Give an example of a function Which is one-one but not onto.

Asked by Sakshi | 1 year ago |  32

##### Solution :-

Let f: Z → Z given by f(x) = 3x + 2

Let us check one-one condition on f(x) = 3x + 2

Injectivity:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

f (x) = f(y)

⇒ 3x + 2 =3y + 2

⇒ 3x = 3y

⇒ x = y

⇒ f(x) = f(y)

⇒ x = y

So, f is 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).

Let f(x) = y

⇒ 3x + 2 = y

⇒ 3x = y – 2

⇒ x = $$\dfrac{ (y – 2)}{3}$$. It may not be in the domain (Z)

Because if we take y = 3,

x =$$\dfrac{ (y – 2)}{3}$$

$$\dfrac{3-2}{3}$$

= $$\dfrac{1}{3}$$ ∉ domain Z.

So, for every element in the co domain there need not be any element in the domain such that f(x) = y.

Thus, f is not onto.

Answered by Aaryan | 1 year ago

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