Continuity of f(x) = 4 if rational and x^2 if irrational












0














Find the points of discontinuity of the following function



f(x) = 4 if x is a rational number and x^2 if x is an irrational number



I know that the function will only be continuous at +/- 2. But I need to prove it. I am not sue how to go about it. I am sure we need to use the sequential criterion for continuity to show that lim f(xn) <> f(c) when lim xn = c. We can take a sequence of irrationals that converges to a rational number and a sequence of rationals that converges to an irrational. Is this right?






real-analysis







share|cite|improve this question






















  • "I am sure we need to use the sequential criterion for continuity" Not at all. You can also use the $epsilon$-$delta$ criterion, or the topological criterion (dealing with open sets and preimages). There are many ways to the goal here. But your approach seems sound from the looks of it.
    – Arthur
    Nov 27 '18 at 20:46












  • Can you tell me how to do it? Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:56
















0














Find the points of discontinuity of the following function



f(x) = 4 if x is a rational number and x^2 if x is an irrational number



I know that the function will only be continuous at +/- 2. But I need to prove it. I am not sue how to go about it. I am sure we need to use the sequential criterion for continuity to show that lim f(xn) <> f(c) when lim xn = c. We can take a sequence of irrationals that converges to a rational number and a sequence of rationals that converges to an irrational. Is this right?






real-analysis







share|cite|improve this question






















  • "I am sure we need to use the sequential criterion for continuity" Not at all. You can also use the $epsilon$-$delta$ criterion, or the topological criterion (dealing with open sets and preimages). There are many ways to the goal here. But your approach seems sound from the looks of it.
    – Arthur
    Nov 27 '18 at 20:46












  • Can you tell me how to do it? Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:56














0












0








0







Find the points of discontinuity of the following function



f(x) = 4 if x is a rational number and x^2 if x is an irrational number



I know that the function will only be continuous at +/- 2. But I need to prove it. I am not sue how to go about it. I am sure we need to use the sequential criterion for continuity to show that lim f(xn) <> f(c) when lim xn = c. We can take a sequence of irrationals that converges to a rational number and a sequence of rationals that converges to an irrational. Is this right?






real-analysis







share|cite|improve this question













Find the points of discontinuity of the following function



f(x) = 4 if x is a rational number and x^2 if x is an irrational number



I know that the function will only be continuous at +/- 2. But I need to prove it. I am not sue how to go about it. I am sure we need to use the sequential criterion for continuity to show that lim f(xn) <> f(c) when lim xn = c. We can take a sequence of irrationals that converges to a rational number and a sequence of rationals that converges to an irrational. Is this right?






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 27 '18 at 20:30









Aishwarya Deore

344




344












  • "I am sure we need to use the sequential criterion for continuity" Not at all. You can also use the $epsilon$-$delta$ criterion, or the topological criterion (dealing with open sets and preimages). There are many ways to the goal here. But your approach seems sound from the looks of it.
    – Arthur
    Nov 27 '18 at 20:46












  • Can you tell me how to do it? Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:56


















  • "I am sure we need to use the sequential criterion for continuity" Not at all. You can also use the $epsilon$-$delta$ criterion, or the topological criterion (dealing with open sets and preimages). There are many ways to the goal here. But your approach seems sound from the looks of it.
    – Arthur
    Nov 27 '18 at 20:46












  • Can you tell me how to do it? Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:56
















"I am sure we need to use the sequential criterion for continuity" Not at all. You can also use the $epsilon$-$delta$ criterion, or the topological criterion (dealing with open sets and preimages). There are many ways to the goal here. But your approach seems sound from the looks of it.
– Arthur
Nov 27 '18 at 20:46






"I am sure we need to use the sequential criterion for continuity" Not at all. You can also use the $epsilon$-$delta$ criterion, or the topological criterion (dealing with open sets and preimages). There are many ways to the goal here. But your approach seems sound from the looks of it.
– Arthur
Nov 27 '18 at 20:46














Can you tell me how to do it? Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
– Aishwarya Deore
Nov 27 '18 at 20:56




Can you tell me how to do it? Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
– Aishwarya Deore
Nov 27 '18 at 20:56










2 Answers
2






active

oldest

votes


















0














note: it is not necessary to use sequential approach as pointed out in the comments. Perhaps you may have been asked to use that approach so here is what you can try.



hint



Suppose we want to check continuity at $x=c$. Choose a sequence $r_n$ of rational numbers that converge to $c$. Then for continuity, $f(r_n) rightarrow f(c)$. Similarly if we choose a sequence $t_n$ of irrational numbers that converge to $c$. Then for continuity, $f(t_n) rightarrow f(c)$.



In the case of rational sequence, $f(r_n)=4$. So this means $f(c)=4$. So if $c$ is a rational number, then we are all set because $f(c)=4$ by the definition of $f$. But then we will also need $t_n^2 rightarrow 4$. This means $c=...$



Now ask yourself, what if $c$ is irrational?






share|cite|improve this answer





















  • Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:58










  • @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
    – Anurag A
    Nov 27 '18 at 21:02












  • <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
    – Aishwarya Deore
    Nov 27 '18 at 21:07



















0














Let $a$ be a real.



$$lim_{xto a,xinBbb Q}f(x)=4$$



$$lim_{xto a,xnotin Bbb Q}f(x)=a^2$$



the function is continuous only at $x=pm 2$.






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016266%2fcontinuity-of-fx-4-if-rational-and-x2-if-irrational%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    note: it is not necessary to use sequential approach as pointed out in the comments. Perhaps you may have been asked to use that approach so here is what you can try.



    hint



    Suppose we want to check continuity at $x=c$. Choose a sequence $r_n$ of rational numbers that converge to $c$. Then for continuity, $f(r_n) rightarrow f(c)$. Similarly if we choose a sequence $t_n$ of irrational numbers that converge to $c$. Then for continuity, $f(t_n) rightarrow f(c)$.



    In the case of rational sequence, $f(r_n)=4$. So this means $f(c)=4$. So if $c$ is a rational number, then we are all set because $f(c)=4$ by the definition of $f$. But then we will also need $t_n^2 rightarrow 4$. This means $c=...$



    Now ask yourself, what if $c$ is irrational?






    share|cite|improve this answer





















    • Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
      – Aishwarya Deore
      Nov 27 '18 at 20:58










    • @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
      – Anurag A
      Nov 27 '18 at 21:02












    • <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
      – Aishwarya Deore
      Nov 27 '18 at 21:07
















    0














    note: it is not necessary to use sequential approach as pointed out in the comments. Perhaps you may have been asked to use that approach so here is what you can try.



    hint



    Suppose we want to check continuity at $x=c$. Choose a sequence $r_n$ of rational numbers that converge to $c$. Then for continuity, $f(r_n) rightarrow f(c)$. Similarly if we choose a sequence $t_n$ of irrational numbers that converge to $c$. Then for continuity, $f(t_n) rightarrow f(c)$.



    In the case of rational sequence, $f(r_n)=4$. So this means $f(c)=4$. So if $c$ is a rational number, then we are all set because $f(c)=4$ by the definition of $f$. But then we will also need $t_n^2 rightarrow 4$. This means $c=...$



    Now ask yourself, what if $c$ is irrational?






    share|cite|improve this answer





















    • Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
      – Aishwarya Deore
      Nov 27 '18 at 20:58










    • @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
      – Anurag A
      Nov 27 '18 at 21:02












    • <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
      – Aishwarya Deore
      Nov 27 '18 at 21:07














    0












    0








    0






    note: it is not necessary to use sequential approach as pointed out in the comments. Perhaps you may have been asked to use that approach so here is what you can try.



    hint



    Suppose we want to check continuity at $x=c$. Choose a sequence $r_n$ of rational numbers that converge to $c$. Then for continuity, $f(r_n) rightarrow f(c)$. Similarly if we choose a sequence $t_n$ of irrational numbers that converge to $c$. Then for continuity, $f(t_n) rightarrow f(c)$.



    In the case of rational sequence, $f(r_n)=4$. So this means $f(c)=4$. So if $c$ is a rational number, then we are all set because $f(c)=4$ by the definition of $f$. But then we will also need $t_n^2 rightarrow 4$. This means $c=...$



    Now ask yourself, what if $c$ is irrational?






    share|cite|improve this answer












    note: it is not necessary to use sequential approach as pointed out in the comments. Perhaps you may have been asked to use that approach so here is what you can try.



    hint



    Suppose we want to check continuity at $x=c$. Choose a sequence $r_n$ of rational numbers that converge to $c$. Then for continuity, $f(r_n) rightarrow f(c)$. Similarly if we choose a sequence $t_n$ of irrational numbers that converge to $c$. Then for continuity, $f(t_n) rightarrow f(c)$.



    In the case of rational sequence, $f(r_n)=4$. So this means $f(c)=4$. So if $c$ is a rational number, then we are all set because $f(c)=4$ by the definition of $f$. But then we will also need $t_n^2 rightarrow 4$. This means $c=...$



    Now ask yourself, what if $c$ is irrational?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 '18 at 20:54









    Anurag A

    25.6k12249




    25.6k12249












    • Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
      – Aishwarya Deore
      Nov 27 '18 at 20:58










    • @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
      – Anurag A
      Nov 27 '18 at 21:02












    • <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
      – Aishwarya Deore
      Nov 27 '18 at 21:07


















    • Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
      – Aishwarya Deore
      Nov 27 '18 at 20:58










    • @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
      – Anurag A
      Nov 27 '18 at 21:02












    • <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
      – Aishwarya Deore
      Nov 27 '18 at 21:07
















    Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:58




    Let us say there is a irrational sequence xn that converges to a rational number c. f(xn) = xn^2 so lim f(xn) = c^2. But f(c) = 4. So lim f(xn) <> f(c) as xn approaches c unless xn is very close to 2. Is this right? And similarly for a rational seqence converging to an irrational number
    – Aishwarya Deore
    Nov 27 '18 at 20:58












    @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
    – Anurag A
    Nov 27 '18 at 21:02






    @AishwaryaDeore I am not sure what you mean when you use the notation <> ? The idea is $x_n^2 rightarrow c^2$ but by continuity, $x_n^2$ also should converge to $f(c)=4$. So $c^2=4$.
    – Anurag A
    Nov 27 '18 at 21:02














    <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
    – Aishwarya Deore
    Nov 27 '18 at 21:07




    <> means "does not equal". Yes that is exactly what I had in mind. And c^2 = 4 only when c = +/-2
    – Aishwarya Deore
    Nov 27 '18 at 21:07











    0














    Let $a$ be a real.



    $$lim_{xto a,xinBbb Q}f(x)=4$$



    $$lim_{xto a,xnotin Bbb Q}f(x)=a^2$$



    the function is continuous only at $x=pm 2$.






    share|cite|improve this answer


























      0














      Let $a$ be a real.



      $$lim_{xto a,xinBbb Q}f(x)=4$$



      $$lim_{xto a,xnotin Bbb Q}f(x)=a^2$$



      the function is continuous only at $x=pm 2$.






      share|cite|improve this answer
























        0












        0








        0






        Let $a$ be a real.



        $$lim_{xto a,xinBbb Q}f(x)=4$$



        $$lim_{xto a,xnotin Bbb Q}f(x)=a^2$$



        the function is continuous only at $x=pm 2$.






        share|cite|improve this answer












        Let $a$ be a real.



        $$lim_{xto a,xinBbb Q}f(x)=4$$



        $$lim_{xto a,xnotin Bbb Q}f(x)=a^2$$



        the function is continuous only at $x=pm 2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 21:06









        hamam_Abdallah

        38k21634




        38k21634






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3016266%2fcontinuity-of-fx-4-if-rational-and-x2-if-irrational%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Ellipse (mathématiques)

            Image
            Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
            Read more

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more