

⇒ R^-1= B × A

⇒ R^-1= { (2,3), (2,5), (2,7)}

Relation R on a set A is said to be a reflexive relation on A if:

⇒ (a,a) ∈ R ∀ a ∈ A

Eg A={1,2,3}

⇒ R={(1,1),(2,2),(3,3)}