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**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 = {(T _{1} , T_{2}) : T_{1} is similar to T_{2} } is equivalence relation. Consider three right angle triangles T_{1} with sides 3, 4, 5, T_{2} with sides 5, 12, 13 and T_{3} with sides 6, 8,10. Which triangles among T_{1}, T_{2} and T_{3} are related?**

R = { (T_{1} , T_{2} ) : T_{1} is similar to T_{2}}

R is reflexive since every triangle is similar to itself. Further, if (T_{1} , T_{2} ) R , then T_{1} is similar to T_{2} .

T_{2} is similar to T_{1}

(T_{2} , T_{1} ) R

Therefore R is symmetric.

Now , Let (T_{1} , T_{2} ) , (T_{2} , T_{3}) R

T_{1} is similar to T_{2} and T_{2} is similar to T_{3}.

T_{1} is similar to T_{3}.

(T_{1} , T_{3} ) 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 T_{1} and T_{3} are in the same ratio. Then , triangle T_{1} is similar to triangle T_{3}.

Hence , T_{1} is related to T_{3}.

**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) = x ^{2}**

f : R → R is given by, f (x) = x^{2}

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) = x^{2} = -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 (a_{1}, b_{1} ), (a_{2}, b_{2}) A × B such that f (a_{1}, b_{1}) = f (a_{2} , b_{2})

(b_{1}, a_{1} ) = (b_{2}, a_{2})

b_{1} = b_{2 } and a_{1} = a_{2}

(a_{1}, b_{1}) = (a_{2}, b_{2})

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) = x ^{4}. 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) = x^{4}.

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

x^{4} = y^{4}

x = ± y

Therefore f (x_{1}) = f (x_{2}) does not imply that x_{1} = x_{2}.

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.

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Plus 2 Computer Science

Kerala (English Medium)