Bijective ? AQ1.9 Q6 - Unable to understand the solution

[Vijayanandam Vm]

How the given solution is Bijective for Q6 in AQ1.9 

1 point

Suppose f:Z→Zf:Z→Z is a function defined by f(x)=Ax+Bf(x)=Ax+B. For which of the following integer values of AA and BB, is the given function bijective?

A={−1,1},B={z|z∈Z}

[Anand Iyer]

Case 1:

f = -x + b

This is injective since this is a linear function.  It's surjective for the same reason; no two x-values can generate the same f-value.

Case 2:

f = x + b

f is injective and surjective, for the same reason as in Case 1.

Thus, it's bijective in both cases.

[Vijayanandam Vm]

Thanks Anand 

[Malabika Guha Mustafi]

@ Anand, It's surjective for the same reason; no two x-values can generate the same f-value.

are you taking this condition to prove surjectivity or injectivity ?

[Anand Iyer]

I meant surjectivity, but the statement got mixed up.  Sorry.

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Cheers,

Anand

[Anand Iyer]

Most linear functions are injective and surjective unless the domain is restricted...

Hope that's understandable @Vijayanandam Vm 

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Cheers,

Anand

[Malabika Guha Mustafi]

Ok, no problem :).

@ Vijayanandam

To check surjectivity, easiest way check whether range( the actual output of function means y, if y is a function of x)= codomain ( the domain of y is defined). Range should cover the codomain to be a surjective function.If codomain is natural number then range should take all natural numbers not few of them.

[Mohit]

In the same problem where we have a function 
f(x)=Ax+B, we have another option 

B=0,A={z|z∈Z}  

Why is this NOT bijective. 

[Malabika Guha Mustafi]

When B =0 , but when B is not equal 0? 

[Anand Iyer]

Consider one instance of this, y = 2x.  (Note specifcally that the domain/co-domain is Z.)

it's injective alright.  

but, this is not surjective.  because y can only be even numbers as per this definition.  So, the range of y doesn't cover the entire co-domain.

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Cheers,

Anand

[Malabika Guha Mustafi]

@ Anand,

 Good catch :)