Can we remove the absolute value from inequality $|a-b|<ε$?












2














So currently I have $|S_n-S|<ε$ where $S_n$ is a sequence and $ε>0$.



I have $bigl||S_n|-|S|bigr|leq|S_n-S|<ε$. Hence, $bigl||S_n|-|S|bigr|<ε$. Since $ε>0$, can I just remove the main absolute value and say $|S_n|-|S|<ε$?






inequality







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  • 1




    Hello @Sashin Chetty, for better questions and answer, here you have to use Math Jax to write math equations... here is a resume math.meta.stackexchange.com/questions/5020/…
    – Robson
    Nov 24 at 4:01






  • 1




    Thanks for the advice!
    – Sashin Chetty
    Nov 24 at 7:41
















2














So currently I have $|S_n-S|<ε$ where $S_n$ is a sequence and $ε>0$.



I have $bigl||S_n|-|S|bigr|leq|S_n-S|<ε$. Hence, $bigl||S_n|-|S|bigr|<ε$. Since $ε>0$, can I just remove the main absolute value and say $|S_n|-|S|<ε$?






inequality







share|cite|improve this question




















  • 1




    Hello @Sashin Chetty, for better questions and answer, here you have to use Math Jax to write math equations... here is a resume math.meta.stackexchange.com/questions/5020/…
    – Robson
    Nov 24 at 4:01






  • 1




    Thanks for the advice!
    – Sashin Chetty
    Nov 24 at 7:41














2












2








2







So currently I have $|S_n-S|<ε$ where $S_n$ is a sequence and $ε>0$.



I have $bigl||S_n|-|S|bigr|leq|S_n-S|<ε$. Hence, $bigl||S_n|-|S|bigr|<ε$. Since $ε>0$, can I just remove the main absolute value and say $|S_n|-|S|<ε$?






inequality







share|cite|improve this question















So currently I have $|S_n-S|<ε$ where $S_n$ is a sequence and $ε>0$.



I have $bigl||S_n|-|S|bigr|leq|S_n-S|<ε$. Hence, $bigl||S_n|-|S|bigr|<ε$. Since $ε>0$, can I just remove the main absolute value and say $|S_n|-|S|<ε$?






inequality




inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 4:10









Saad

19.7k92252




19.7k92252










asked Nov 24 at 3:54









Sashin Chetty

222




222








  • 1




    Hello @Sashin Chetty, for better questions and answer, here you have to use Math Jax to write math equations... here is a resume math.meta.stackexchange.com/questions/5020/…
    – Robson
    Nov 24 at 4:01






  • 1




    Thanks for the advice!
    – Sashin Chetty
    Nov 24 at 7:41














  • 1




    Hello @Sashin Chetty, for better questions and answer, here you have to use Math Jax to write math equations... here is a resume math.meta.stackexchange.com/questions/5020/…
    – Robson
    Nov 24 at 4:01






  • 1




    Thanks for the advice!
    – Sashin Chetty
    Nov 24 at 7:41








1




1




Hello @Sashin Chetty, for better questions and answer, here you have to use Math Jax to write math equations... here is a resume math.meta.stackexchange.com/questions/5020/…
– Robson
Nov 24 at 4:01




Hello @Sashin Chetty, for better questions and answer, here you have to use Math Jax to write math equations... here is a resume math.meta.stackexchange.com/questions/5020/…
– Robson
Nov 24 at 4:01




1




1




Thanks for the advice!
– Sashin Chetty
Nov 24 at 7:41




Thanks for the advice!
– Sashin Chetty
Nov 24 at 7:41










2 Answers
2






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4














Yes, you always can do that because $|x|< a iff x<a$ and $x>-a$, so in particular you can remove absolute value.



Also observe that is always true that $x<|x|$, hence if $|x|<a$ we conclude that $x<a$






share|cite|improve this answer





























    3














    Just to spell out Robson's argument:



    If $|S_n| -|S|$ is positive then it's equal to $bigg||S_n| -|S| bigg|
    <varepsilon$     and if $|S_n| -|S|$ is negative then it's certainly less than $epsilon$ which is positive.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      Yes, you always can do that because $|x|< a iff x<a$ and $x>-a$, so in particular you can remove absolute value.



      Also observe that is always true that $x<|x|$, hence if $|x|<a$ we conclude that $x<a$






      share|cite|improve this answer


























        4














        Yes, you always can do that because $|x|< a iff x<a$ and $x>-a$, so in particular you can remove absolute value.



        Also observe that is always true that $x<|x|$, hence if $|x|<a$ we conclude that $x<a$






        share|cite|improve this answer
























          4












          4








          4






          Yes, you always can do that because $|x|< a iff x<a$ and $x>-a$, so in particular you can remove absolute value.



          Also observe that is always true that $x<|x|$, hence if $|x|<a$ we conclude that $x<a$






          share|cite|improve this answer












          Yes, you always can do that because $|x|< a iff x<a$ and $x>-a$, so in particular you can remove absolute value.



          Also observe that is always true that $x<|x|$, hence if $|x|<a$ we conclude that $x<a$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 4:02









          Robson

          771221




          771221























              3














              Just to spell out Robson's argument:



              If $|S_n| -|S|$ is positive then it's equal to $bigg||S_n| -|S| bigg|
              <varepsilon$               and if $|S_n| -|S|$ is negative then it's certainly less than $epsilon$ which is positive.






              share|cite|improve this answer


























                3














                Just to spell out Robson's argument:



                If $|S_n| -|S|$ is positive then it's equal to $bigg||S_n| -|S| bigg|
                <varepsilon$                 and if $|S_n| -|S|$ is negative then it's certainly less than $epsilon$ which is positive.






                share|cite|improve this answer
























                  3












                  3








                  3






                  Just to spell out Robson's argument:



                  If $|S_n| -|S|$ is positive then it's equal to $bigg||S_n| -|S| bigg|
                  <varepsilon$                   and if $|S_n| -|S|$ is negative then it's certainly less than $epsilon$ which is positive.






                  share|cite|improve this answer












                  Just to spell out Robson's argument:



                  If $|S_n| -|S|$ is positive then it's equal to $bigg||S_n| -|S| bigg|
                  <varepsilon$                   and if $|S_n| -|S|$ is negative then it's certainly less than $epsilon$ which is positive.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 at 4:07









                  Mason

                  1,9341530




                  1,9341530






























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