R = {(P1, P2): P1 and P2 have same the number of sides}
R is reflexive since (P1, P1) ∈ R as the same polygon has the same number of sides with itself.
Let (P1, P2) ∈ R.
⇒ P1 and P2 have the same number of sides.
⇒ P2 and P1 have the same number of sides.
⇒ (P2, P1) ∈ R
∴R is symmetric.
Now,
Let (P1, P2), (P2, P3) ∈ R.
⇒ P1 and P2 have the same number of sides. Also, P2 and P3 have the same number of sides.
⇒ P1 and P3 have the same number of sides.
⇒ (P1, P3) ∈ R
∴R is transitive.
Hence, R is an equivalence relation.
The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are those polygons which have 3 sides (since T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.
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 than or equal to x.
Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x2
(ii) f : Z → Z given by f(x) = x2
(iii) f : R → R given by f(x) = x2
(iv) f : N → N given by f(x) = x3
(v) f : Z → Z given by f(x) = x3
Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.
Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x2
(ii) f : Z → Z given by f(x) = x2
(iii) f : R → R given by f(x) = x2
(iv) f : N → N given by f(x) = x3
(v) f : Z → Z given by f(x) = x3
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 than or equal to x.
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.
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x2
Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.