IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - Super tough Question on limits Experts please try it too

oneyeartogo

let f be a function defined in (0,1) given by

f(x) = 0 if x is irrational

and f(x) = 1/q if x is a rational of the form p/q (p,q are positive integers and are coprime)

(a) f(x) is continuous at rational points in (0,1)

(b) f(x) is discontinuous at rational points in (0,1)

(c) f(x) is continuous at irrational points in (0,1)

(d) f(x) is discontinuous at irrational points in (0,1)

Note: More than 1 option maybe correct.

Please give an explanation of your answer.

astatine19

(b) and (d). I dont know how to explain this because, quite frankly, this question is very easy. Besides, I do recall seing this before...

oneyeartogo

No that is not correct.

I also thought that that this is the answer (this came in a test given by my teacher).

naughtygal

Is de ans (b) n (c)

oneyeartogo

I'm not saying its right or wrong unless you post the explanation

naughtygal

look I've just gueesed it .on de basis of point function.......in de ques u've written p/q form where p,q are positive integers and are coprime.which means p n q are 1 2 3 5 7 11 n so on wen u'll divide dem u'll get an irrational no ......hence my ans......now tell were I'm wrong n wat's de right ans........

ankurgupta91

see , btw two rational pts there r infinte irrational pts
at rational pts btw (0,1)
there is a break in the graph
so that lhl doesnt equal to the value of the func at that point

thus, it is discontinous at rational pts
bt continuous at irrational pts

hope u gt it .....

hsbhatt

The basic concept is that Dedekind (if I remember right) defined irrational numbers as being the limit of a series of rational numbers.

Next, there is a definition of Heine Continuity of a function, that a function f is continuous if for every sequence $x_1, x_2, x_3,..., x_n \rightarrow x$ , then $f(x_1), f(x_2), f(x_3),..., f(x_n) \rightarrow f(x)$

So, from this, if you consider a sequence of rational numbers that in the limit is an irrational number, the limit of the sequence is definitely non-zero. However we are given that f(x) = 0 if x is an irrational number.

From the above definition, the function becomes discontinuous at irrational points.

The rational points one, I am not sure, gotta think

hsbhatt

This is a shot in the dark, but consider that a rational number p/q is the limit of the sequence

$\frac{p}{2q} \left( 1 + \frac{1}{2} + \frac{1}{2^2} + ...\infty} \right)$

The numerator of the bracketed expression is an odd number and the denominator is 2ⁿ.

So, again, we see that the denominator does not tend to q for the members of the sequence.

So, for rational points also it is discontinuous.

sachinguptaiit

a very simple and very logical solution:- Answer is (b) and (d)

there are infinite rational and irrational numbers between two rational numbers,

there are infinite irrational and rational numbers between two irrational numbers

So,this kind of function cannot be continuous as the LHL RHL and f ( number ) values will be different

from LHL limit will be jumping between 1/q and 0

from RHL limit will also be jumping between 1/q and 0

cheers,,,,

oneyeartogo

@ naughtygal

coprime means that the GCD of p& q is 1. eg. 3 and 4 are coprime. 19 and 102 are coprime

@ ankurgupta91

how is it continuous at irrational.

@hsbhatt sir

"So, from this, if you consider a sequence of rational numbers that in the limit is an irrational number, the limit of the sequence is definitely non-zero"

Sir how can you say it will be non zero.

"The numerator of the bracketed expression is an odd number and the denominator is 2ⁿ.

So, again, we see that the denominator does not tend to q for the members of the sequence."

Sir please could you explain this part I did not understand it.

@sachinguptaiit

but q is not constant it mqaybe q is very large so 1/q ----------> 0

hsbhatt

Qn 1. Say you choose a sequence that has as its limit.

Obviously none of the denominators will be even tending to zero.

Qn 2. See, wehn you take LCM, every numerator except the last will be multiplied by an even number. The last numerator is 1. So, the numerator is odd.

the denom will be in general 2^kq where k is large. So, the sequence of denominators does not tend to q.

oneyeartogo

I understood how you proved that it is discontinuous at rational numbers. Its a superb proof.

but I did not understand your other post

hsbhatt

Sorry for the delay in response.

First, a small correction: the definition of irrational numbers as the limit of a sequence of rational numbers is due to Cantor (who also proved that there are infinitely more irrational numbers than rational numbers - in other words the set of real numbers consists mostly of irrational numbers)

An example is $\sqrt 2$

For this generally the sequence considered is the limit of

$1, \frac{14}{10}, \frac{141}{100}, \frac{1414}{10000}, ...$

(for $\frac{1}{\sqrt 2}$ of course the consider the reciprocals)

More precisely it is considered as the limit of the successive "convergents" obtained from the continued fraction:

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"So, from this, if you consider a sequence of rational numbers that in the limit is an irrational number, the limit of the sequence is definitely non-zero"

"The numerator of the bracketed expression is an odd number and the denominator is 2n.So, again, we see that the denominator does not tend to q for the members of the sequence."

"The numerator of the bracketed expression is an odd number and the denominator is 2ⁿ.

So, again, we see that the denominator does not tend to q for the members of the sequence."