# How to find onto, one to one and everywhere defined from a formula?

I was reading functions, I came across this question, 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 rest –  Fahad Uddin Aug 25 '11 at 19:30
Well, the "one-to-one" part is easily demonstrated... you know what that means, right? –  Ｊ. Ｍ. Aug 25 '11 at 19:30
@J.M: Yes, that I get it now –  Fahad Uddin Aug 25 '11 at 19:32

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.

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Sure, for every c, I can come up with a b. I did not got your one to one part/ –  Fahad Uddin Aug 25 '11 at 19:46
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. –  Tanner Swett Aug 25 '11 at 19:54
I would be very thankful if anyone of you could give me a virtual representation through the diagram. –  Fahad Uddin Sep 11 '11 at 15:15
@fahad: see my addition. –  Ross Millikan Sep 11 '11 at 16:12
@Ross: Thanks a lot for that :) –  Fahad Uddin Sep 11 '11 at 16:18
Presumably when the problem says $A=B=Z$ it means that $A$ and $B$ are each the set of (positive, zero and negative) integers.