Order in which to take max and square

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I am pretty sure that $(displaystylemax_i)^2=max_i^2$ where $i=1,ldots, n$ and $a_iinmathbbR^n$. Could someone please confirm?








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    I am pretty sure that $(displaystylemax_i)^2=max_i^2$ where $i=1,ldots, n$ and $a_iinmathbbR^n$. Could someone please confirm?








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      I am pretty sure that $(displaystylemax_i)^2=max_i^2$ where $i=1,ldots, n$ and $a_iinmathbbR^n$. Could someone please confirm?








      share|cite|improve this question














      I am pretty sure that $(displaystylemax_i)^2=max_i^2$ where $i=1,ldots, n$ and $a_iinmathbbR^n$. Could someone please confirm?










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      edited Jul 21 at 3:00









      Math Lover

      12.4k21232




      12.4k21232









      asked Jul 21 at 2:14









      kmiyazaki

      31611




      31611




















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          Yes, this is because $x mapsto x^2$ is an increasing function for $x > 0$.






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          • Does any increasing function commute with the operation of taking max?
            – kmiyazaki
            Jul 21 at 2:49










          • Yes; try to show this, it's not too hard.
            – Marcus M
            Jul 21 at 2:50










          • I will. Thank you very much.
            – kmiyazaki
            Jul 21 at 2:56










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          1 Answer
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          1 Answer
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          up vote
          0
          down vote













          Yes, this is because $x mapsto x^2$ is an increasing function for $x > 0$.






          share|cite|improve this answer





















          • Does any increasing function commute with the operation of taking max?
            – kmiyazaki
            Jul 21 at 2:49










          • Yes; try to show this, it's not too hard.
            – Marcus M
            Jul 21 at 2:50










          • I will. Thank you very much.
            – kmiyazaki
            Jul 21 at 2:56














          up vote
          0
          down vote













          Yes, this is because $x mapsto x^2$ is an increasing function for $x > 0$.






          share|cite|improve this answer





















          • Does any increasing function commute with the operation of taking max?
            – kmiyazaki
            Jul 21 at 2:49










          • Yes; try to show this, it's not too hard.
            – Marcus M
            Jul 21 at 2:50










          • I will. Thank you very much.
            – kmiyazaki
            Jul 21 at 2:56












          up vote
          0
          down vote










          up vote
          0
          down vote









          Yes, this is because $x mapsto x^2$ is an increasing function for $x > 0$.






          share|cite|improve this answer













          Yes, this is because $x mapsto x^2$ is an increasing function for $x > 0$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 21 at 2:46









          Marcus M

          8,1731847




          8,1731847











          • Does any increasing function commute with the operation of taking max?
            – kmiyazaki
            Jul 21 at 2:49










          • Yes; try to show this, it's not too hard.
            – Marcus M
            Jul 21 at 2:50










          • I will. Thank you very much.
            – kmiyazaki
            Jul 21 at 2:56
















          • Does any increasing function commute with the operation of taking max?
            – kmiyazaki
            Jul 21 at 2:49










          • Yes; try to show this, it's not too hard.
            – Marcus M
            Jul 21 at 2:50










          • I will. Thank you very much.
            – kmiyazaki
            Jul 21 at 2:56















          Does any increasing function commute with the operation of taking max?
          – kmiyazaki
          Jul 21 at 2:49




          Does any increasing function commute with the operation of taking max?
          – kmiyazaki
          Jul 21 at 2:49












          Yes; try to show this, it's not too hard.
          – Marcus M
          Jul 21 at 2:50




          Yes; try to show this, it's not too hard.
          – Marcus M
          Jul 21 at 2:50












          I will. Thank you very much.
          – kmiyazaki
          Jul 21 at 2:56




          I will. Thank you very much.
          – kmiyazaki
          Jul 21 at 2:56












           

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