f(x)=0/0 is not undefined per se... see Hospital rule

0/0 is undefined. L''Hopital''s rule is for finding limits, but we are not talking about limits here.

The AP way up above gave the answer you''re looking for, although they didn''t explain it particularly clearly, which may be why it was overlooked.

A function f is onto if and only if, for all y in the range, there exists an x in the domain that maps to y through f .

Thus, to prove that f is onto , you must show that for all y satisfying x =f -1(y ), the resultant x satisfies y =f (x ). Given that there is some tedious rearrangement required for this proof, given this example, I''ll leave that up to you!

Cheers,

Timkin

However, as pacman2003 says, for A=2 and y=-1, there is no x such that f(x)=y (unless you cancel out the factor (x-1)).

When A=2, x=1 cannot be considered to be part of the domain, because f(x=1) is not defined. Thus, it''s irrelevant unless you find a value for y such that f-1(y) = 1... then you have proved that f is NOT ''onto'', because that point y maps onto a point that is not a part of the domain.

It seems to me that there is a fundamental misunderstanding of what ''onto'' means in this thread and how you must deal with points that are and are not in the domain. Think of it this way...

D & C are manifolds. f maps points on D to points on C. f is 1-1 if, obviously, each point in D is mapped to exactly 1 exclusive point in C. f is ''onto'' IF all points in C map to points in D via the mapping f-1 (if its 1-1 and onto then each point in D has exactly 1 point in C that it maps to and all points on both manifolds are connected via f... a straight line in any number of dimensions is a perfect example of this). It is completely irrelevant that D may be a subset of R (the reals) in that some points in R may not exist in D (which is the case in this problem for A=2,x=1).

I hope this helps clear things up...

Timkin

quote:Original post by Timkin
When A=2, x=1 cannot be considered to be part of the domain, because f(x=1) is not defined. Thus, it''s irrelevant unless you find a value for y such that f^-1(y) = 1... then you have proved that f is NOT ''onto'', because that point y maps onto a point that is not a part of the domain.[/qoute]

That sounds weird. Can you give an example of a function that is not ''onto'' in that sense? As I understand it being ''onto'' is the same as being a surjection. Would you agree with that?

quote:Original post by davepermen
it''s not required, that, for every input, you get an output. it''s only required that every possible output has to be reached with some input.

What do you mean by "every possible output"? If you mean the output you can actually get from the function (its range), then what you say is tautologically true for every function. If you by "every possible output" mean all of the real numbers then it is not true for y=-1 above.

I''m sorry to contradict most people in this thread, but an object that is not a function cannot be an onto function. A function f from X to Y is, by definition, a set of pairs (x,y), x in X, y in Y such that every element in X appears in one and only one pair. Then we introduce convinient notation, like f(x) to denote the element of Y that appears in the pair in which x appears.

If f(x) is not defined for some real value x, that means that f is not a function from R to R, so it can''t possibly be an onto function from R to R.

We can change the problem a little bit and say that a function f:A -> R, where A is a subset of R, is given by some formula for all real numbers in which the formula makes sense.

The problem in the book was just poorly worded.

quote:Original post by alvaro
I'm sorry to contradict most people in this thread, but an object that is not a function cannot be an onto function.

I don't believe anyone is talking about non-functions being onto... we're discussing functions being onto... if you want to raise the point that the original question, as posed, does not present a function, that's fine... but don't accuse people of being wrong without citing specifics please.

quote:Original post by Anonymous Poster
As I understand it being 'onto' is the same as being a surjection. Would you agree with that?

Yes, the words are used interchangeably.

quote:Original post by Anonymous Poster
That sounds weird. Can you give an example of a function that is not 'onto' in that sense?

It sounds weird probably because it's not the typical way in which something is described as being 'onto' (or not 'onto'). I was trying to point out that if such an unusual situation arose, you would know that the function was not onto. Typically, one would say that a function is not 'onto' if there exists values 'a' in the domain and 'b' in the range for which f(a)=b and for which f-1(b)!=a (I'm using != to mean not equal to). A function would also not be onto if there were values a and b such that f-1(b)=a and 'b' was an element of the range but 'a' was not an element of the domain. Of course, f wouldn't be a function in this case either, so I probably chose a bad way to explain what I meant. My apologies if it caused confusion. What I was trying to get at was that we can ignore points that are not in the domain when we are considering whether a mapping (function) is 'onto'.

Cheers,

Timkin
Sure can...



[edited by - Timkin on January 28, 2004 7:55:54 PM]

quote:Original post by Timkin
Typically, one would say that a function is not ''onto'' if there exists values ''a'' in the domain and ''b'' in the range for which f(a)=b and for which f^-1(b)!=a (I''m using != to mean not equal to).

That still sounds weird to me. If there exists an inverse for f(x), I can''t see how that condition could ever be met, as it seems to contradict the very definition of an inverse function. And if an inverse doesn''t exist (as for most functions), the condition would not even be applicable. Could you give an example of a function that is not ''onto'' according to that condition?