If $xin[a,b],$ then $|x|leq max{|a|,|b|}$












2












$begingroup$


The following seems to be quite known:




If $xin[a,b],$ then $|x|leq max{|a|,|b|}$




I do use it while treating problems on series of functions but how does it come about? Any hint please?






real-analysis sequences-and-series absolute-value







share|cite|improve this question









$endgroup$












  • $begingroup$
    Sorry, can you tell me why you're voting for a close? Perhaps, I'll change somethings.
    $endgroup$
    – Omojola Micheal
    Dec 29 '18 at 15:50






  • 5




    $begingroup$
    $|x|$ is the distance of $x$ from $0$, it seems obvious that the distance of $x$ from $0$ would be less than the maximum of distance of $a$ and $b$ from $0$.
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:50
















2












$begingroup$


The following seems to be quite known:




If $xin[a,b],$ then $|x|leq max{|a|,|b|}$




I do use it while treating problems on series of functions but how does it come about? Any hint please?






real-analysis sequences-and-series absolute-value







share|cite|improve this question









$endgroup$












  • $begingroup$
    Sorry, can you tell me why you're voting for a close? Perhaps, I'll change somethings.
    $endgroup$
    – Omojola Micheal
    Dec 29 '18 at 15:50






  • 5




    $begingroup$
    $|x|$ is the distance of $x$ from $0$, it seems obvious that the distance of $x$ from $0$ would be less than the maximum of distance of $a$ and $b$ from $0$.
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:50














2












2








2


1



$begingroup$


The following seems to be quite known:




If $xin[a,b],$ then $|x|leq max{|a|,|b|}$




I do use it while treating problems on series of functions but how does it come about? Any hint please?






real-analysis sequences-and-series absolute-value







share|cite|improve this question









$endgroup$




The following seems to be quite known:




If $xin[a,b],$ then $|x|leq max{|a|,|b|}$




I do use it while treating problems on series of functions but how does it come about? Any hint please?






real-analysis sequences-and-series absolute-value




real-analysis sequences-and-series absolute-value






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 15:37









Omojola MichealOmojola Micheal

1,753324




1,753324












  • $begingroup$
    Sorry, can you tell me why you're voting for a close? Perhaps, I'll change somethings.
    $endgroup$
    – Omojola Micheal
    Dec 29 '18 at 15:50






  • 5




    $begingroup$
    $|x|$ is the distance of $x$ from $0$, it seems obvious that the distance of $x$ from $0$ would be less than the maximum of distance of $a$ and $b$ from $0$.
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:50


















  • $begingroup$
    Sorry, can you tell me why you're voting for a close? Perhaps, I'll change somethings.
    $endgroup$
    – Omojola Micheal
    Dec 29 '18 at 15:50






  • 5




    $begingroup$
    $|x|$ is the distance of $x$ from $0$, it seems obvious that the distance of $x$ from $0$ would be less than the maximum of distance of $a$ and $b$ from $0$.
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:50
















$begingroup$
Sorry, can you tell me why you're voting for a close? Perhaps, I'll change somethings.
$endgroup$
– Omojola Micheal
Dec 29 '18 at 15:50




$begingroup$
Sorry, can you tell me why you're voting for a close? Perhaps, I'll change somethings.
$endgroup$
– Omojola Micheal
Dec 29 '18 at 15:50




5




5




$begingroup$
$|x|$ is the distance of $x$ from $0$, it seems obvious that the distance of $x$ from $0$ would be less than the maximum of distance of $a$ and $b$ from $0$.
$endgroup$
– Jakobian
Dec 29 '18 at 15:50




$begingroup$
$|x|$ is the distance of $x$ from $0$, it seems obvious that the distance of $x$ from $0$ would be less than the maximum of distance of $a$ and $b$ from $0$.
$endgroup$
– Jakobian
Dec 29 '18 at 15:50










2 Answers
2






active

oldest

votes


















4












$begingroup$

If $xge0$ then $bge0$, hence $$|x|=xle b=|b|lemax(|a|,|b|).$$Similarly if $x<0$ then $a<0$, hence $$|x|=-xle-a=|a|lemax(|a|,|b|).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can't it be $xin [-5,4]$ for example, which means that $x$ can also be positive and negative as well ?
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:50












  • $begingroup$
    @Rebellos Huh???? Saying $xin[-5,4]$ does not say that $x$ is both positive and negative - that's total nonsense.
    $endgroup$
    – David C. Ullrich
    Dec 29 '18 at 15:52












  • $begingroup$
    Very straightforward proof +1
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:56










  • $begingroup$
    @DavidC.Ullrich You hadn't added the second line when I commented :) Still, it's a correct proof. +1 now.
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:56






  • 2




    $begingroup$
    @Rebellos $x$ is fixed
    $endgroup$
    – Jakobian
    Dec 29 '18 at 16:14



















1












$begingroup$

Since $x in [a,b]$, it is $a<b$ (or even $a leq b$ if you want to be more strict for a trivial case).



If $text{sgn}(a) = text{sgn}(b)=1$, then $max{|a|,|b|} = |b| implies |x| leq max{|a|,|b|}$.



If $text{sgn}(a) = text{sgn}(b)=-1$, then $max{|a|,|b|} = |a| implies |x| leq max{|a|,|b|}$.



In the case of $text{sgn}(a) neq text{sgn}(b)$, that means that $a$ would be negative. But, in that case, it can be $|a| > |b|$. Then, for $|x|$, it would be : $|x| leq |a| implies |x| leq max{|a|,|b|}$.



So, for all cases, we have that : $|x| leq max{|a|,|b|}$.



Edit (Alternativelly) : The comment of Jakobian is really nice to consider for a really straightforward proof. What the absolute value $|x|$ means is the distance of $x$ from $0$. But since $x in [a,b]$, then this distance cannot be larger than the distance of $a$ from zero $(|a|)$ or the distance of $b$ from zero $(|b|)$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I do not understand your first statement. If I have $a=-3$ and $b=-2$, then the maximum of $|a|$ and $|b|$ would be $|a|=3$, not $|b|=2$.
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 15:45










  • $begingroup$
    @NobleMushtak Sure, edited. I missed a line.
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:46













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

If $xge0$ then $bge0$, hence $$|x|=xle b=|b|lemax(|a|,|b|).$$Similarly if $x<0$ then $a<0$, hence $$|x|=-xle-a=|a|lemax(|a|,|b|).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can't it be $xin [-5,4]$ for example, which means that $x$ can also be positive and negative as well ?
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:50












  • $begingroup$
    @Rebellos Huh???? Saying $xin[-5,4]$ does not say that $x$ is both positive and negative - that's total nonsense.
    $endgroup$
    – David C. Ullrich
    Dec 29 '18 at 15:52












  • $begingroup$
    Very straightforward proof +1
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:56










  • $begingroup$
    @DavidC.Ullrich You hadn't added the second line when I commented :) Still, it's a correct proof. +1 now.
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:56






  • 2




    $begingroup$
    @Rebellos $x$ is fixed
    $endgroup$
    – Jakobian
    Dec 29 '18 at 16:14
















4












$begingroup$

If $xge0$ then $bge0$, hence $$|x|=xle b=|b|lemax(|a|,|b|).$$Similarly if $x<0$ then $a<0$, hence $$|x|=-xle-a=|a|lemax(|a|,|b|).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can't it be $xin [-5,4]$ for example, which means that $x$ can also be positive and negative as well ?
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:50












  • $begingroup$
    @Rebellos Huh???? Saying $xin[-5,4]$ does not say that $x$ is both positive and negative - that's total nonsense.
    $endgroup$
    – David C. Ullrich
    Dec 29 '18 at 15:52












  • $begingroup$
    Very straightforward proof +1
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:56










  • $begingroup$
    @DavidC.Ullrich You hadn't added the second line when I commented :) Still, it's a correct proof. +1 now.
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:56






  • 2




    $begingroup$
    @Rebellos $x$ is fixed
    $endgroup$
    – Jakobian
    Dec 29 '18 at 16:14














4












4








4





$begingroup$

If $xge0$ then $bge0$, hence $$|x|=xle b=|b|lemax(|a|,|b|).$$Similarly if $x<0$ then $a<0$, hence $$|x|=-xle-a=|a|lemax(|a|,|b|).$$






share|cite|improve this answer











$endgroup$



If $xge0$ then $bge0$, hence $$|x|=xle b=|b|lemax(|a|,|b|).$$Similarly if $x<0$ then $a<0$, hence $$|x|=-xle-a=|a|lemax(|a|,|b|).$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 15:50

























answered Dec 29 '18 at 15:47









David C. UllrichDavid C. Ullrich

59.3k43893




59.3k43893












  • $begingroup$
    Can't it be $xin [-5,4]$ for example, which means that $x$ can also be positive and negative as well ?
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:50












  • $begingroup$
    @Rebellos Huh???? Saying $xin[-5,4]$ does not say that $x$ is both positive and negative - that's total nonsense.
    $endgroup$
    – David C. Ullrich
    Dec 29 '18 at 15:52












  • $begingroup$
    Very straightforward proof +1
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:56










  • $begingroup$
    @DavidC.Ullrich You hadn't added the second line when I commented :) Still, it's a correct proof. +1 now.
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:56






  • 2




    $begingroup$
    @Rebellos $x$ is fixed
    $endgroup$
    – Jakobian
    Dec 29 '18 at 16:14


















  • $begingroup$
    Can't it be $xin [-5,4]$ for example, which means that $x$ can also be positive and negative as well ?
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:50












  • $begingroup$
    @Rebellos Huh???? Saying $xin[-5,4]$ does not say that $x$ is both positive and negative - that's total nonsense.
    $endgroup$
    – David C. Ullrich
    Dec 29 '18 at 15:52












  • $begingroup$
    Very straightforward proof +1
    $endgroup$
    – Jakobian
    Dec 29 '18 at 15:56










  • $begingroup$
    @DavidC.Ullrich You hadn't added the second line when I commented :) Still, it's a correct proof. +1 now.
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:56






  • 2




    $begingroup$
    @Rebellos $x$ is fixed
    $endgroup$
    – Jakobian
    Dec 29 '18 at 16:14
















$begingroup$
Can't it be $xin [-5,4]$ for example, which means that $x$ can also be positive and negative as well ?
$endgroup$
– Rebellos
Dec 29 '18 at 15:50






$begingroup$
Can't it be $xin [-5,4]$ for example, which means that $x$ can also be positive and negative as well ?
$endgroup$
– Rebellos
Dec 29 '18 at 15:50














$begingroup$
@Rebellos Huh???? Saying $xin[-5,4]$ does not say that $x$ is both positive and negative - that's total nonsense.
$endgroup$
– David C. Ullrich
Dec 29 '18 at 15:52






$begingroup$
@Rebellos Huh???? Saying $xin[-5,4]$ does not say that $x$ is both positive and negative - that's total nonsense.
$endgroup$
– David C. Ullrich
Dec 29 '18 at 15:52














$begingroup$
Very straightforward proof +1
$endgroup$
– Jakobian
Dec 29 '18 at 15:56




$begingroup$
Very straightforward proof +1
$endgroup$
– Jakobian
Dec 29 '18 at 15:56












$begingroup$
@DavidC.Ullrich You hadn't added the second line when I commented :) Still, it's a correct proof. +1 now.
$endgroup$
– Rebellos
Dec 29 '18 at 15:56




$begingroup$
@DavidC.Ullrich You hadn't added the second line when I commented :) Still, it's a correct proof. +1 now.
$endgroup$
– Rebellos
Dec 29 '18 at 15:56




2




2




$begingroup$
@Rebellos $x$ is fixed
$endgroup$
– Jakobian
Dec 29 '18 at 16:14




$begingroup$
@Rebellos $x$ is fixed
$endgroup$
– Jakobian
Dec 29 '18 at 16:14











1












$begingroup$

Since $x in [a,b]$, it is $a<b$ (or even $a leq b$ if you want to be more strict for a trivial case).



If $text{sgn}(a) = text{sgn}(b)=1$, then $max{|a|,|b|} = |b| implies |x| leq max{|a|,|b|}$.



If $text{sgn}(a) = text{sgn}(b)=-1$, then $max{|a|,|b|} = |a| implies |x| leq max{|a|,|b|}$.



In the case of $text{sgn}(a) neq text{sgn}(b)$, that means that $a$ would be negative. But, in that case, it can be $|a| > |b|$. Then, for $|x|$, it would be : $|x| leq |a| implies |x| leq max{|a|,|b|}$.



So, for all cases, we have that : $|x| leq max{|a|,|b|}$.



Edit (Alternativelly) : The comment of Jakobian is really nice to consider for a really straightforward proof. What the absolute value $|x|$ means is the distance of $x$ from $0$. But since $x in [a,b]$, then this distance cannot be larger than the distance of $a$ from zero $(|a|)$ or the distance of $b$ from zero $(|b|)$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I do not understand your first statement. If I have $a=-3$ and $b=-2$, then the maximum of $|a|$ and $|b|$ would be $|a|=3$, not $|b|=2$.
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 15:45










  • $begingroup$
    @NobleMushtak Sure, edited. I missed a line.
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:46


















1












$begingroup$

Since $x in [a,b]$, it is $a<b$ (or even $a leq b$ if you want to be more strict for a trivial case).



If $text{sgn}(a) = text{sgn}(b)=1$, then $max{|a|,|b|} = |b| implies |x| leq max{|a|,|b|}$.



If $text{sgn}(a) = text{sgn}(b)=-1$, then $max{|a|,|b|} = |a| implies |x| leq max{|a|,|b|}$.



In the case of $text{sgn}(a) neq text{sgn}(b)$, that means that $a$ would be negative. But, in that case, it can be $|a| > |b|$. Then, for $|x|$, it would be : $|x| leq |a| implies |x| leq max{|a|,|b|}$.



So, for all cases, we have that : $|x| leq max{|a|,|b|}$.



Edit (Alternativelly) : The comment of Jakobian is really nice to consider for a really straightforward proof. What the absolute value $|x|$ means is the distance of $x$ from $0$. But since $x in [a,b]$, then this distance cannot be larger than the distance of $a$ from zero $(|a|)$ or the distance of $b$ from zero $(|b|)$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I do not understand your first statement. If I have $a=-3$ and $b=-2$, then the maximum of $|a|$ and $|b|$ would be $|a|=3$, not $|b|=2$.
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 15:45










  • $begingroup$
    @NobleMushtak Sure, edited. I missed a line.
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:46
















1












1








1





$begingroup$

Since $x in [a,b]$, it is $a<b$ (or even $a leq b$ if you want to be more strict for a trivial case).



If $text{sgn}(a) = text{sgn}(b)=1$, then $max{|a|,|b|} = |b| implies |x| leq max{|a|,|b|}$.



If $text{sgn}(a) = text{sgn}(b)=-1$, then $max{|a|,|b|} = |a| implies |x| leq max{|a|,|b|}$.



In the case of $text{sgn}(a) neq text{sgn}(b)$, that means that $a$ would be negative. But, in that case, it can be $|a| > |b|$. Then, for $|x|$, it would be : $|x| leq |a| implies |x| leq max{|a|,|b|}$.



So, for all cases, we have that : $|x| leq max{|a|,|b|}$.



Edit (Alternativelly) : The comment of Jakobian is really nice to consider for a really straightforward proof. What the absolute value $|x|$ means is the distance of $x$ from $0$. But since $x in [a,b]$, then this distance cannot be larger than the distance of $a$ from zero $(|a|)$ or the distance of $b$ from zero $(|b|)$.






share|cite|improve this answer











$endgroup$



Since $x in [a,b]$, it is $a<b$ (or even $a leq b$ if you want to be more strict for a trivial case).



If $text{sgn}(a) = text{sgn}(b)=1$, then $max{|a|,|b|} = |b| implies |x| leq max{|a|,|b|}$.



If $text{sgn}(a) = text{sgn}(b)=-1$, then $max{|a|,|b|} = |a| implies |x| leq max{|a|,|b|}$.



In the case of $text{sgn}(a) neq text{sgn}(b)$, that means that $a$ would be negative. But, in that case, it can be $|a| > |b|$. Then, for $|x|$, it would be : $|x| leq |a| implies |x| leq max{|a|,|b|}$.



So, for all cases, we have that : $|x| leq max{|a|,|b|}$.



Edit (Alternativelly) : The comment of Jakobian is really nice to consider for a really straightforward proof. What the absolute value $|x|$ means is the distance of $x$ from $0$. But since $x in [a,b]$, then this distance cannot be larger than the distance of $a$ from zero $(|a|)$ or the distance of $b$ from zero $(|b|)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 15:54

























answered Dec 29 '18 at 15:43









RebellosRebellos

14.5k31246




14.5k31246








  • 1




    $begingroup$
    I do not understand your first statement. If I have $a=-3$ and $b=-2$, then the maximum of $|a|$ and $|b|$ would be $|a|=3$, not $|b|=2$.
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 15:45










  • $begingroup$
    @NobleMushtak Sure, edited. I missed a line.
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:46
















  • 1




    $begingroup$
    I do not understand your first statement. If I have $a=-3$ and $b=-2$, then the maximum of $|a|$ and $|b|$ would be $|a|=3$, not $|b|=2$.
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 15:45










  • $begingroup$
    @NobleMushtak Sure, edited. I missed a line.
    $endgroup$
    – Rebellos
    Dec 29 '18 at 15:46










1




1




$begingroup$
I do not understand your first statement. If I have $a=-3$ and $b=-2$, then the maximum of $|a|$ and $|b|$ would be $|a|=3$, not $|b|=2$.
$endgroup$
– Noble Mushtak
Dec 29 '18 at 15:45




$begingroup$
I do not understand your first statement. If I have $a=-3$ and $b=-2$, then the maximum of $|a|$ and $|b|$ would be $|a|=3$, not $|b|=2$.
$endgroup$
– Noble Mushtak
Dec 29 '18 at 15:45












$begingroup$
@NobleMushtak Sure, edited. I missed a line.
$endgroup$
– Rebellos
Dec 29 '18 at 15:46






$begingroup$
@NobleMushtak Sure, edited. I missed a line.
$endgroup$
– Rebellos
Dec 29 '18 at 15:46




















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Human spaceflight

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Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
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Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
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File:DeusFollowingSea.jpg

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File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
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