Can we remove the absolute value from inequality $|a-b|<ε$?
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
edited Nov 24 at 4:10
Saad
19.7k92252
asked Nov 24 at 3:54
Sashin Chetty
222
add a comment |
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
edited Nov 24 at 4:10
Saad
19.7k92252
asked Nov 24 at 3:54
Sashin Chetty
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
add a comment |
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
edited Nov 24 at 4:10
Saad
19.7k92252
asked Nov 24 at 3:54
Sashin Chetty
222
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
edited Nov 24 at 4:10
Saad
19.7k92252
asked Nov 24 at 3:54
Sashin Chetty
222
edited Nov 24 at 4:10
Saad
19.7k92252
asked Nov 24 at 3:54
Sashin Chetty
222
edited Nov 24 at 4:10
Saad
19.7k92252
edited Nov 24 at 4:10
Saad
19.7k92252
edited Nov 24 at 4:10
Saad
19.7k92252
19.7k92252
asked Nov 24 at 3:54
Sashin Chetty
222
asked Nov 24 at 3:54
Sashin Chetty
222
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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 $x<a$
answered Nov 24 at 4:02
Robson
771221
add a comment |
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.
answered Nov 24 at 4:07
Mason
1,9341530
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
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votes
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 $x<a$
answered Nov 24 at 4:02
Robson
771221
add a comment |
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 $x<a$
answered Nov 24 at 4:02
Robson
771221
add a comment |
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 $x<a$
answered Nov 24 at 4:02
Robson
771221
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 $x<a$
answered Nov 24 at 4:02
Robson
771221
answered Nov 24 at 4:02
Robson
771221
answered Nov 24 at 4:02
Robson
771221
answered Nov 24 at 4:02
Robson
771221
771221
add a comment |
add a comment |
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.
answered Nov 24 at 4:07
Mason
1,9341530
add a comment |
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.
answered Nov 24 at 4:07
Mason
1,9341530
add a comment |
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.
answered Nov 24 at 4:07
Mason
1,9341530
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.
answered Nov 24 at 4:07
Mason
1,9341530
answered Nov 24 at 4:07
Mason
1,9341530
answered Nov 24 at 4:07
Mason
1,9341530
answered Nov 24 at 4:07
Mason
1,9341530
1,9341530
add a comment |
add a comment |
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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