Proving that $g(x)$ defined as $x^2$ on the rationals and $x^4$ on irrationals, is discontinuous at $2$












1












$begingroup$


Define $g: mathbb{R} to mathbb{R}$ by $$g(x) =begin{cases} x^2 & text{if } x text{ is rational}, \\ x^4 & text{if } x text{ is irrational}. end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.



Solution Attempt:



Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_{xto2}g(x)=4$;



so taking $epsilon=1$, there is a $delta>0$ such that if $0<|x-2|<delta$, then $big|g(x)-4big|<1$.



Therefore if $x$ is irrational and $2-delta<x<2$, then
$big|x^4-4big|<1$



$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac{1}{4} implies x < 2$ Contradiction






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  • 1




    $begingroup$
    Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
    $endgroup$
    – Julian Mejia
    20 hours ago








  • 1




    $begingroup$
    $x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
    $endgroup$
    – Julian Mejia
    20 hours ago
















1












$begingroup$


Define $g: mathbb{R} to mathbb{R}$ by $$g(x) =begin{cases} x^2 & text{if } x text{ is rational}, \\ x^4 & text{if } x text{ is irrational}. end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.



Solution Attempt:



Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_{xto2}g(x)=4$;



so taking $epsilon=1$, there is a $delta>0$ such that if $0<|x-2|<delta$, then $big|g(x)-4big|<1$.



Therefore if $x$ is irrational and $2-delta<x<2$, then
$big|x^4-4big|<1$



$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac{1}{4} implies x < 2$ Contradiction






real-analysis







share|cite|improve this question









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SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    $begingroup$
    Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
    $endgroup$
    – Julian Mejia
    20 hours ago








  • 1




    $begingroup$
    $x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
    $endgroup$
    – Julian Mejia
    20 hours ago














1












1








1





$begingroup$


Define $g: mathbb{R} to mathbb{R}$ by $$g(x) =begin{cases} x^2 & text{if } x text{ is rational}, \\ x^4 & text{if } x text{ is irrational}. end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.



Solution Attempt:



Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_{xto2}g(x)=4$;



so taking $epsilon=1$, there is a $delta>0$ such that if $0<|x-2|<delta$, then $big|g(x)-4big|<1$.



Therefore if $x$ is irrational and $2-delta<x<2$, then
$big|x^4-4big|<1$



$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac{1}{4} implies x < 2$ Contradiction






real-analysis







share|cite|improve this question









New contributor




SHajs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$




Define $g: mathbb{R} to mathbb{R}$ by $$g(x) =begin{cases} x^2 & text{if } x text{ is rational}, \\ x^4 & text{if } x text{ is irrational}. end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.



Solution Attempt:



Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $;;displaystyle lim_{xto2}g(x)=4$;



so taking $epsilon=1$, there is a $delta>0$ such that if $0<|x-2|<delta$, then $big|g(x)-4big|<1$.



Therefore if $x$ is irrational and $2-delta<x<2$, then
$big|x^4-4big|<1$



$-1 <x^4-4<1implies 3<x^4<5implies x^4>3 implies x = 3^frac{1}{4} implies x < 2$ Contradiction






real-analysis




real-analysis






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edited 18 hours ago









Asaf Karagila♦

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asked 20 hours ago









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  • 1




    $begingroup$
    Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
    $endgroup$
    – Julian Mejia
    20 hours ago








  • 1




    $begingroup$
    $x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
    $endgroup$
    – Julian Mejia
    20 hours ago














  • 1




    $begingroup$
    Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
    $endgroup$
    – Julian Mejia
    20 hours ago








  • 1




    $begingroup$
    $x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
    $endgroup$
    – Julian Mejia
    20 hours ago








1




1




$begingroup$
Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
$endgroup$
– Julian Mejia
20 hours ago






$begingroup$
Instead of delta-epsilon definition. See the definition using limit of sequences. Note that if you approximate 2 by a sequence of rational numbers you get g(x) approaches 2^2=4. But if you approximate 2 by a sequence of irrational numbers then g(x) approaches 2^4=16. Since the limits of g(x) w.r.t. these sequences don't match then g is discontinuous at 2. (In fact same argument proves that g is discontinuous everywhere, except at 0)
$endgroup$
– Julian Mejia
20 hours ago






1




1




$begingroup$
$x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
$endgroup$
– Julian Mejia
20 hours ago




$begingroup$
$x<2$ is not a contradiction. What you should note is that you can choose $delta$ arbitrarily small, so choose $delta$ such that $3^{1/4}<2-delta$. Then you get a contradiction.
$endgroup$
– Julian Mejia
20 hours ago










4 Answers
4






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$begingroup$

Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
$$-sqrt[4]{5} < x < sqrt[4]{5}.$$
If you had chosen instead an irrational $x$ such that
$$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



    Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




    Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.







    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      An option?



      Sequential definition of continuity .



      $x_n=2(4n)^{(1/(4n))}.$



      $lim_{n rightarrow infty}x_n=2.$



      $lim_{n rightarrow infty}g(x_n)=$



      $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



      $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



      $=2^4 not =g(2)$.






      share|cite|improve this answer











      $endgroup$





















        0












        $begingroup$

        You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



        $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



        Observe the above sequence is irrational:



        $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



        yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...






        share|cite|improve this answer









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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
          $$-sqrt[4]{5} < x < sqrt[4]{5}.$$
          If you had chosen instead an irrational $x$ such that
          $$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
          then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.






          share|cite|improve this answer









          $endgroup$


















            1












            $begingroup$

            Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
            $$-sqrt[4]{5} < x < sqrt[4]{5}.$$
            If you had chosen instead an irrational $x$ such that
            $$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
            then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.






            share|cite|improve this answer









            $endgroup$
















              1












              1








              1





              $begingroup$

              Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
              $$-sqrt[4]{5} < x < sqrt[4]{5}.$$
              If you had chosen instead an irrational $x$ such that
              $$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
              then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.






              share|cite|improve this answer









              $endgroup$



              Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
              $$-sqrt[4]{5} < x < sqrt[4]{5}.$$
              If you had chosen instead an irrational $x$ such that
              $$2 - min{delta, 2 - sqrt[4]{5}} < x < 2,$$
              then you'd have $x > 2 - (2 - sqrt[4]{5}) = sqrt[4]{5}$, a contradiction.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 20 hours ago









              Theo BenditTheo Bendit

              20.5k12354




              20.5k12354























                  1












                  $begingroup$

                  I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



                  Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




                  Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.







                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



                    Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




                    Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.







                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



                      Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




                      Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.







                      share|cite|improve this answer









                      $endgroup$



                      I think a simpe way is to construct two sequences $a_n$ and $b_n$, where $a_n$ are rational and $b_n$ are irrational, such that $lim_{ntoinfty}g(a_n)=2^2$ and $lim_{ntoinfty}g(b_n)=2^4$.



                      Then we can say $lim_{xto2^+}g(x)$ does not exist. Hence $g$ is discontinuous at $x=2$.




                      Put $a_n = 2 + frac{1}{n}$ and $b_n=2+frac{pi}{n}$.








                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 19 hours ago









                      zongxiang yizongxiang yi

                      352110




                      352110























                          1












                          $begingroup$

                          An option?



                          Sequential definition of continuity .



                          $x_n=2(4n)^{(1/(4n))}.$



                          $lim_{n rightarrow infty}x_n=2.$



                          $lim_{n rightarrow infty}g(x_n)=$



                          $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



                          $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



                          $=2^4 not =g(2)$.






                          share|cite|improve this answer











                          $endgroup$


















                            1












                            $begingroup$

                            An option?



                            Sequential definition of continuity .



                            $x_n=2(4n)^{(1/(4n))}.$



                            $lim_{n rightarrow infty}x_n=2.$



                            $lim_{n rightarrow infty}g(x_n)=$



                            $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



                            $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



                            $=2^4 not =g(2)$.






                            share|cite|improve this answer











                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              An option?



                              Sequential definition of continuity .



                              $x_n=2(4n)^{(1/(4n))}.$



                              $lim_{n rightarrow infty}x_n=2.$



                              $lim_{n rightarrow infty}g(x_n)=$



                              $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



                              $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



                              $=2^4 not =g(2)$.






                              share|cite|improve this answer











                              $endgroup$



                              An option?



                              Sequential definition of continuity .



                              $x_n=2(4n)^{(1/(4n))}.$



                              $lim_{n rightarrow infty}x_n=2.$



                              $lim_{n rightarrow infty}g(x_n)=$



                              $lim_{n rightarrow infty} 2^4(4n)^{(1/n)}=$



                              $lim_{n rightarrow infty}2^4 cdot 4^{(1/n)} cdot n^{(1/n)}=$



                              $=2^4 not =g(2)$.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 19 hours ago

























                              answered 19 hours ago









                              Peter SzilasPeter Szilas

                              11.8k2822




                              11.8k2822























                                  0












                                  $begingroup$

                                  You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



                                  $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



                                  Observe the above sequence is irrational:



                                  $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



                                  yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



                                    $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



                                    Observe the above sequence is irrational:



                                    $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



                                    yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



                                      $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



                                      Observe the above sequence is irrational:



                                      $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



                                      yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...






                                      share|cite|improve this answer









                                      $endgroup$



                                      You haven't reached a contradiction, as commented...but the general idea is correct. Try this, instead: take



                                      $$left{a_n:=2pifrac n{pi n+1}right}_{ninBbb N}$$



                                      Observe the above sequence is irrational:



                                      $$a_nnotinBbb Qimplies g(a_n)=a_n^4,,,,text{so};;lim_{ntoinfty}g(a_n)=lim_{ntoinfty}16pi^4frac{n^4}{(npi+1)^4}=16pi^4frac1{pi^4}=16$$



                                      yet $;g(2)=2^2=4;$ , so the function is not continuous at $;2;$ ...







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 19 hours ago









                                      DonAntonioDonAntonio

                                      180k1494233




                                      180k1494233






















                                          SHajs is a new contributor. Be nice, and check out our Code of Conduct.










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                                          How to label and detect the document text images

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                                          1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C...
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                                          Vallis Paradisi

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                                          Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para...
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                                          Tabula Rosettana

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                                          Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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