Proof $xin R^times wedge bin R^times Rightarrow abin R^times$ [closed]












0












$begingroup$



Let $R$ be a communitative ring. Prove $ain R^times wedge bin R^times >Rightarrow abin R^times$ with $R^times := {xin R | x text{ >invertible}}.$




Do you have any ideas and tips on how I could prove this?
I know that this isn't really difficult but as so often in
proofs concerning algebraic groups you need the right beginning
and I can't get it...



Thanks!






algebraic-groups







share|cite|improve this question











$endgroup$



closed as off-topic by Adrian Keister, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus Nov 28 '18 at 6:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    What do you know about $a$?
    $endgroup$
    – Fakemistake
    Nov 27 '18 at 19:42










  • $begingroup$
    sorry, misspelling. Edited it
    $endgroup$
    – kjwemke13
    Nov 27 '18 at 19:43










  • $begingroup$
    Consider the element $a^{-1}b^{-1}$
    $endgroup$
    – Fakemistake
    Nov 27 '18 at 19:54
















0












$begingroup$



Let $R$ be a communitative ring. Prove $ain R^times wedge bin R^times >Rightarrow abin R^times$ with $R^times := {xin R | x text{ >invertible}}.$




Do you have any ideas and tips on how I could prove this?
I know that this isn't really difficult but as so often in
proofs concerning algebraic groups you need the right beginning
and I can't get it...



Thanks!






algebraic-groups







share|cite|improve this question











$endgroup$



closed as off-topic by Adrian Keister, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus Nov 28 '18 at 6:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    What do you know about $a$?
    $endgroup$
    – Fakemistake
    Nov 27 '18 at 19:42










  • $begingroup$
    sorry, misspelling. Edited it
    $endgroup$
    – kjwemke13
    Nov 27 '18 at 19:43










  • $begingroup$
    Consider the element $a^{-1}b^{-1}$
    $endgroup$
    – Fakemistake
    Nov 27 '18 at 19:54














0












0








0


1



$begingroup$



Let $R$ be a communitative ring. Prove $ain R^times wedge bin R^times >Rightarrow abin R^times$ with $R^times := {xin R | x text{ >invertible}}.$




Do you have any ideas and tips on how I could prove this?
I know that this isn't really difficult but as so often in
proofs concerning algebraic groups you need the right beginning
and I can't get it...



Thanks!






algebraic-groups







share|cite|improve this question











$endgroup$





Let $R$ be a communitative ring. Prove $ain R^times wedge bin R^times >Rightarrow abin R^times$ with $R^times := {xin R | x text{ >invertible}}.$




Do you have any ideas and tips on how I could prove this?
I know that this isn't really difficult but as so often in
proofs concerning algebraic groups you need the right beginning
and I can't get it...



Thanks!






algebraic-groups




algebraic-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 23:00









greedoid

40.2k114799




40.2k114799










asked Nov 27 '18 at 19:40









kjwemke13kjwemke13

72




72




closed as off-topic by Adrian Keister, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus Nov 28 '18 at 6:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Adrian Keister, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus Nov 28 '18 at 6:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, Saad, Brahadeesh, Lord Shark the Unknown, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    What do you know about $a$?
    $endgroup$
    – Fakemistake
    Nov 27 '18 at 19:42










  • $begingroup$
    sorry, misspelling. Edited it
    $endgroup$
    – kjwemke13
    Nov 27 '18 at 19:43










  • $begingroup$
    Consider the element $a^{-1}b^{-1}$
    $endgroup$
    – Fakemistake
    Nov 27 '18 at 19:54


















  • $begingroup$
    What do you know about $a$?
    $endgroup$
    – Fakemistake
    Nov 27 '18 at 19:42










  • $begingroup$
    sorry, misspelling. Edited it
    $endgroup$
    – kjwemke13
    Nov 27 '18 at 19:43










  • $begingroup$
    Consider the element $a^{-1}b^{-1}$
    $endgroup$
    – Fakemistake
    Nov 27 '18 at 19:54
















$begingroup$
What do you know about $a$?
$endgroup$
– Fakemistake
Nov 27 '18 at 19:42




$begingroup$
What do you know about $a$?
$endgroup$
– Fakemistake
Nov 27 '18 at 19:42












$begingroup$
sorry, misspelling. Edited it
$endgroup$
– kjwemke13
Nov 27 '18 at 19:43




$begingroup$
sorry, misspelling. Edited it
$endgroup$
– kjwemke13
Nov 27 '18 at 19:43












$begingroup$
Consider the element $a^{-1}b^{-1}$
$endgroup$
– Fakemistake
Nov 27 '18 at 19:54




$begingroup$
Consider the element $a^{-1}b^{-1}$
$endgroup$
– Fakemistake
Nov 27 '18 at 19:54










3 Answers
3






active

oldest

votes


















1












$begingroup$

Here's a beginning: “Suppose that $a,b in R^times$.”



Here's the end: “Therefore, $ab in R^times$.



The middle is up to you. For inspiration, you might try examples with $R=mathbb{Z}$ or $mathbb{Q}$. What is the inverse of a product of numbers? Is it expressible in terms of the inverses of the factors?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
    $endgroup$
    – kjwemke13
    Nov 27 '18 at 19:48












  • $begingroup$
    The inverse elements of $a,b$ exists by assumption!
    $endgroup$
    – Fakemistake
    Nov 27 '18 at 20:03










  • $begingroup$
    @kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
    $endgroup$
    – Matthew Leingang
    Nov 27 '18 at 20:19



















0












$begingroup$

With



$a, b in R^times, tag 1$



we have



$c, d in R^times tag 2$



with



$ac = bd = 1_R, tag 3$



where $1_R$ is the multiplicative identity of $R$; then



$(ab)(cd) = a(bc)d = a(cb)d = (ac)(bd) = 1_R 1_R = 1_R, tag 4$



that is,



$ab, cd in R^times. tag 5$



$OEDelta$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Since $a,bin R^times $, exsist $a^{-1}$ and $b^{-1}$ in $R$. Let $c = b^{-1}a^{-1}$, then $$(ab)cdot c = aunderbrace {bcdot b^{-1}}_{=e}a^{-1} = a cdot a^{-1} = e$$



    so $abin R^times $.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
      $endgroup$
      – Gykonik
      Nov 27 '18 at 20:38












    • $begingroup$
      $(ab)c =e$ what does that say about $ab$? @Gykonik
      $endgroup$
      – greedoid
      Nov 27 '18 at 20:40












    • $begingroup$
      ehm, that $(ab)=c^{-1}$?
      $endgroup$
      – Gykonik
      Nov 27 '18 at 20:43










    • $begingroup$
      So...............
      $endgroup$
      – greedoid
      Nov 27 '18 at 20:44










    • $begingroup$
      $c^{-1}$ and $c$ has to be in $R^times$?
      $endgroup$
      – Gykonik
      Nov 27 '18 at 20:44




















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Here's a beginning: “Suppose that $a,b in R^times$.”



    Here's the end: “Therefore, $ab in R^times$.



    The middle is up to you. For inspiration, you might try examples with $R=mathbb{Z}$ or $mathbb{Q}$. What is the inverse of a product of numbers? Is it expressible in terms of the inverses of the factors?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
      $endgroup$
      – kjwemke13
      Nov 27 '18 at 19:48












    • $begingroup$
      The inverse elements of $a,b$ exists by assumption!
      $endgroup$
      – Fakemistake
      Nov 27 '18 at 20:03










    • $begingroup$
      @kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
      $endgroup$
      – Matthew Leingang
      Nov 27 '18 at 20:19
















    1












    $begingroup$

    Here's a beginning: “Suppose that $a,b in R^times$.”



    Here's the end: “Therefore, $ab in R^times$.



    The middle is up to you. For inspiration, you might try examples with $R=mathbb{Z}$ or $mathbb{Q}$. What is the inverse of a product of numbers? Is it expressible in terms of the inverses of the factors?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
      $endgroup$
      – kjwemke13
      Nov 27 '18 at 19:48












    • $begingroup$
      The inverse elements of $a,b$ exists by assumption!
      $endgroup$
      – Fakemistake
      Nov 27 '18 at 20:03










    • $begingroup$
      @kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
      $endgroup$
      – Matthew Leingang
      Nov 27 '18 at 20:19














    1












    1








    1





    $begingroup$

    Here's a beginning: “Suppose that $a,b in R^times$.”



    Here's the end: “Therefore, $ab in R^times$.



    The middle is up to you. For inspiration, you might try examples with $R=mathbb{Z}$ or $mathbb{Q}$. What is the inverse of a product of numbers? Is it expressible in terms of the inverses of the factors?






    share|cite|improve this answer









    $endgroup$



    Here's a beginning: “Suppose that $a,b in R^times$.”



    Here's the end: “Therefore, $ab in R^times$.



    The middle is up to you. For inspiration, you might try examples with $R=mathbb{Z}$ or $mathbb{Q}$. What is the inverse of a product of numbers? Is it expressible in terms of the inverses of the factors?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 '18 at 19:44









    Matthew LeingangMatthew Leingang

    16.3k12244




    16.3k12244












    • $begingroup$
      Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
      $endgroup$
      – kjwemke13
      Nov 27 '18 at 19:48












    • $begingroup$
      The inverse elements of $a,b$ exists by assumption!
      $endgroup$
      – Fakemistake
      Nov 27 '18 at 20:03










    • $begingroup$
      @kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
      $endgroup$
      – Matthew Leingang
      Nov 27 '18 at 20:19


















    • $begingroup$
      Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
      $endgroup$
      – kjwemke13
      Nov 27 '18 at 19:48












    • $begingroup$
      The inverse elements of $a,b$ exists by assumption!
      $endgroup$
      – Fakemistake
      Nov 27 '18 at 20:03










    • $begingroup$
      @kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
      $endgroup$
      – Matthew Leingang
      Nov 27 '18 at 20:19
















    $begingroup$
    Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
    $endgroup$
    – kjwemke13
    Nov 27 '18 at 19:48






    $begingroup$
    Do you mean $(ab)^{-1}=a^{-1} cdot {b^-1}$ and therefore $a$ and $b$ has to be in $R^times$ because of the definition of a ring?
    $endgroup$
    – kjwemke13
    Nov 27 '18 at 19:48














    $begingroup$
    The inverse elements of $a,b$ exists by assumption!
    $endgroup$
    – Fakemistake
    Nov 27 '18 at 20:03




    $begingroup$
    The inverse elements of $a,b$ exists by assumption!
    $endgroup$
    – Fakemistake
    Nov 27 '18 at 20:03












    $begingroup$
    @kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
    $endgroup$
    – Matthew Leingang
    Nov 27 '18 at 20:19




    $begingroup$
    @kjwemke13: By your definition, $R^times$ is the set of invertible elements of $R$. So you have to show that $ab$ is invertible.
    $endgroup$
    – Matthew Leingang
    Nov 27 '18 at 20:19











    0












    $begingroup$

    With



    $a, b in R^times, tag 1$



    we have



    $c, d in R^times tag 2$



    with



    $ac = bd = 1_R, tag 3$



    where $1_R$ is the multiplicative identity of $R$; then



    $(ab)(cd) = a(bc)d = a(cb)d = (ac)(bd) = 1_R 1_R = 1_R, tag 4$



    that is,



    $ab, cd in R^times. tag 5$



    $OEDelta$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      With



      $a, b in R^times, tag 1$



      we have



      $c, d in R^times tag 2$



      with



      $ac = bd = 1_R, tag 3$



      where $1_R$ is the multiplicative identity of $R$; then



      $(ab)(cd) = a(bc)d = a(cb)d = (ac)(bd) = 1_R 1_R = 1_R, tag 4$



      that is,



      $ab, cd in R^times. tag 5$



      $OEDelta$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        With



        $a, b in R^times, tag 1$



        we have



        $c, d in R^times tag 2$



        with



        $ac = bd = 1_R, tag 3$



        where $1_R$ is the multiplicative identity of $R$; then



        $(ab)(cd) = a(bc)d = a(cb)d = (ac)(bd) = 1_R 1_R = 1_R, tag 4$



        that is,



        $ab, cd in R^times. tag 5$



        $OEDelta$.






        share|cite|improve this answer









        $endgroup$



        With



        $a, b in R^times, tag 1$



        we have



        $c, d in R^times tag 2$



        with



        $ac = bd = 1_R, tag 3$



        where $1_R$ is the multiplicative identity of $R$; then



        $(ab)(cd) = a(bc)d = a(cb)d = (ac)(bd) = 1_R 1_R = 1_R, tag 4$



        that is,



        $ab, cd in R^times. tag 5$



        $OEDelta$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 '18 at 19:55









        Robert LewisRobert Lewis

        45.4k23065




        45.4k23065























            0












            $begingroup$

            Since $a,bin R^times $, exsist $a^{-1}$ and $b^{-1}$ in $R$. Let $c = b^{-1}a^{-1}$, then $$(ab)cdot c = aunderbrace {bcdot b^{-1}}_{=e}a^{-1} = a cdot a^{-1} = e$$



            so $abin R^times $.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
              $endgroup$
              – Gykonik
              Nov 27 '18 at 20:38












            • $begingroup$
              $(ab)c =e$ what does that say about $ab$? @Gykonik
              $endgroup$
              – greedoid
              Nov 27 '18 at 20:40












            • $begingroup$
              ehm, that $(ab)=c^{-1}$?
              $endgroup$
              – Gykonik
              Nov 27 '18 at 20:43










            • $begingroup$
              So...............
              $endgroup$
              – greedoid
              Nov 27 '18 at 20:44










            • $begingroup$
              $c^{-1}$ and $c$ has to be in $R^times$?
              $endgroup$
              – Gykonik
              Nov 27 '18 at 20:44


















            0












            $begingroup$

            Since $a,bin R^times $, exsist $a^{-1}$ and $b^{-1}$ in $R$. Let $c = b^{-1}a^{-1}$, then $$(ab)cdot c = aunderbrace {bcdot b^{-1}}_{=e}a^{-1} = a cdot a^{-1} = e$$



            so $abin R^times $.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
              $endgroup$
              – Gykonik
              Nov 27 '18 at 20:38












            • $begingroup$
              $(ab)c =e$ what does that say about $ab$? @Gykonik
              $endgroup$
              – greedoid
              Nov 27 '18 at 20:40












            • $begingroup$
              ehm, that $(ab)=c^{-1}$?
              $endgroup$
              – Gykonik
              Nov 27 '18 at 20:43










            • $begingroup$
              So...............
              $endgroup$
              – greedoid
              Nov 27 '18 at 20:44










            • $begingroup$
              $c^{-1}$ and $c$ has to be in $R^times$?
              $endgroup$
              – Gykonik
              Nov 27 '18 at 20:44
















            0












            0








            0





            $begingroup$

            Since $a,bin R^times $, exsist $a^{-1}$ and $b^{-1}$ in $R$. Let $c = b^{-1}a^{-1}$, then $$(ab)cdot c = aunderbrace {bcdot b^{-1}}_{=e}a^{-1} = a cdot a^{-1} = e$$



            so $abin R^times $.






            share|cite|improve this answer









            $endgroup$



            Since $a,bin R^times $, exsist $a^{-1}$ and $b^{-1}$ in $R$. Let $c = b^{-1}a^{-1}$, then $$(ab)cdot c = aunderbrace {bcdot b^{-1}}_{=e}a^{-1} = a cdot a^{-1} = e$$



            so $abin R^times $.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 27 '18 at 19:57









            greedoidgreedoid

            40.2k114799




            40.2k114799












            • $begingroup$
              didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
              $endgroup$
              – Gykonik
              Nov 27 '18 at 20:38












            • $begingroup$
              $(ab)c =e$ what does that say about $ab$? @Gykonik
              $endgroup$
              – greedoid
              Nov 27 '18 at 20:40












            • $begingroup$
              ehm, that $(ab)=c^{-1}$?
              $endgroup$
              – Gykonik
              Nov 27 '18 at 20:43










            • $begingroup$
              So...............
              $endgroup$
              – greedoid
              Nov 27 '18 at 20:44










            • $begingroup$
              $c^{-1}$ and $c$ has to be in $R^times$?
              $endgroup$
              – Gykonik
              Nov 27 '18 at 20:44




















            • $begingroup$
              didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
              $endgroup$
              – Gykonik
              Nov 27 '18 at 20:38












            • $begingroup$
              $(ab)c =e$ what does that say about $ab$? @Gykonik
              $endgroup$
              – greedoid
              Nov 27 '18 at 20:40












            • $begingroup$
              ehm, that $(ab)=c^{-1}$?
              $endgroup$
              – Gykonik
              Nov 27 '18 at 20:43










            • $begingroup$
              So...............
              $endgroup$
              – greedoid
              Nov 27 '18 at 20:44










            • $begingroup$
              $c^{-1}$ and $c$ has to be in $R^times$?
              $endgroup$
              – Gykonik
              Nov 27 '18 at 20:44


















            $begingroup$
            didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
            $endgroup$
            – Gykonik
            Nov 27 '18 at 20:38






            $begingroup$
            didn't you proof that $(ab)c in R^times$ I don't see how this is a proof for $ab in R^times$
            $endgroup$
            – Gykonik
            Nov 27 '18 at 20:38














            $begingroup$
            $(ab)c =e$ what does that say about $ab$? @Gykonik
            $endgroup$
            – greedoid
            Nov 27 '18 at 20:40






            $begingroup$
            $(ab)c =e$ what does that say about $ab$? @Gykonik
            $endgroup$
            – greedoid
            Nov 27 '18 at 20:40














            $begingroup$
            ehm, that $(ab)=c^{-1}$?
            $endgroup$
            – Gykonik
            Nov 27 '18 at 20:43




            $begingroup$
            ehm, that $(ab)=c^{-1}$?
            $endgroup$
            – Gykonik
            Nov 27 '18 at 20:43












            $begingroup$
            So...............
            $endgroup$
            – greedoid
            Nov 27 '18 at 20:44




            $begingroup$
            So...............
            $endgroup$
            – greedoid
            Nov 27 '18 at 20:44












            $begingroup$
            $c^{-1}$ and $c$ has to be in $R^times$?
            $endgroup$
            – Gykonik
            Nov 27 '18 at 20:44






            $begingroup$
            $c^{-1}$ and $c$ has to be in $R^times$?
            $endgroup$
            – Gykonik
            Nov 27 '18 at 20:44





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            How to change which sound is reproduced for terminal bell?

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            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
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            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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            Can I use Tabulator js library in my java Spring + Thymeleaf project?

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            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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