EPTDoes there exist a function f:R -> R that is continuous at each rational but discontinuous at each irrational?
Does anyone have an idea of what to do? I swear math wasn't this hard in high school
[H] Calculus (realy!)
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infinity21
Canada6683 Posts
Does there exist a function f:R -> R that is continuous at each rational but discontinuous at each irrational? Does anyone have an idea of what to do? I swear math wasn't this hard in high school | ||
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fight_or_flight
United States3988 Posts
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infinity21
Canada6683 Posts
R = ![[image loading]](http://upload.wikimedia.org/math/6/9/a/69a45f1e602cd2b2c2e67e41811fd226.png) | ||
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infinity21
Canada6683 Posts
speaking of which, can a mod change that? thanks x_x | ||
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fight_or_flight
United States3988 Posts
http://www.google.com/search?hl=en&q=continuous at each rational but discontinuous at each irrational?&btnG=Google Search http://www.jstor.org/view/0025570x/di021166/02p0055m/0 The answer is no. However, if you switch the words "irrational" and "rational" then the answer is yes. edit: btw this seems pretty bizarre, and interesting. What class is this? | ||
Chill
Calgary25969 Posts
On October 18 2007 11:57 infinity21 wrote: LOL i mispelled really ;; speaking of which, can a mod change that? thanks x_x Nope =] | ||
Kau
Canada3500 Posts
Edit: So I guess I'd say that such a function doesn't exist. | ||
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infinity21
Canada6683 Posts
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infinity21
Canada6683 Posts
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fight_or_flight
United States3988 Posts
On October 18 2007 12:58 infinity21 wrote: I find it interesting that when my internet's down, the only things that still work are the university website, MSN, and teamliquid.net =/ Well that sucks. Maybe this will help: + Show Spoiler + ![[image loading]](http://www.jstor.org/jstor/gifcvtdir/di001758/0025570x/di021166/02p0055m_l.1.gif?jstor) ![[image loading]](http://www.jstor.org/jstor/gifcvtdir/di001758/0025570x/di021166/02p0055m_l.2.gif?jstor) ![[image loading]](http://www.jstor.org/jstor/gifcvtdir/di001758/0025570x/di021166/02p0055m_l.3.gif?jstor) | ||
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infinity21
Canada6683 Posts
On October 18 2007 13:04 fight_or_flight wrote: Well that sucks. Maybe this will help: + Show Spoiler + Thanks for trying but it's not working too well. I'll just take a look at it when my internet starts to work again. I'm not in a hurry -- this question is a bonus question and we are allowed to submit this at any time of the term.  | ||
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Ender
United States294 Posts
I'll work on your particular problem later, but let's say we had the function g(x) = xf(x) where f(x) was 1 for x rational and 0 for x irrational. Let's say we wanted to prove that the limit as x approaches 0 of xf(x) = 0. This means for |xf(x)-0|<epsilon, there exists a 0<|x|<delta. Note that since f(x) can be 1 at the most, then we can choose delta = epsilon and our limit is proved. Now for continuity, we need to show that for each epsilon>0 there exists a delta>0 such that if 0<|x-c|<delta, then |f(x)-f(c)|<epsilon. Hope that starts you off a little. | ||
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infinity21
Canada6683 Posts
On October 18 2007 13:10 Ender wrote: Ok to show a limit at a point we must show that lim as x approaches c of f(x) = L. Now in rigorous terms this means for any epsilon>0, there exists a delta>0 such that if 0<|x-c|<delta then |f(x)-L|<epsilon. I'll work on your particular problem later, but let's say we had the function g(x) = xf(x) where f(x) was 1 for x rational and 0 for x irrational. Let's say we wanted to prove that the limit as x approaches 0 of xf(x) = 0. This means for |xf(x)-0|<epsilon, there exists a 0<|x|<delta. Note that since f(x) can be 1 at the most, then we can choose delta = epsilon and our limit is proved. Now for continuity, we need to show that for each epsilon>0 there exists a delta>0 such that if 0<|x-c|<delta, then |f(x)-f(c)|<epsilon. Hope that starts you off a little. I'm not entirely sure I understood what you said. Does f(x) simply look like y = 1 and y = 0 and g(x) look like y = x and y = 0 when graphed? (albeit with infinitely many holes) Would I have to come up with a function f(x) where f(a) is defined for all a in rationals and is not defined for all a in irrationals? | ||
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Ender
United States294 Posts
Now consider irrational numbers. Here, L = c(0) = 0. So, for any epsilon |xf(x)|<epsilon implies that |x-c|<delta. hmmm...this one's tricky...yeah according to the links other people have been giving, it seems that the irrational numbers must be continuous so there must be an epsilon that works here but i have an exam tomorrow so ill do more stuff later. | ||
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oneofthem
Cayman Islands24199 Posts
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r0ar
Australia24 Posts
there are no restrictions on the function first of all there is an irrational number between any two rational numbers so i dont really get what they mean by continous on the rationals, maybe defined on the rationals assuming that i would suggest x if x is rational x= 1/0 if x is irrational have you been given a rigorous defn of continuity yet? | ||
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LxRogue
United States1415 Posts
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infinity21
Canada6683 Posts
On October 18 2007 15:26 r0ar wrote: ok there are no restrictions on the function first of all there is an irrational number between any two rational numbers so i dont really get what they mean by continous on the rationals, maybe defined on the rationals assuming that i would suggest x if x is rational x= 1/0 if x is irrational have you been given a rigorous defn of continuity yet? fairly rigorous, I'd say. | ||
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infinity21
Canada6683 Posts
On October 18 2007 15:54 LxRogue wrote: 2^x i think qualifies? How so? It's continuous for all real numbers, including the irrationals. | ||
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LxRogue
United States1415 Posts
Well i guess that only means it doesn't exist at irrationals...not really saying anything about ccontinuity. | ||
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infinity21
Canada6683 Posts
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LxRogue
United States1415 Posts
Say for simplicity it represents root 2 as 7/5. It then finds the 5th root of 2^7. The only way to calculate this is when the exponent can be expressed as a ratio of integers, meaning it's rational. | ||
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GeneralCash
Croatia346 Posts
in mechanics, the phase space of 2 pendulums (or any oscilator with 2 degrees of freedom) in action/angle coordinates is represented by a torus where one pendulum's phase portret is the circle around the center of the torus and the other's is the outline of the vertical diametrical cross-section. representation, the pendulums orbit around the center of the torus and around the center of the cross-section forming a bent spiral. hope you understand what i'm saying, it seems my english sucks at this. anyway, it's easy to prove that if the ratio of the frequencies is rational, the trajectory in the phase space is closed, and it's not if the ratio is irrational. so a simple line itegral over the whole trajectory with both bondries at the same point should converge if the ratio is rational (divided by some system-dependant factor, it should give the number of periods it takes for for pendulums to end up in the same point of the phase space, it takes it infinite number of periods if the ratio irrational). simply put f(x)=i where i is the integral and x is the ratio of the frequencies. i have no idea how to calculate that integral or if it's even possible... meh, forget the pendulums if it's confusing, simply imagine a line moving in a spiral motion over the surface of a torus, making A turns around the center of the torus and B turns around the center of the vertical cross-section. the function is a simple superposition of sinusoidal terms (sullutions for 2 different x"+kx=0 equasions). the line integral from a to a (a being the point on the spiral) repersents the lenght of the spiral before it reaches the same point in 3d space again. obviosly, disregard the sollution that is 0 and the ones representing more than one full circle. that integral should be your function as it infinite if the ratio is irrational and periodic if it's not. hope i helped. this is the final extent of both my math and my english and i don't feel like i've helped a lot. fuck it. gonna ask around the colledge when i actualy manage to drag my ass there. and 2^x is still continuous at irrational numbers. it's expand a(n), n being the order of expansion used is a cauchy sequence (<-- corect term??) that converges within rationals. it has no discontinuities. ps. oh how i hate math. oh how i hate mechanics. man, this brought back a lot of painful memories. sure hope it was worth it... | ||
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betaben
681 Posts
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Zanno
United States1484 Posts
On October 18 2007 12:51 Kau wrote: an infinite number of polynomials is called a taylor series and those functions collapse into regular functions provided they can exist.I think that theoretically you could have a function that is discontinuous at every irrational number, but it would require an infinite amount of polynomials. For example if you have the function 1/x, it is continuous at every number except 0. If you have the function 1/(x-pi), your function is continuous at every number except the irrational number pi. Now suppose you did that for every irrational number (we'll call an irrational number U). You'd have 1/(x-U_1)(x-U_2)...(x-U_n), thus being continuous at every rational number, but discontinuous at every irrational number. The only problem is that there are infinitely many irrational numbers. Edit: So I guess I'd say that such a function doesn't exist. | ||
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Muirhead
United States556 Posts
The standard proof involves first showing that the set of points at which f is continuous is a G_delta set (i.e. a countable intersection of open sets) To do this, let K_n be the set of points x for which there exists some delta such that |f(x)-f(y)|<1/n whenever |x-y|<delta Each K_n is open and the set of points where f is continuous is the intersection of the K_n Now, you must show that the rationals are not a G_delta set. First of all, notice that the set of irrationals is a G_delta set ( it is the intersection over all rational q of {all reals number except for q} ) The problem is reduced to the following general fact: If the set of all real numbers is the disjoint union of X and Y, where both X and Y are dense, then at most one of X,Y is G_delta. This follows directly from the Baire Category Theorem, which is very well-known and you can look up online. I hope this helps... ask if you need any clarification | ||
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