#### Topics

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.

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 = b 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.

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