Question

Let f be the function defined on the real line by f(x) = x/2 if x is rational and f(x) = x/3 if x is irrational. If D is the set of points of discontinuity of f, then D is the:

empty set
set of rational numbers
set of irrational numbers
set of nonzero real numbers
set of real numbers

Answer 1

It seems to me that the whole thing should be discontinuous, but that is purely from a simple view of not really knowing too rigorous a definition of continuous. I have issues with understanding continuity because it seems entirely unintuitive.

Answer 2

im inclining more toward nonzero real numbers

Answer 3

as x->0, both rational and irrational numbers make f(x) converge toward 0

Answer 4

That's the right answer, i put that the answer should be only the set of real numbers, why is 0 excluded from the set of points of discontinuity? Is it not considered rational or irrational somehow, that's sort of weird since it seems undefined.

Answer 5

*make f(x) converge to 0

Answer 6

Ahhh I see, so you're using the definition \[\Large \lim_{x \to 0}f(x) = f(0)\] then it is continuous?

Answer 7

Yes that looks simple and works! but i was thinking of sequences earlier..

Answer 8

No show me the more complicated way because I don't at all like this definition and I don't know how to do calculus, specifically limits, with sequences and series but I should really know just so I am comfortable with both examples

Answer 9

Well I guess I could try to find something online before I just ask you to teach me haha.

Answer 10

need help?

Answer 11

that sequence definition belongs to real analysis domain which you dont like lol

Answer 12

i saw eliassaab answering a similar question few days ago with nice explanation.. let me see if i can pull it up..

Answer 13

Awesome, thanks.

Answer 14

no one answered my question? :(

Answer 15

i said do you need help? @Kainui

Answer 16

Well kind of, I need help understanding continuity since by definition all we require is that the limit as we approach that point exists, which doesn't in my mind seem very continuous at all, like this example. This example basically makes the function look like two lines intersecting at x=0.

Answer 17

okay lemme try to work this out ;) @Kainui

Answer 18

one sec

Answer 19

I am actually fairly good with the concepts of epsilon neighborhoods, closed and open set definitions and simple stuff in real analysis, so I'm looking to make sure I have these definitions very precisely memorized before I take the mathematics subject test GRE.

Answer 20

so do you know what the answer is or....do you not just understand?

Answer 21

@chosenmatt have you read the discussion?

Answer 22

the definition of continuity isn't just that the limit exists ....

The limit has to be the same as the actual value of the function

Answer 23

Right, my mistake I meant to say that, I just wrote it wrong.

Answer 24

yrS @PaxPolaris is right

Answer 25

Kai - i like this explanation of @eliassaab on a similar question
http://openstudy.com/users/ganeshie8#/updates/5482a8fae4b0edc680fd4565

Answer 26

My problem with "continuity" is the fact that if our function, say f(x)=x^2 maps to the real numbers then the limit and function exist at every point.

But we can take this exact same function and take the compliment of the rationals in the reals (take all the rational numbers out) and the function is still continuous. Seems like we're taking out an infinite amount of points in our graph, so I don't really like to call it continuous.

Answer 27

i think that explains at what points the limit exists..
in our case it is easy to see that the value of limit equals the value of function at x = 0

Answer 28

Yeah, I just hate to think of continuity as being where two "ghost graphs" intersect. It just seems wrong. Oh well.

Answer 29

Exactly im in number theory mode from last couple of days.. i dont really remember much of analysis to explain a sensitive topic like continuity hmm

Answer 30

Oh well, this isn't really my biggest priority or anything, but at least this sort of got me thinking about it enough that I think I can catch myself on errors in the future with the simple limit definition. Thanks everyone. =)