Let $a,b,c$ be real numbers and denote by $textmid(a,b,c)$ the 'value in the middle'.

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Let $a,b,c$ be real numbers and denote by mid$(a,b,c)$ the 'value in the middle'.



Show that $textmid(a,b,c) = infsupa,b, supa,c, supb,c$.




Suppose that $a<b<c$. Then, it follows that $infb,c,c= b$, so it works, but I don't know how to formally prove this.




real-analysis




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    The value in the middle? The average value? Or the median?
    – Calvin Khor
    Aug 2 at 7:36






  • 1




    @CalvinKhor The one of the three values which lies between the other two. So, I guess the median would be appropriate.
    – Arthur
    Aug 2 at 7:40











  • That's called the median
    – Calvin Khor
    Aug 2 at 7:40














up vote
2
down vote

favorite













Let $a,b,c$ be real numbers and denote by mid$(a,b,c)$ the 'value in the middle'.



Show that $textmid(a,b,c) = infsupa,b, supa,c, supb,c$.




Suppose that $a<b<c$. Then, it follows that $infb,c,c= b$, so it works, but I don't know how to formally prove this.




real-analysis




share|cite|improve this question

















  • 1




    The value in the middle? The average value? Or the median?
    – Calvin Khor
    Aug 2 at 7:36






  • 1




    @CalvinKhor The one of the three values which lies between the other two. So, I guess the median would be appropriate.
    – Arthur
    Aug 2 at 7:40











  • That's called the median
    – Calvin Khor
    Aug 2 at 7:40












up vote
2
down vote

favorite









up vote
2
down vote

favorite












Let $a,b,c$ be real numbers and denote by mid$(a,b,c)$ the 'value in the middle'.



Show that $textmid(a,b,c) = infsupa,b, supa,c, supb,c$.




Suppose that $a<b<c$. Then, it follows that $infb,c,c= b$, so it works, but I don't know how to formally prove this.




real-analysis




share|cite|improve this question














Let $a,b,c$ be real numbers and denote by mid$(a,b,c)$ the 'value in the middle'.



Show that $textmid(a,b,c) = infsupa,b, supa,c, supb,c$.




Suppose that $a<b<c$. Then, it follows that $infb,c,c= b$, so it works, but I don't know how to formally prove this.





real-analysis





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edited Aug 2 at 12:18









David G. Stork

7,3302728




7,3302728









asked Aug 2 at 7:26









Sihyun Kim

701210




701210







  • 1




    The value in the middle? The average value? Or the median?
    – Calvin Khor
    Aug 2 at 7:36






  • 1




    @CalvinKhor The one of the three values which lies between the other two. So, I guess the median would be appropriate.
    – Arthur
    Aug 2 at 7:40











  • That's called the median
    – Calvin Khor
    Aug 2 at 7:40












  • 1




    The value in the middle? The average value? Or the median?
    – Calvin Khor
    Aug 2 at 7:36






  • 1




    @CalvinKhor The one of the three values which lies between the other two. So, I guess the median would be appropriate.
    – Arthur
    Aug 2 at 7:40











  • That's called the median
    – Calvin Khor
    Aug 2 at 7:40







1




1




The value in the middle? The average value? Or the median?
– Calvin Khor
Aug 2 at 7:36




The value in the middle? The average value? Or the median?
– Calvin Khor
Aug 2 at 7:36




1




1




@CalvinKhor The one of the three values which lies between the other two. So, I guess the median would be appropriate.
– Arthur
Aug 2 at 7:40





@CalvinKhor The one of the three values which lies between the other two. So, I guess the median would be appropriate.
– Arthur
Aug 2 at 7:40













That's called the median
– Calvin Khor
Aug 2 at 7:40




That's called the median
– Calvin Khor
Aug 2 at 7:40










2 Answers
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Without loss of generality $ale ble c$ so the equation reduces to the trivial $b=infb,,c,,c$.






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    down vote













    There are (at least) two options that you can use to prove it:



    The first one is likely to be quite tedious. You consider all six possible orders that $a, b, c$ can be in, and prove the equation in each case.



    The second is similar: You still consider a specific ordering of $a, b, c$ and prove the equation under that assumption. However, you don't consider every possible order. Instead, you notice that swapping any two variables in the equation doesn't change the value of either side of the equation. e.g. If you swap the values of $a$ and $b$, then the left hand side becomes
    $$ operatornamemid(b, a, c) = operatornamemid(a, b, c) $$ and the right hand side similarly remains the same. Thus you can reorder $a, b, c$ as you please, and so it is enough to prove the equation for the specific case $a leq b leq c$, which you have already done.






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      2 Answers
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      2 Answers
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      up vote
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      Without loss of generality $ale ble c$ so the equation reduces to the trivial $b=infb,,c,,c$.






      share|cite|improve this answer

























        up vote
        4
        down vote













        Without loss of generality $ale ble c$ so the equation reduces to the trivial $b=infb,,c,,c$.






        share|cite|improve this answer























          up vote
          4
          down vote










          up vote
          4
          down vote









          Without loss of generality $ale ble c$ so the equation reduces to the trivial $b=infb,,c,,c$.






          share|cite|improve this answer













          Without loss of generality $ale ble c$ so the equation reduces to the trivial $b=infb,,c,,c$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 2 at 7:39









          J.G.

          12.9k11423




          12.9k11423




















              up vote
              0
              down vote













              There are (at least) two options that you can use to prove it:



              The first one is likely to be quite tedious. You consider all six possible orders that $a, b, c$ can be in, and prove the equation in each case.



              The second is similar: You still consider a specific ordering of $a, b, c$ and prove the equation under that assumption. However, you don't consider every possible order. Instead, you notice that swapping any two variables in the equation doesn't change the value of either side of the equation. e.g. If you swap the values of $a$ and $b$, then the left hand side becomes
              $$ operatornamemid(b, a, c) = operatornamemid(a, b, c) $$ and the right hand side similarly remains the same. Thus you can reorder $a, b, c$ as you please, and so it is enough to prove the equation for the specific case $a leq b leq c$, which you have already done.






              share|cite|improve this answer

























                up vote
                0
                down vote













                There are (at least) two options that you can use to prove it:



                The first one is likely to be quite tedious. You consider all six possible orders that $a, b, c$ can be in, and prove the equation in each case.



                The second is similar: You still consider a specific ordering of $a, b, c$ and prove the equation under that assumption. However, you don't consider every possible order. Instead, you notice that swapping any two variables in the equation doesn't change the value of either side of the equation. e.g. If you swap the values of $a$ and $b$, then the left hand side becomes
                $$ operatornamemid(b, a, c) = operatornamemid(a, b, c) $$ and the right hand side similarly remains the same. Thus you can reorder $a, b, c$ as you please, and so it is enough to prove the equation for the specific case $a leq b leq c$, which you have already done.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  There are (at least) two options that you can use to prove it:



                  The first one is likely to be quite tedious. You consider all six possible orders that $a, b, c$ can be in, and prove the equation in each case.



                  The second is similar: You still consider a specific ordering of $a, b, c$ and prove the equation under that assumption. However, you don't consider every possible order. Instead, you notice that swapping any two variables in the equation doesn't change the value of either side of the equation. e.g. If you swap the values of $a$ and $b$, then the left hand side becomes
                  $$ operatornamemid(b, a, c) = operatornamemid(a, b, c) $$ and the right hand side similarly remains the same. Thus you can reorder $a, b, c$ as you please, and so it is enough to prove the equation for the specific case $a leq b leq c$, which you have already done.






                  share|cite|improve this answer













                  There are (at least) two options that you can use to prove it:



                  The first one is likely to be quite tedious. You consider all six possible orders that $a, b, c$ can be in, and prove the equation in each case.



                  The second is similar: You still consider a specific ordering of $a, b, c$ and prove the equation under that assumption. However, you don't consider every possible order. Instead, you notice that swapping any two variables in the equation doesn't change the value of either side of the equation. e.g. If you swap the values of $a$ and $b$, then the left hand side becomes
                  $$ operatornamemid(b, a, c) = operatornamemid(a, b, c) $$ and the right hand side similarly remains the same. Thus you can reorder $a, b, c$ as you please, and so it is enough to prove the equation for the specific case $a leq b leq c$, which you have already done.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Aug 2 at 7:50









                  Dylan

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