Question:

Let A and B be sets. Show that f:A×B→B×A such that f(a,b)=(b,a) is bijective function.

Answer:

1. To show that f is a bijective function, we must first show that it is both injective and surjective.

2. To show that f is injective, we must show that for all a1, a2 ∈ A, and for all b1, b2 ∈ B, if f(a1, b1) = f(a2, b2) then a1 = a2 and b1 = b2.

3. Suppose f(a1, b1) = f(a2, b2). Then (b1, a1) = (b2, a2). Since the order of the elements in a set does not matter, this implies that b1 = b2 and a1 = a2. Therefore, f is injective.

4. To show that f is surjective, we must show that for all b1, b2 ∈ B, there exists a1, a2 ∈ A such that f(a1, b1) = f(a2, b2).

5. Let b1, b2 ∈ B. Then we can define a1 = b2 and a2 = b1. Then f(a1, b1) = f(b2, b1) = (b1, b2). Therefore, f is surjective.

6. Since f is both injective and surjective, it is a bijective function.

Question:

Show that the relation R in the set A of all the books in a library of a college, given by R={(x,y):x and yhave same number of pages} is an equivalence relation.

Answer:

1. To show that R is an equivalence relation, we must first show that it is reflexive, symmetric, and transitive.

2. Reflexivity: For any book x in A, x has the same number of pages as itself. Therefore, (x,x) is in R. This shows that R is reflexive.

3. Symmetry: If (x,y) is in R, then x and y have the same number of pages. This implies that y and x also have the same number of pages, so (y,x) is also in R. This shows that R is symmetric.

4. Transitivity: If (x,y) and (y,z) are in R, then x and y have the same number of pages, and y and z have the same number of pages. This implies that x and z also have the same number of pages, so (x,z) is in R. This shows that R is transitive.

5. Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

Question:

Find gof and fog, if (i) f(x)=∣x∣ and g(x)=∣5x−2∣ ii)f(x)=8x³ and g(x)=x^1/3

Answer:

i) f(g(x)) = f(|5x-2|) = | |5x-2| | = |5x-2|

g(f(x)) = g(|x|) = |5|x| -2| = |5x - 2|

Question:

Show that the relation R defined in the set A of all polygons as R={(P₁,P₂):P₁ and P₂ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3,4 and 5?

Answer:

1. To show that R is an equivalence relation, we must show that it is reflexive, symmetric, and transitive.

2. Reflexive: A relation is reflexive if every element in the set is related to itself. In this case, any polygon P in the set A is related to itself, since it has the same number of sides as itself. Therefore, R is reflexive.

3. Symmetric: A relation is symmetric if, for every pair of elements (x,y), if x is related to y then y is related to x. In this case, if two polygons P₁ and P₂ have the same number of sides, then P₂ has the same number of sides as P₁, and vice versa. Therefore, R is symmetric.

4. Transitive: A relation is transitive if, for every pair of elements (x,y) and (y,z), if x is related to y and y is related to z, then x is related to z. In this case, if two polygons P₁ and P₂ have the same number of sides, and another polygon P₃ has the same number of sides as P₂, then P₁ has the same number of sides as P₃. Therefore, R is transitive.

5. The set of all elements in A related to the right angle triangle T with sides 3, 4, and 5 is the set of all right angle triangles with the same number of sides.

Question:

Show that the relative R in the set {1,2,3} given by R={(1,2),(2,1)} is symmetric but neither reflexive nor transitive.

Answer:

1. To show that R is symmetric, we need to show that for every (a,b)∈R, (b,a)∈R.

2. Let’s take (1,2)∈R. Then (2,1)∈R, which means that R is symmetric.

3. To show that R is not reflexive, we need to show that for every a∈{1,2,3}, (a,a)∉R.

4. Let’s take a=1. Then (1,1)∉R, which means that R is not reflexive.

5. To show that R is not transitive, we need to show that for every (a,b)∈R and (b,c)∈R, (a,c)∉R.

6. Let’s take (1,2)∈R and (2,3)∈R. Then (1,3)∉R, which means that R is not transitive.

Question:

Show that the modulus function f:R→R given by f(x)=∣x∣, is neither one-one nor onto, where ∣x∣ is x, if x is positive or 0 and ∣x∣ is −x, if x is negative.

Answer:

Step 1: To show that the modulus function f:R→R is neither one-one nor onto, we need to find a counterexample.

Step 2: Consider x=2 and x=-2.

Step 3: Since f(2)=|2|=2 and f(-2)=|-2|=-2, we can see that f(2)=f(-2).

Step 4: Hence, the modulus function f:R→R is not one-one as there exists two different inputs that lead to the same output.

Step 5: Since the modulus function f:R→R is not one-one, it cannot be onto.

Question:

Show that the function f:R⋅→R⋅ defined by f(x)=x1​ is one-one and onto, where ‘R⋅’ is the set of all non-zero real numbers. Is the result true, if the domain ‘R⋅’ is replaced by N with co-domain being same as ‘R⋅’?

Answer:

To show that the function f:R⋅→R⋅ defined by f(x)=x1​ is one-one and onto:

Step 1: Show that f is one-one:

Let x1 and x2 be two elements of the domain R⋅.

If f(x1)=f(x2), then x1^1=x2^1

Since both x1 and x2 are in R⋅, then x1=x2

Therefore, f is one-one.

Step 2: Show that f is onto:

Let y be any element of the co-domain R⋅.

Then, we can find an element x in the domain R⋅ such that f(x)=y.

Since x1=y, then x=y^1.

Therefore, f is onto.

The result is true, if the domain R⋅ is replaced by N with co-domain being same as R⋅.

Question:

Let f:R→R be defined as f(x)=3x. Choose the correct answer. A f is one-one onto B f is many-one onto C f is one-one but not onto D f is neither one-one nor onto

Answer:

Answer: C f is one-one but not onto

Question:

Let f:R→R be defined as f(x)=x⁴. Choose the correct answer. A f is one-one onto B f is many-one onto C f is one-one but not onto D f is neither one-one nor onto

Answer:

C f is one-one but not onto

Question:

Let f:1,3,4→1,2,5 and g:1,2,5→1,3 be given by f=(1,2),(3,5),(4,1) and g=(1,3),(2,3),(5,1). Write down gof.

Answer:

gof=(1,3),(3,3),(4,1)

Question:

Check whether the relative R in R defined by R={(a,b):a≤b³} is reflexive, symmetric or transitive.

Answer:

1. To check if R is reflexive, we must determine if (a,a)∈R for all a∈A. In this case, A={a:a∈ℝ}.

2. Since a≤b3 for all a,b∈A, (a,a)∈R for all a∈A, so R is reflexive.

3. To check if R is symmetric, we must determine if (a,b)∈R implies (b,a)∈R for all a,b∈A.

4. Since a≤b3 for all a,b∈A, (a,b)∈R implies (b,a)∈R for all a,b∈A, so R is symmetric.

5. To check if R is transitive, we must determine if (a,b)∈R and (b,c)∈R implies (a,c)∈R for all a,b,c∈A.

6. Since a≤b3 for all a,b∈A, (a,b)∈R and (b,c)∈R implies a≤c3, so (a,c)∈R for all a,b,c∈A, so R is transitive.

Question:

Show that the Signum function f:R→R, given by f(x)={{{1, if x>00, if x=0−1, if x<0​}}} is neither one-one nor onto.

Answer:

1. To show that the Signum function is not one-one, consider the inputs x = -1 and x = 1. Both of these inputs have the same output, so the function is not one-one.

2. To show that the Signum function is not onto, consider the output y = 0. There is no input x that can produce this output, so the function is not onto.

Question:

Prove that the Greatest Integer Function f:R→R given by f(x)=[x], is neither one-one nor onto, where [x] denotes the greatest integer less that or equal to x

Answer:

Proof:

1. To prove that f(x) is not one-one, consider x = 1 and x = 1.5. We have f(1) = [1] = 1 and f(1.5) = [1.5] = 1. Thus, f(1) = f(1.5) which implies that f(x) is not one-one.

2. To prove that f(x) is not onto, consider the range of f(x) which is {[x] : x ∈ R}. Since there are no integers greater than [x], the range of f(x) is bounded above by [x] and therefore, f(x) is not onto.

Question:

Show that the relative R in the set A={1,2,3,4,5} given by R={(a,b):∣a−b∣ is even}, is an equivalence relation. Show that all the elements of {1,3,5} are related to each other and all the elements of {2,4} are related to each other. But no element of {1,3,5} is related to any element of {2,4}.

Answer:

1. Show that the relative R in the set A={1,2,3,4,5} given by R={(a,b):∣a−b∣ is even}, is an equivalence relation.

To show that R is an equivalence relation, we must show that it is reflexive, symmetric, and transitive.

Reflexive: Let a be an element of A. Then, a-a=0, which is an even number. Therefore, (a,a)∈R, which shows that R is reflexive.

Symmetric: Let (a,b)∈R. Then, |a-b| is an even number. Since an even number is equal to its opposite, |b-a| is also an even number. Therefore, (b,a)∈R, which shows that R is symmetric.

Transitive: Let (a,b)∈R and (b,c)∈R. Then, |a-b| and |b-c| are both even numbers. Therefore, |a-c|=|a-b|+|b-c| is also an even number. Therefore, (a,c)∈R, which shows that R is transitive.

Therefore, R is an equivalence relation.

2. Show that all the elements of {1,3,5} are related to each other and all the elements of {2,4} are related to each other.

Let a and b be elements of {1,3,5}. Then, |a-b|=|1-3|=2, which is an even number. Therefore, (a,b)∈R. Similarly, let a and b be elements of {2,4}. Then, |a-b|=|2-4|=2, which is an even number. Therefore, (a,b)∈R.

3. But no element of {1,3,5} is related to any element of {2,4}.

Let a be an element of {1,3,5} and b be an element of {2,4}. Then, |a-b|=|1-2|=1, which is an odd number. Therefore, (a,b)∉R.

Question:

Show that the relation R in the set A of points in a plane given by R={(P,Q):distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all point related to a point P=(0,0) is the circle passing through P with origin as centre.

Answer:

Solution: Step 1: Show that R is an equivalence relation

To show that R is an equivalence relation, we need to show that it is reflexive, symmetric and transitive.

Reflexive: For any point P in the set A, the distance of the point P from the origin is same as the distance of the point P from the origin. Therefore, (P,P)∈R, which shows that R is reflexive.

Symmetric: For any two points P and Q in the set A, if the distance of the point P from the origin is same as the distance of the point Q from the origin, then the distance of the point Q from the origin is same as the distance of the point P from the origin. Therefore, (P,Q)∈R implies (Q,P)∈R, which shows that R is symmetric.

Transitive: For any three points P, Q and R in the set A, if the distance of the point P from the origin is same as the distance of the point Q from the origin and the distance of the point Q from the origin is same as the distance of the point R from the origin, then the distance of the point P from the origin is same as the distance of the point R from the origin. Therefore, (P,Q)∈R and (Q,R)∈R implies (P,R)∈R, which shows that R is transitive.

Hence, R is an equivalence relation.

Step 2: Show that the set of all point related to a point P=(0,0) is the circle passing through P with origin as centre.

For any point P in the set A, the distance of the point P from the origin is same as the distance of the point (0,0) from the origin. Therefore, (P, (0,0))∈R.

This implies that all the points related to (0,0) are the points at the same distance from the origin as (0,0).

Hence, the set of all points related to (0,0) is the circle passing through (0,0) with origin as centre.

Question:

Let f:N→N be defined by f(n)={n+1​/2,if n is odd n​/2,if n is even​} for all n∈N. State whether the function f is bijective. Justify your answer.

Answer:

Answer: No, the function f is not bijective.

To prove this, we need to show that either the function is not injective or not surjective.

To show that the function is not injective, we can take two distinct elements in the domain, say n=3 and n=4. We see that f(3)=2 and f(4)=2, thus showing that f(3) = f(4). This means that the function is not injective.

To show that the function is not surjective, we can take any element in the codomain, say y=2. We see that there is no element in the domain such that f(n)=2. This means that the function is not surjective.

Therefore, the function f is not bijective.

Question:

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. Show that f is one-one.

Answer:

Step 1: First, we need to define what it means for a function to be one-one. A function f is one-one if for any two elements a and b in the domain of f, if f(a) = f(b) then a = b.

Step 2: Next, we will look at the function f given in the question, which maps elements of A to elements of B. We can see that for any element in A, there is a unique element in B that it is mapped to. For example, 1 is mapped to 4, 2 is mapped to 5, and 3 is mapped to 6.

Step 3: Since the mapping is unique, we can conclude that the function f is one-one.

Question:

Let L be the set of all lines in XY plane and R be the relation in L defined as R={(L1​,L2​):L1​ is parallel to L₂ }. Show that R is an equivalence relation. Find the set of all lines related to the line y=2x+4.

Answer:

Answer: R is an equivalence relation because it is reflexive, symmetric, and transitive.

Reflexive: For any line L1 in L, L₁ is parallel to itself. Symmetric: If L1 is parallel to L₂, then L₂ is parallel to L₁. Transitive: If L1 is parallel to L₂ and L₂ is parallel to L₃, then L₁ is parallel to L₃.

The set of all lines related to the line y=2x+4 is all lines in the XY plane that are parallel to the line y=2x+4.

Question:

Let A=R−3 and B=R−1. Consider the function f:A→B defined by f(x)=(x−2​/x−3). Is f one-one and onto? Justify your answer

Answer:

Answer: No, f is not one-one and onto.

To show that f is not one-one, let’s consider two elements x₁ and x₂ in A such that f(x₁)=f(x₂).

We have: f(x₁)=(x₁−2​/x₁−3) and f(x₂)=(x₂−2​/x₂−3).

Since f(x₁)=f(x₂), we have (x₁−2​/x₁−3)=(x₂−2​/x₂−3).

This implies that x₁=x₂, which shows that f is not one-one.

To show that f is not onto, let’s consider an element y in B.

We have: y=R−1.

Since f(x)=(x−2​/x−3), we have f(x)=y if and only if (x−2​/x−3)=R−1.

This implies that x=2, which is not an element of A, since A=R−3.

Hence, f is not onto.

Therefore, f is not one-one and onto.

Question:

Show that the relation R defined in the set A of all triangles as R={(T₁,T₂):T₁ is similar to T₂}, is equivalence relation. Consider three right angle triangles T1​ with sides 3,4,5,T₂ with sides 5,12,13 and T3​ with sides 6,8,10. Which triangles among T₁,T₂ and T₃ are related?

Answer:

Solution: Step 1: Show that the relation R is reflexive. R is reflexive as for every element T in A, (T,T) is an element of R, i.e. every triangle is similar to itself.

Step 2: Show that the relation R is symmetric. R is symmetric as if (T₁,T₂) is an element of R, then (T₂,T₁) is also an element of R, i.e. if T₁ is similar to T₂, then T₂ is similar to T₁.

Step 3: Show that the relation R is transitive. R is transitive as if (T₁,T₂) and (T₂,T₃) are elements of R, then (T₁,T₃) is also an element of R, i.e. if T1 is similar to T₂ and T₂ is similar to T₃, then T₁ is similar to T₃.

Step 4: Consider three right angle triangles T₁ with sides 3,4,5,T₂ with sides 5,12,13 and T₃ with sides 6,8,10. T₁, T₂ and T₃ are all right angle triangles and therefore are all similar to each other.

Step 5: Which triangles among T₁,T₂ and T₃ are related? All three triangles, T₁, T₂ and T₃, are related as they are all similar to each other.

Question:

Show that the relation R in the set R of real numbers, defined as R={(a,b):a≤b₂} is neither reflexive nor symmetric nor transitive.

Answer:

Reflexive: A relation R is reflexive if (a,a)∈R for all a∈A, where A is the set of elements.

To show that R is not reflexive, we need to find an a∈A such that (a,a)∉R.

Let a=2. Then, (2,2)∉R since 2≠2₂. Therefore, R is not reflexive.

Symmetric: A relation R is symmetric if (a,b)∈R implies (b,a)∈R for all a,b∈A.

To show that R is not symmetric, we need to find an a,b∈A such that (a,b)∈R and (b,a)∉R.

Let a=2 and b=3. Then, (2,3)∈R since 2≤3^2. However, (3,2)∉R since 3≠2₂. Therefore, R is not symmetric.

Transitive: A relation R is transitive if (a,b)∈R and (b,c)∈R implies (a,c)∈R for all a,b,c∈A.

To show that R is not transitive, we need to find an a,b,c∈A such that (a,b)∈R, (b,c)∈R, but (a,c)∉R.

Let a=2, b=3, and c=4. Then, (2,3)∈R since 2≤3^2 and (3,4)∈R since 3≤4^2. However, (2,4)∉R since 2≠4₂. Therefore, R is not transitive.

Question:

Show that the relative R in R defined as R={(a,b):a≤b}, is reflexive and transitive but not symmetric.

Answer:

Reflexive: R={(a,b):a≤b} Let a=b Therefore, (a,a)∈R Since (a,a)∈R, R is reflexive.

Transitive: Let (a,b)∈R and (b,c)∈R Therefore, a≤b and b≤c Since a≤b and b≤c, a≤c Therefore, (a,c)∈R Since (a,c)∈R, R is transitive.

Not Symmetric: Let (a,b)∈R Therefore, a≤b Let (b,a)∈R Therefore, b≤a Since a≤b and b≤a, a≠b Therefore, (a,b)∉R Since (a,b)∉R, R is not symmetric.

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.

Answer:

Reflexive: A relation R is reflexive if (a,a) is present in the relation for all a in the set.

To check if this relation is reflexive, we need to check if (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6) are present in the relation.

Since none of these pairs are present in the relation, R is not reflexive.

Symmetric: A relation R is symmetric if (a,b) is present in the relation for all a and b in the set and (b,a) is also present in the relation.

To check if this relation is symmetric, we need to check if (1,2), (2,3), (3,4), (4,5), (5,6) and (6,7) are present in the relation.

Since none of these pairs are present in the relation, R is not symmetric.

Transitive: A relation R is transitive if (a,b) and (b,c) are present in the relation for all a, b and c in the set and (a,c) is also present in the relation.

To check if this relation is transitive, we need to check if (1,2) and (2,3), (2,3) and (3,4), (3,4) and (4,5), (4,5) and (5,6) and (5,6) and (6,7) are present in the relation.

Since none of these pairs are present in the relation, R is not transitive.