1. Determine whether each of the following relations are reflexive, symmetric and transitive? Relation R in the set A = { 1 , 2 , 3 , 4 , 5 , 6 } as R = {(x , y) : y is  divisible by x }

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

R = {(x , y) : y is divisible by x }

We know that any number (x) is divisible by itself.

   (x , x)  R.

Therefore R is reflexive.

Now , (2 , 4)  R [ as 4 is divisible by 2]

But , (4, 2) [ as 2 is not divisible by 4 ]

Therefore R is not symmetric.

Let (x , y) , (y , z) R. Then , y is divisible by x and z is divisible by y.

Therefore  z is divisible by x.

(x , z ) R

Therefore R is transitive.

Hence , R is reflexive and transitive but not symmetric.

2. 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?

R = { (T₁ , T₂ ) : T₁ is similar to T₂}

R is reflexive since every triangle is similar to itself. Further, if (T₁ , T₂ ) R , then T₁ is similar to T₂ .

  T₂ is similar to T₁

      (T₂ , T₁ ) R

Therefore R is symmetric.

Now , Let (T₁ , T₂ ) , (T₂ , T₃) R

  T₁ is similar to T₂ and T₂ is similar to T₃.

  T₁ is similar to T₃.

   (T₁ , T₃ ) R

Therefore R is transitive. Thus, R is an equivalence relation. Now, we can observe that : ³/₆ = ⁴/₈ = ⁵/₁₀  (=½ )

Therefore the corresponding sides of triangles T₁ and T₃ are in the same ratio. Then , triangle T₁ is similar to triangle T₃.

Hence , T₁ is related to T₃.

3. Let R be the relation in the set {1, 2, 3, 4 } given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2) } choose the correct answer? (a) R is reflexive and symmetric but not transitive. (b) R is reflexive and transitive but not symmetric. (c) R is symmetric and transitive but not reflexive. (d) D is an equivalence relation.

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

It is seen that (a , a) R, for every a {1, 2, 3, 4 }

Therefore  R is reflexive.

It is seen that (1, 2) R, but (2, 1) R.

Therefore  R is not symmetric.

Also, it is observed that (a, b), (b, c) R

  (a, c) R for all a, b, c   { 1, 2, 3, 4 }

Therefore  R is transitive.

Hence, R is reflexive and transitive but not symmetric.

The correct answer is B.

4. Check the injectivity and surjectivity of f : R ? R given by f(x) = x²

f : R → R is given by, f (x) = x²

It is seen that f(-1) = f(1) = 1, but -1 ≠ 1

Therefore f is not injective.

Now, -2 R. But, there does not exist any element x R such that f (x) = x² = -2

Therefore f is not surjective.

Hence, function f is neither injective nor surjective.

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

f : R  → R  is given by f (x) = [x]

It is seen that f (1.2) = [1.2] = 1

f (1.9) = [1.9] = 1

Therefore  f (1.2) = f(1.9), but 1.2 ≠ 1.9

Therefore f is not one - one.

Now, consider 0.7 R.

It is known that f (x) = [x] is always an integer. Thus, there does not exist any element x R such that f (x) = 0.7

Therefore  f is not onto.

Hence, the greatest integer function is neither one - one nor onto.

6. 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?

f : R R is given by

It is seen that f (-1) = |-1| = 1

f (1) = |1| = 1

Therefore   f(-1) = f (1), but -1 ? 1

Therefore f is not one - one.

Now, consider -1 R.

It is known that f (x) = |x| is always non - negative.

Thus, there does not exist any element x in domain R such that f(x) = |x| = -1.

Therefore f is not onto.

Hence , the modulus function is neither one - one nor onto.

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

A × B → B × A is defined as f (a, b) = (b, a).

Let (a₁, b₁ ), (a₂, b₂) A × B such that f (a₁, b₁) = f (a₂ , b₂)

(b₁, a₁ ) = (b₂,  a₂)

b₁ = b₂  and  a₁ = a₂

(a₁, b₁) = (a₂, b₂)

Therefore f is one - one.

Now, let (b, a) B × A be any element.

Then, there exists (a, b) A × B such that f (a, b) = (b, a)

Therefore f is onto.

Hence, f is bijective.

8. 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.

f : R → R is defined as f(x) = x⁴.

Let x, y R such that f (x) = f (y)

x⁴ = y⁴

x = ± y

Therefore  f (x₁) = f (x₂) does not imply that x₁ = x₂.

For instance,

f (1) = f (-1) = 1

Therefore f is not one - one.

Consider an element 2 in co - domain R. It is clear that there does not exist any x in domain R such that f (x) = 2

Therefore f is not onto.

Hence, function f is neither one - one nor onto.

The correct answer is D.