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I was reading functions, I came across this question, enter image description here Next, the author has given an exercise to find out 3 things from the example,

 1. Onto  2. Everywhere defined  3. One to one 

I am stuck with how do I come to know if it has these there qualities? I mean if I had values I could have come up with an answer easily but with just a function formula, I don't get how to proceed. Please guide me.

thanks

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    I see that the domain is fully occupied so everywhere defined is true. I am stuck with the rest2011-08-25
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    Well, the "one-to-one" part is easily demonstrated... you know what that means, right?2011-08-25
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    @J.M: Yes, that I get it now2011-08-25

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A function is "onto" if for every element $c$ in the range there is an element $b$ in the domain such that $f(b)=c$. So can you determine whether this is true? If I give you a number $c$, can you find an appropriate $b$?

A function is "one to one" if $f(c)=f(d)$ implies $c=d$. $g(x)=x^2$ is not one to one because $g(2)=g(-2)$. What happens here?

Added: If the sets are finite, you can draw a diagram. You have some points that are set A and some points that are set B. If you draw a line from each point in A to its image in B,

  • Onto means that every point in B is the image of at least one point in A. So there must be a line to all the points of B
  • Everywhere defined means there is a line from each point in A. Sometimes people require that a function be everywhere defined, reducing the set A as necessary so every element has a function value.
  • One to one means that each point in B is connected to at most one point in A.

function diagrams

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    Sure, for every c, I can come up with a b. I did not got your one to one part/2011-08-25
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    A function $f$ is one-to-one if $f(c) = f(d)$ implies $c = d$. Therefore, assume that $f(c) = f(d)$, and see if you can prove that $c = d$; if so, this means that $f$ is one-to-one.2011-08-25
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    I would be very thankful if anyone of you could give me a virtual representation through the diagram.2011-09-11
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    @fahad: see my addition.2011-09-11
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    @Ross: Thanks a lot for that :)2011-09-11
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Presumably when the problem says $A=B=Z$ it means that $A$ and $B$ are each the set of (positive, zero and negative) integers.

So you now have the values and can test each of the three properties for this function.