Question -
Answer -
(i) Letf: Z → Z given by f(x) = 3x + 2
Let us check one-onecondition on f(x) = 3x + 2
Injectivity:
Let x and y be any two elements in the domain (Z), suchthat f(x) = f(y).
f (x)= f(y)
⇒ 3x + 2=3y + 2
⇒ 3x = 3y
⇒ x = y
⇒ f(x)= f(y)
⇒ x = y
So, f isone-one.
Surjectivity:
Let y be any element in the co-domain (Z), such that f(x)= y for some element x in Z(domain).
Let f(x) = y
⇒ 3x + 2 = y
⇒ 3x = y –2
⇒ x = (y –2)/3. It may not be in the domain (Z)
Because if we take y = 3,
x = (y – 2)/3 =(3-2)/3 = 1/3 ∉ domain Z.
So, for every elementin the co domain there need not be any element in the domain suchthat f(x) = y.
Thus, f is not onto.
(ii) Example for thefunction which is not one-one but onto
Letf: Z → N ∪{0} given by f(x) = |x|
Injectivity:
Let x and y be any two elements in the domain (Z),
Such that f(x)= f(y).
⇒ |x| = |y|
⇒ x = ± y
So, different elementsof domain f may give the same image.
So, f is not one-one.
Surjectivity:
Let y be any element in the co domain (Z), such that f(x)= y for some element x in Z (domain).
f(x) = y
⇒ |x| = y
⇒ x = ± y
Which is an elementin Z (domain).
So, for every element in the co-domain, there exists a pre-image in the domain.
Thus, f is onto.
(iii) Example for thefunction which is neither one-one nor onto.
Letf: Z → Z given by f(x) = 2x2 + 1
Injectivity:
Let x and y be any two elements in the domain (Z), suchthat f(x) = f(y).
f(x) = f(y)
⇒ 2x2+1 = 2y2+1
⇒ 2x2 = 2y2
⇒ x2 = y2
⇒ x = ± y
So, different elementsof domain f may give the same image.
Thus, f is not one-one.
Surjectivity:
Let y be any element in the co-domain (Z), such that f(x)= y for some element x in Z (domain).
f (x) = y
⇒ 2x2+1=y
⇒ 2x2= y − 1
⇒ x2 =(y-1)/2
⇒ x = √((y-1)/2) ∉ Z always.
For example, if we take, y = 4,
x = ± √ ((y-1)/2)
= ± √ ((4-1)/2)
= ± √ (3/2) ∉ Z
So, x may not be in Z (domain).
Thus, f isnot onto.