In order to determine which of the relations listed is reflexive, we must recall that a relation R on a set A is reflexive if for every element x ∈ A, the pair (x, x) is in R. This means that x must be related to itself. Let's examine each of the relations listed: (a) R = {(x, y) : x > y, x, y ∈Read more
In order to determine which of the relations listed is reflexive, we must recall that a relation R on a set A is reflexive if for every element x ∈ A, the pair (x, x) is in R. This means that x must be related to itself.
Let’s examine each of the relations listed:
(a) R = {(x, y) : x > y, x, y ∈ ℕ}
For this relation to be reflexive, we must have x > x for all x ∈ ℕ. However, this is never the case since x is never greater than itself. So this relation is not reflexive.
(b) R = {(x, y) : x + y = 10, x, y ∈ ℕ}
For this relation to be reflexive, we require that x + x = 10 for all x ∈ ℕ. This yields 2x = 10, or x = 5. So, the only element that satisfies this is x = 5. Hence, the relation is **not reflexive** for all elements of ℕ, but only for x = 5.
(c) R = {(x, y) : xy is a square number, x, y ∈ ℕ}
For this relation to be reflexive, we must have x * x to be a square number for all x ∈ ℕ. Since x * x = x² is always a square number for all natural numbers x, this relation is reflexive.
(d) R = {(x, y) : x + 4y = 10, x, y ∈ ℕ}
For this relation to be reflexive, we require x + 4x = 10 for all x ∈ ℕ. This simplifies to 5x = 10, or x = 2. Hence, the only element satisfying this condition is x = 2, so this relation is not reflexive for all elements of ℕ, only for x = 2.
Conclusion:
The only reflexive relation is:
– (c) R = {(x, y) : xy is a square number, x, y ∈ ℕ}.
We are given the relation R on the set A = {x ∈ ℤ : 0 ≤ x ≤ 12}, defined as: R = {(a, b) : |a - b| is a multiple of 4} This means a and b are connected if the absolute difference between them is a multiple of 4. We now have to determine the equivalence class of 1, denoted by [1]. This is going to beRead more
We are given the relation R on the set A = {x ∈ ℤ : 0 ≤ x ≤ 12}, defined as:
R = {(a, b) : |a – b| is a multiple of 4}
This means a and b are connected if the absolute difference between them is a multiple of 4. We now have to determine the equivalence class of 1, denoted by [1]. This is going to be all elements b ∈ A such that |1 – b| is a multiple of 4.
Step 1: Determine what values of b make |1 – b| a multiple of 4.
The possible values of b such that |1 – b| is a multiple of 4 are those for which:
|1 – b| = 4k for some integer k.
This gives us the following conditions:
1 – b = 4k or b – 1 = 4k
Thus, b = 1 + 4k for some integer k. Now let’s check the values of b in the set A = {0, 1, 2, 3,., 12}.
For k = 0, b = 1.
For k = 1, b = 1 + 4 = 5.
For k = 2, b = 1 + 8 = 9.
For k = -1, b = 1 – 4 = -3 (which is outside of A).
For k = -2, b = 1 – 8 = -7 (which is outside of A).
Thus, the equivalence class of 1, [1], includes the elements 1, 5, and 9.
Conclusion:
The equivalence class that contains 1 is {1, 5, 9}.
We are given the relation R on ℤ, defined by: aRb if and only if a² - 7ab + 6b² = 0 We need to find out the properties of this relation: whether it is reflexive, symmetric, and/or transitive. Step 1: Check if the relation is reflexive. A relation R is reflexive if for every element a ∈ ℤ, we have aRRead more
We are given the relation R on ℤ, defined by:
aRb if and only if a² – 7ab + 6b² = 0
We need to find out the properties of this relation: whether it is reflexive, symmetric, and/or transitive.
Step 1: Check if the relation is reflexive.
A relation R is reflexive if for every element a ∈ ℤ, we have aRa. That is, we must check whether the equation a² – 7a ⋅ a + 6a² = 0 holds for every integer a.
Substitute a = b into the relation:
a² – 7a ⋅ a + 6a² = a² – 7a² + 6a² = 0
This reduces to:
0 = 0
Thus, the relation is reflexive because the equation holds for all a ∈ ℤ.
Step 2: Check if the relation is symmetric.
A relation R is symmetric if for every pair (a, b) ∈ ℤ, whenever aRb, we also have bRa. That is, if a² – 7ab + 6b² = 0 we need to see if b² – 7ba + 6a² = 0 also holds.
Since the relation aRb is defined by the equation a² – 7ab + 6b² = 0, we simply reverse a and b in the equation:
b² – 7ba + 6a² = 0
This is exactly the same as a² – 7ab + 6b² = 0 (just switching a and b does not affect the equation because it is symmetric in a and b).
Thus, the relation is symmetric.
Step 3: Determine if the relation is transitive.
A relation R is transitive if for every a, b, c ∈ ℤ, whenever aRb and bRc, we must have aRc. To check transitivity, we would need to check the equation a² – 7ab + 6b² = 0 and b² – 7bc + 6c² = 0, and then determine if a² – 7ac + 6c² = 0 holds. This is however complicated and does not ensure the transitivity simply by observing the structure of the equation (an algebraic check would be necessary).
Since dividing by zero is undefined, f: ℝ → ℝ, given by f(x) = 1/x, is not defined for x = 0. Since the function is not defined at x = 0, it is not well-defined on the entire set ℝ (since the domain of the function is not the whole ℝ). - (d) not defined. Click for more: https://www.tiwariacademy.comRead more
Since dividing by zero is undefined, f: ℝ → ℝ, given by f(x) = 1/x, is not defined for x = 0.
Since the function is not defined at x = 0, it is not well-defined on the entire set ℝ (since the domain of the function is not the whole ℝ).
– (d) not defined.
We have a function f: ℝ → ℝ defined as f(x) = 4 + 3 cos(x) Whether the given function is bijective, one-one, onto, or neither can be determined. Step 1 Check if the given function is one-one (injective). A function is said to be one-one (injective) if the different elements in the domain of a functiRead more
We have a function f: ℝ → ℝ defined as
f(x) = 4 + 3 cos(x)
Whether the given function is bijective, one-one, onto, or neither can be determined.
Step 1 Check if the given function is one-one (injective).
A function is said to be one-one (injective) if the different elements in the domain of a function go to different elements in the range i.e. f(x₁) = f(x₂) => x₁ = x₂.
Let’s see if the function is injective. Assume that f(x₁) = f(x₂), that is:
4 + 3 cos(x₁) = 4 + 3 cos(x₂)
We simplify this to obtain:
3 cos(x₁) = 3 cos(x₂)
cos(x₁) = cos(x₂)
The cosine function is periodic and not injective. That means that there may be different values of x₁ and x₂, like x₁ = 0 and x₂ = 2π, for which cos(x₁) = cos(x₂). In other words, the function is not injective.
Step 2: Determine whether the function is onto (surjective).
A function is onto (surjective) if for every element y ∈ ℝ, there exists an x ∈ ℝ such that f(x) = y.
For f(x) = 4 + 3 cos(x), the range of the cosine function is between -1 and 1, so the range of f(x) will be:
f(x) = 4 + 3 cos(x)
Since cos(x) lies between -1 and 1, the values of f(x) will lie between:
f(x) = 4 + 3(-1) = 1 and f(x) = 4 + 3(1) = 7
So, f(x) can only be taken in the interval [1, 7]. Since f(x) cannot take any value outside that interval, it is not onto because no values of x map to values of y that are outside that interval.
Conclusion:
The function f(x) = 4 + 3 cos(x) is neither one-one nor onto. So the correct option is
(d) neither one-one nor onto.
The function f: ℝ → ℝ defined by f(x) = x³ is a function from the real numbers to the real numbers. Analyzing the function: 1. One-to-one (Injective): A function is one-to-one if different inputs map to different outputs. Suppose f(x₁) = f(x₂), i.e., x₁³ = x₂³. Then, x₁ = x₂. This demonstrates thatRead more
The function f: ℝ → ℝ defined by f(x) = x³ is a function from the real numbers to the real numbers.
Analyzing the function:
1. One-to-one (Injective):
A function is one-to-one if different inputs map to different outputs.
Suppose f(x₁) = f(x₂), i.e., x₁³ = x₂³.
Then, x₁ = x₂.
This demonstrates that f(x) = x³ is one-to-one (injective).
2. Onto (Surjective): A function is onto if every element in the target set, here ℝ, has a pre-image in the domain. For any real number y ∈ ℝ, we can find an x ∈ ℝ such that f(x) = x³ = y.
Specifically, x = ∛y. This demonstrates that the function f(x) = x³ is onto (surjective).
Conclusion: The function f(x) = x³ is one-to-one and onto, so the correct answer is: – One-one and onto.
In order to find the number of one-one and onto mappings or bijective functions from set A to set B, consider the following: 1. One-one mapping: Each element of A should map to a unique element of B. 2. Onto mapping: Every element of B should be associated with at least one element of A. ObservationRead more
In order to find the number of one-one and onto mappings or bijective functions from set A to set B, consider the following:
1. One-one mapping: Each element of A should map to a unique element of B.
2. Onto mapping: Every element of B should be associated with at least one element of A.
Observations:
Set A has 5 elements (|A| = 5), and set B has 6 elements (|B| = 6).
– For a bijection (one-one and onto), the no. of elements in both the sets must be equal, that is, |A| = |B|.
Now, |A| ≠ |B| hence it is not possible to have a one-one and onto mapping.
Conclusion:
The number of one-one and onto mappings from A to B is:
0.
We have the relation R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} on the set A = {1, 2, 3, 4}. Definitions: 1. Function: A relation R is a function if each element of A occurs as the first component in at most one pair of R. 2. Transitive: A relation R is transitive if for all a, b, c ∈ A, wheneverRead more
We have the relation R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} on the set A = {1, 2, 3, 4}.
Definitions:
1. Function: A relation R is a function if each element of A occurs as the first component in at most one pair of R.
2. Transitive: A relation R is transitive if for all a, b, c ∈ A, whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
3. Symmetric: A relation R is symmetric if for all a, b ∈ A, whenever (a, b) ∈ R, then (b, a) ∈ R.
4. Reflexive: A relation R is reflexive if for all a ∈ A, (a, a) ∈ R.
Analysis:
1. Is R a function?
– The element 2 occurs as the first element in both (2, 4) and (2, 3), which goes against the definition of a function. Thus, R is not a function.
2. Is R transitive?
– Since (1, 3) ∈ R and (3, 1) ∈ R, we would need (1, 1) ∈ R for it to be transitive, but it isn’t there.
– For (2, 4) ∈ R and (4, 2) ∈ R, we would need (2, 2) ∈ R, which is absent too.
– Thus, R is not transitive.
3. Is R symmetric?
– Do we have (a, b) ∈ R implies (b, a) ∈ R?
– (1, 3) ∈ R, and (3, 1) ∈ R (symmetric pair).
– (4, 2) ∈ R, but (2, 4) ∈ R (symmetric pair).
– All needed symmetric pairs are there. So, R is symmetric.
4. Is R reflexive?
– See if (a, a) ∈ R for every a ∈ A:
– (1, 1), (2, 2), (3, 3), and (4, 4) are not in R.
– Hence, R is not reflexive.
Conclusion:
The relation R is:
– Not a function
– Not transitive
– Symmetric
– Not reflexive
Let A = {1, 2, 3} and B = {4, 5, 6, 7}. Let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. 1. Injective (One-to-one): Each element of A goes to a distinct element in B. There are no two elements of A going to the same element in B. Therefore, f is injective. 2. Surjective (Onto): Not all elRead more
Let A = {1, 2, 3} and B = {4, 5, 6, 7}.
Let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B.
1. Injective (One-to-one): Each element of A goes to a distinct element in B. There are no two elements of A going to the same element in B.
Therefore, f is injective.
2. Surjective (Onto):
Not all elements of B are covered by f. For instance, 7 is not mapped by any element of A.
Therefore, f is not surjective.
3. Bijective (One-to-one and Onto):
Since f is not surjective, it cannot be bijective.
4. Function:
f is a function since every element of A is related to exactly one element of B.
Therefore, the correct answer is: Injective function.
We are given the relation R on the set N, defined as: R = {(a, b) : a = b - 2, b > 6} This is to say, for every pair (a, b), a should be b - 2, and b must be greater than 6. (6, 8) ∈ R - a = 6 and b = 8. - a = b - 2 gives 6 = 8 - 2, which is true. - b = 8 > 6, so the condition is satisfied.Read more
We are given the relation R on the set N, defined as:
R = {(a, b) : a = b – 2, b > 6}
This is to say, for every pair (a, b), a should be b – 2, and b must be greater than 6.
(6, 8) ∈ R
– a = 6 and b = 8.
– a = b – 2 gives 6 = 8 – 2, which is true.
– b = 8 > 6, so the condition is satisfied.
– Therefore, (6, 8) ∈ R.
A relation R is defined on N. Which of the following is the reflexive relation?
In order to determine which of the relations listed is reflexive, we must recall that a relation R on a set A is reflexive if for every element x ∈ A, the pair (x, x) is in R. This means that x must be related to itself. Let's examine each of the relations listed: (a) R = {(x, y) : x > y, x, y ∈Read more
In order to determine which of the relations listed is reflexive, we must recall that a relation R on a set A is reflexive if for every element x ∈ A, the pair (x, x) is in R. This means that x must be related to itself.
Let’s examine each of the relations listed:
(a) R = {(x, y) : x > y, x, y ∈ ℕ}
For this relation to be reflexive, we must have x > x for all x ∈ ℕ. However, this is never the case since x is never greater than itself. So this relation is not reflexive.
(b) R = {(x, y) : x + y = 10, x, y ∈ ℕ}
For this relation to be reflexive, we require that x + x = 10 for all x ∈ ℕ. This yields 2x = 10, or x = 5. So, the only element that satisfies this is x = 5. Hence, the relation is **not reflexive** for all elements of ℕ, but only for x = 5.
(c) R = {(x, y) : xy is a square number, x, y ∈ ℕ}
For this relation to be reflexive, we must have x * x to be a square number for all x ∈ ℕ. Since x * x = x² is always a square number for all natural numbers x, this relation is reflexive.
(d) R = {(x, y) : x + 4y = 10, x, y ∈ ℕ}
For this relation to be reflexive, we require x + 4x = 10 for all x ∈ ℕ. This simplifies to 5x = 10, or x = 2. Hence, the only element satisfying this condition is x = 2, so this relation is not reflexive for all elements of ℕ, only for x = 2.
Conclusion:
The only reflexive relation is:
– (c) R = {(x, y) : xy is a square number, x, y ∈ ℕ}.
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Let the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by R = {(a, b) : |a – b| is a multiple of 4}. Then [1], the equivalence class containing 1, is:
We are given the relation R on the set A = {x ∈ ℤ : 0 ≤ x ≤ 12}, defined as: R = {(a, b) : |a - b| is a multiple of 4} This means a and b are connected if the absolute difference between them is a multiple of 4. We now have to determine the equivalence class of 1, denoted by [1]. This is going to beRead more
We are given the relation R on the set A = {x ∈ ℤ : 0 ≤ x ≤ 12}, defined as:
R = {(a, b) : |a – b| is a multiple of 4}
This means a and b are connected if the absolute difference between them is a multiple of 4. We now have to determine the equivalence class of 1, denoted by [1]. This is going to be all elements b ∈ A such that |1 – b| is a multiple of 4.
Step 1: Determine what values of b make |1 – b| a multiple of 4.
The possible values of b such that |1 – b| is a multiple of 4 are those for which:
|1 – b| = 4k for some integer k.
This gives us the following conditions:
1 – b = 4k or b – 1 = 4k
Thus, b = 1 + 4k for some integer k. Now let’s check the values of b in the set A = {0, 1, 2, 3,., 12}.
For k = 0, b = 1.
For k = 1, b = 1 + 4 = 5.
For k = 2, b = 1 + 8 = 9.
For k = -1, b = 1 – 4 = -3 (which is outside of A).
For k = -2, b = 1 – 8 = -7 (which is outside of A).
Thus, the equivalence class of 1, [1], includes the elements 1, 5, and 9.
Conclusion:
The equivalence class that contains 1 is {1, 5, 9}.
Hence, the correct answer is:
– (a) {1, 5, 9}.
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A relation R is defined on Z as: aRb if and only if a² – 7ab + 6b² = 0 Then, R is
We are given the relation R on ℤ, defined by: aRb if and only if a² - 7ab + 6b² = 0 We need to find out the properties of this relation: whether it is reflexive, symmetric, and/or transitive. Step 1: Check if the relation is reflexive. A relation R is reflexive if for every element a ∈ ℤ, we have aRRead more
We are given the relation R on ℤ, defined by:
aRb if and only if a² – 7ab + 6b² = 0
We need to find out the properties of this relation: whether it is reflexive, symmetric, and/or transitive.
Step 1: Check if the relation is reflexive.
A relation R is reflexive if for every element a ∈ ℤ, we have aRa. That is, we must check whether the equation a² – 7a ⋅ a + 6a² = 0 holds for every integer a.
Substitute a = b into the relation:
a² – 7a ⋅ a + 6a² = a² – 7a² + 6a² = 0
This reduces to:
0 = 0
Thus, the relation is reflexive because the equation holds for all a ∈ ℤ.
Step 2: Check if the relation is symmetric.
A relation R is symmetric if for every pair (a, b) ∈ ℤ, whenever aRb, we also have bRa. That is, if a² – 7ab + 6b² = 0 we need to see if b² – 7ba + 6a² = 0 also holds.
Since the relation aRb is defined by the equation a² – 7ab + 6b² = 0, we simply reverse a and b in the equation:
b² – 7ba + 6a² = 0
This is exactly the same as a² – 7ab + 6b² = 0 (just switching a and b does not affect the equation because it is symmetric in a and b).
Thus, the relation is symmetric.
Step 3: Determine if the relation is transitive.
A relation R is transitive if for every a, b, c ∈ ℤ, whenever aRb and bRc, we must have aRc. To check transitivity, we would need to check the equation a² – 7ab + 6b² = 0 and b² – 7bc + 6c² = 0, and then determine if a² – 7ac + 6c² = 0 holds. This is however complicated and does not ensure the transitivity simply by observing the structure of the equation (an algebraic check would be necessary).
For now, let us focus on the properties that have been given, namely, the reflexive and symmetric.
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Let f : R → R be defined by f(x) = 1/x, for all x ∈ R. Then, f is
Since dividing by zero is undefined, f: ℝ → ℝ, given by f(x) = 1/x, is not defined for x = 0. Since the function is not defined at x = 0, it is not well-defined on the entire set ℝ (since the domain of the function is not the whole ℝ). - (d) not defined. Click for more: https://www.tiwariacademy.comRead more
Since dividing by zero is undefined, f: ℝ → ℝ, given by f(x) = 1/x, is not defined for x = 0.
Since the function is not defined at x = 0, it is not well-defined on the entire set ℝ (since the domain of the function is not the whole ℝ).
– (d) not defined.
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The function f : R → R defined by f(x) = 4 + 3 cos x is
We have a function f: ℝ → ℝ defined as f(x) = 4 + 3 cos(x) Whether the given function is bijective, one-one, onto, or neither can be determined. Step 1 Check if the given function is one-one (injective). A function is said to be one-one (injective) if the different elements in the domain of a functiRead more
We have a function f: ℝ → ℝ defined as
f(x) = 4 + 3 cos(x)
Whether the given function is bijective, one-one, onto, or neither can be determined.
Step 1 Check if the given function is one-one (injective).
A function is said to be one-one (injective) if the different elements in the domain of a function go to different elements in the range i.e. f(x₁) = f(x₂) => x₁ = x₂.
Let’s see if the function is injective. Assume that f(x₁) = f(x₂), that is:
4 + 3 cos(x₁) = 4 + 3 cos(x₂)
We simplify this to obtain:
3 cos(x₁) = 3 cos(x₂)
cos(x₁) = cos(x₂)
The cosine function is periodic and not injective. That means that there may be different values of x₁ and x₂, like x₁ = 0 and x₂ = 2π, for which cos(x₁) = cos(x₂). In other words, the function is not injective.
Step 2: Determine whether the function is onto (surjective).
A function is onto (surjective) if for every element y ∈ ℝ, there exists an x ∈ ℝ such that f(x) = y.
For f(x) = 4 + 3 cos(x), the range of the cosine function is between -1 and 1, so the range of f(x) will be:
f(x) = 4 + 3 cos(x)
Since cos(x) lies between -1 and 1, the values of f(x) will lie between:
f(x) = 4 + 3(-1) = 1 and f(x) = 4 + 3(1) = 7
So, f(x) can only be taken in the interval [1, 7]. Since f(x) cannot take any value outside that interval, it is not onto because no values of x map to values of y that are outside that interval.
Conclusion:
The function f(x) = 4 + 3 cos(x) is neither one-one nor onto. So the correct option is
(d) neither one-one nor onto.
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The function f R → R defined as f(x) = x³ is:
The function f: ℝ → ℝ defined by f(x) = x³ is a function from the real numbers to the real numbers. Analyzing the function: 1. One-to-one (Injective): A function is one-to-one if different inputs map to different outputs. Suppose f(x₁) = f(x₂), i.e., x₁³ = x₂³. Then, x₁ = x₂. This demonstrates thatRead more
The function f: ℝ → ℝ defined by f(x) = x³ is a function from the real numbers to the real numbers.
Analyzing the function:
1. One-to-one (Injective):
A function is one-to-one if different inputs map to different outputs.
Suppose f(x₁) = f(x₂), i.e., x₁³ = x₂³.
Then, x₁ = x₂.
This demonstrates that f(x) = x³ is one-to-one (injective).
2. Onto (Surjective): A function is onto if every element in the target set, here ℝ, has a pre-image in the domain. For any real number y ∈ ℝ, we can find an x ∈ ℝ such that f(x) = x³ = y.
Specifically, x = ∛y. This demonstrates that the function f(x) = x³ is onto (surjective).
Conclusion: The function f(x) = x³ is one-to-one and onto, so the correct answer is: – One-one and onto.
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If a set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is
In order to find the number of one-one and onto mappings or bijective functions from set A to set B, consider the following: 1. One-one mapping: Each element of A should map to a unique element of B. 2. Onto mapping: Every element of B should be associated with at least one element of A. ObservationRead more
In order to find the number of one-one and onto mappings or bijective functions from set A to set B, consider the following:
1. One-one mapping: Each element of A should map to a unique element of B.
2. Onto mapping: Every element of B should be associated with at least one element of A.
Observations:
Set A has 5 elements (|A| = 5), and set B has 6 elements (|B| = 6).
– For a bijection (one-one and onto), the no. of elements in both the sets must be equal, that is, |A| = |B|.
Now, |A| ≠ |B| hence it is not possible to have a one-one and onto mapping.
Conclusion:
The number of one-one and onto mappings from A to B is:
0.
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Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. Then relation R is
We have the relation R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} on the set A = {1, 2, 3, 4}. Definitions: 1. Function: A relation R is a function if each element of A occurs as the first component in at most one pair of R. 2. Transitive: A relation R is transitive if for all a, b, c ∈ A, wheneverRead more
We have the relation R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} on the set A = {1, 2, 3, 4}.
Definitions:
1. Function: A relation R is a function if each element of A occurs as the first component in at most one pair of R.
2. Transitive: A relation R is transitive if for all a, b, c ∈ A, whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
3. Symmetric: A relation R is symmetric if for all a, b ∈ A, whenever (a, b) ∈ R, then (b, a) ∈ R.
4. Reflexive: A relation R is reflexive if for all a ∈ A, (a, a) ∈ R.
Analysis:
1. Is R a function?
– The element 2 occurs as the first element in both (2, 4) and (2, 3), which goes against the definition of a function. Thus, R is not a function.
2. Is R transitive?
– Since (1, 3) ∈ R and (3, 1) ∈ R, we would need (1, 1) ∈ R for it to be transitive, but it isn’t there.
– For (2, 4) ∈ R and (4, 2) ∈ R, we would need (2, 2) ∈ R, which is absent too.
– Thus, R is not transitive.
3. Is R symmetric?
– Do we have (a, b) ∈ R implies (b, a) ∈ R?
– (1, 3) ∈ R, and (3, 1) ∈ R (symmetric pair).
– (4, 2) ∈ R, but (2, 4) ∈ R (symmetric pair).
– All needed symmetric pairs are there. So, R is symmetric.
4. Is R reflexive?
– See if (a, a) ∈ R for every a ∈ A:
– (1, 1), (2, 2), (3, 3), and (4, 4) are not in R.
– Hence, R is not reflexive.
Conclusion:
The relation R is:
– Not a function
– Not transitive
– Symmetric
– Not reflexive
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Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Based on the given information, f is best defined as:
Let A = {1, 2, 3} and B = {4, 5, 6, 7}. Let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. 1. Injective (One-to-one): Each element of A goes to a distinct element in B. There are no two elements of A going to the same element in B. Therefore, f is injective. 2. Surjective (Onto): Not all elRead more
Let A = {1, 2, 3} and B = {4, 5, 6, 7}.
Let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B.
1. Injective (One-to-one): Each element of A goes to a distinct element in B. There are no two elements of A going to the same element in B.
Therefore, f is injective.
2. Surjective (Onto):
Not all elements of B are covered by f. For instance, 7 is not mapped by any element of A.
Therefore, f is not surjective.
3. Bijective (One-to-one and Onto):
Since f is not surjective, it cannot be bijective.
4. Function:
f is a function since every element of A is related to exactly one element of B.
Therefore, the correct answer is: Injective function.
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Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b – 2, b > 6}, then
We are given the relation R on the set N, defined as: R = {(a, b) : a = b - 2, b > 6} This is to say, for every pair (a, b), a should be b - 2, and b must be greater than 6. (6, 8) ∈ R - a = 6 and b = 8. - a = b - 2 gives 6 = 8 - 2, which is true. - b = 8 > 6, so the condition is satisfied.Read more
We are given the relation R on the set N, defined as:
R = {(a, b) : a = b – 2, b > 6}
This is to say, for every pair (a, b), a should be b – 2, and b must be greater than 6.
(6, 8) ∈ R
– a = 6 and b = 8.
– a = b – 2 gives 6 = 8 – 2, which is true.
– b = 8 > 6, so the condition is satisfied.
– Therefore, (6, 8) ∈ R.
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