Mock test Q5. Surjective function

[Kalpana T J]

Have a question. Can anyone explain how option 4 is "onto"/"surjective" ?

f(n) = n/2 if n is even

        = n+1 otherwise.

Per "onto" concept every y that belongs to codomain of f has a unique pre-image x that belongs to domain of f such that y = f(x)

In option 4, f(1) and f(4) have the same value in codomain. So the pre-image x is not unique for y value of 2. So also, are the pairs f(3) and f(8).

How is this surjective ?

Can someone please help understand ?

Thanks

Kalpana

[Kalpana T J]

On the other hand, option 3 seems to be surjective.

f(n) = n if n is odd,

        = 2n otherwise.

At least this looks like an onto function. Every value in the codomain will have a unique mapping to a value in the domain.

Can someone also explain how this is NOT surjective ?

Thanks

Kalpana

[Subhrajyoti Dutta]

you have confused onto with one-one.

onto means exhausting the whole range.

whereas one-one means every input has a unique output.

[viju...@gmail.com]

Hi Kalpana,

Surjective = Every unique Output has atleast one input (it can have more than one input but it must have atleast one input). Going by this if f(1) and f(4) has output = 2, its a case of Many to One and doesnt violate the requirement of a Surjective function.

Whereas,

Option 3, doesnt return values...for example 2,6,8...which are part of Codomain of the function is NOT equal to its Range...so NOT surjective. 

Hope it helps.

Regards,

Vijender