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I am not sure about my answers, I got:

D= (-infinity, 0) U (0, infinity)

Image= (o, infinity)

Codomain= Set of all reals

I dont think this is right...can anyone help?

- Thread starter jj0424
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- #1

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I am not sure about my answers, I got:

D= (-infinity, 0) U (0, infinity)

Image= (o, infinity)

Codomain= Set of all reals

I dont think this is right...can anyone help?

- #2

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The codomain is what may possibly come out. What might possibly come out when you put 1 over a rational number? Think about it by writing out a few examples. Hint: It's not the set of all Real Numbers? How do you get sqrt(2)?

As for the image... well, certainly any image will be a subset of the codomain (if it actually happened, it has to be possible, right?) However, out of the possible numbers, are there any that don't happen? Why? Can you get 2? Can you get 0?

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- #4

matt grime

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Rant over.

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matt grime

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First, f cannot be a function from Q to Q (rationals to rationals). Obviously we cannot divide by zero. Let Q* be the rational numbers excluding zero.

That f is now a well defined function from Q* to Q. What is its range? (Hint if x is non-zero, is 1/x non-zero?).

- #6

HallsofIvy

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Not so. It is standard that if the domain of a function is NOT given explicitely as "part of its definition", then it is the largest set on which the the given formula can be calculated. If someone were to say "the function f(x)= x

Rant over.

- #7

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I think the domain is part of the definition of a function. However, I also believe that when it is not given, it is implicitly "whatever works". Perhaps the question could be worded in slightly less colloquial terms by asking "what is the largest set of numbers for which the given function is defined?"

I don't know. I think the question is reasonable.

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matt grime

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But Halls, you've just introduced a huge load of subjectivity into the question. There is no such thing as the largest set on which it can be calculated. Shall we take the union of the proper class of such objects? Here, are we supposed to use R, C, Q, A, Z, an extension of F_p, an elliptic curve,... as our starting point? All have notions of rationality. Yes, I am playing devil's advocate, since it is clear what the question *intends*, but that does not stop it being a stupid thing to ask. It is also trivial to make the question well founded.

Students should not be required to make such leaps as to guess what is intended when people make non-mathematical statements. And several generations of students have been taught something that is just plain wrong.

If the domain is not given then it cannot technically be a function.

If you don't believe this to be a problem, then just consider that one of the standard examples they'll have to do will involve square rooting something like 1-x^2. What's the domain now? They'll be required to say [-1,1], only to be told about complex numbers in the next course meaning that the same question now has two answers. Which are they to pick if asked again?

Students should not be required to make such leaps as to guess what is intended when people make non-mathematical statements. And several generations of students have been taught something that is just plain wrong.

If the domain is not given then it cannot technically be a function.

If you don't believe this to be a problem, then just consider that one of the standard examples they'll have to do will involve square rooting something like 1-x^2. What's the domain now? They'll be required to say [-1,1], only to be told about complex numbers in the next course meaning that the same question now has two answers. Which are they to pick if asked again?

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matt grime

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But it is not so clear on what domain the inverse will be continuous

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matt grime

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That is neither here nor there. It is still a function. What you need to restrict to to get a function with some particular property is not the issue.But it is not so clear on what domain the inverse will be continuous

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matt grime

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The answer: domain all of Q, codomain the Riemann sphere is a perfectly respectable answer if you happen to know what the Riemann sphere is. Don't stand for this abuse! Just 'cos it's in the book doesn't mean you have to teach it.

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... I think you're belaboring the point a bit.

- #16

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Right, but usually the thing you're trying to prove is which set the function has a continuous inverse on.That is neither here nor there. It is still a function. What you need to restrict to to get a function with some particular property is not the issue.

I'm not trying to say you're wrong, obviously you can wiggle your way around the problem by defining a bunch of sets, but the point is that the terminology is in use, and apparently some mathematicians like it.

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How would you ask the same question? Just curious.

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