Is $0.sqrt9$ a valid number?












0












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Is $0.sqrt9$ valid number? Are such numbers allowed?



First, I thought the value of the above number can be 0.3 but then it occurred how I would interpret this number: $0.65sqrt2$ or $0.65sqrt229$



Are such numbers valid?






real-numbers







share|cite|improve this question









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  • 1




    $begingroup$
    It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
    $endgroup$
    – Math_QED
    Jan 4 at 18:39






  • 3




    $begingroup$
    I would think that $0.sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$?
    $endgroup$
    – Eleven-Eleven
    Jan 4 at 18:40
















0












$begingroup$


Is $0.sqrt9$ valid number? Are such numbers allowed?



First, I thought the value of the above number can be 0.3 but then it occurred how I would interpret this number: $0.65sqrt2$ or $0.65sqrt229$



Are such numbers valid?






real-numbers







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
    $endgroup$
    – Math_QED
    Jan 4 at 18:39






  • 3




    $begingroup$
    I would think that $0.sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$?
    $endgroup$
    – Eleven-Eleven
    Jan 4 at 18:40














0












0








0





$begingroup$


Is $0.sqrt9$ valid number? Are such numbers allowed?



First, I thought the value of the above number can be 0.3 but then it occurred how I would interpret this number: $0.65sqrt2$ or $0.65sqrt229$



Are such numbers valid?






real-numbers







share|cite|improve this question









$endgroup$




Is $0.sqrt9$ valid number? Are such numbers allowed?



First, I thought the value of the above number can be 0.3 but then it occurred how I would interpret this number: $0.65sqrt2$ or $0.65sqrt229$



Are such numbers valid?






real-numbers




real-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 18:36









user33786user33786

116129




116129








  • 1




    $begingroup$
    It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
    $endgroup$
    – Math_QED
    Jan 4 at 18:39






  • 3




    $begingroup$
    I would think that $0.sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$?
    $endgroup$
    – Eleven-Eleven
    Jan 4 at 18:40














  • 1




    $begingroup$
    It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
    $endgroup$
    – Math_QED
    Jan 4 at 18:39






  • 3




    $begingroup$
    I would think that $0.sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$?
    $endgroup$
    – Eleven-Eleven
    Jan 4 at 18:40








1




1




$begingroup$
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
$endgroup$
– Math_QED
Jan 4 at 18:39




$begingroup$
It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous.
$endgroup$
– Math_QED
Jan 4 at 18:39




3




3




$begingroup$
I would think that $0.sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$?
$endgroup$
– Eleven-Eleven
Jan 4 at 18:40




$begingroup$
I would think that $0.sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$?
$endgroup$
– Eleven-Eleven
Jan 4 at 18:40










2 Answers
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$begingroup$

They're valid or invalid depending on the definition you use. What do you take these expressions to mean? If you divide each digit (treating surds as honorary digits) by successively larger powers of $10$, then sum, you'll have a definition that generalises the usual no-surds-allowed one. I sincerely doubt it'll catch on, though; you're getting expressions of the form $a+bsqrt{c}$, so you might as well just write that.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I asked myself a similar question a while ago about a supposed function $f(x) = 0.x$. It turns out you can express this function in terms of a more widely accepted notation as



    $$f(x) = 10^{-lfloor log_{10}(x)+1 rfloor}x.$$



    Using this definition, then indeed it turns out



    $$f(sqrt{9}) = 0.3.$$



    $65 sqrt 2$ is commonly interpreted as $65 cdot sqrt 2 = 91.9239...$ as opposed to the concatenation of the two, $65 || sqrt 2 = 651.41421...$. In either case that you mean, we have
    $$f(65 cdot sqrt 2) = frac{13}{10sqrt 2}, f(65||sqrt2) = frac{650+sqrt 2}{1000}$$



    and so on.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      3












      $begingroup$

      They're valid or invalid depending on the definition you use. What do you take these expressions to mean? If you divide each digit (treating surds as honorary digits) by successively larger powers of $10$, then sum, you'll have a definition that generalises the usual no-surds-allowed one. I sincerely doubt it'll catch on, though; you're getting expressions of the form $a+bsqrt{c}$, so you might as well just write that.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        They're valid or invalid depending on the definition you use. What do you take these expressions to mean? If you divide each digit (treating surds as honorary digits) by successively larger powers of $10$, then sum, you'll have a definition that generalises the usual no-surds-allowed one. I sincerely doubt it'll catch on, though; you're getting expressions of the form $a+bsqrt{c}$, so you might as well just write that.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          They're valid or invalid depending on the definition you use. What do you take these expressions to mean? If you divide each digit (treating surds as honorary digits) by successively larger powers of $10$, then sum, you'll have a definition that generalises the usual no-surds-allowed one. I sincerely doubt it'll catch on, though; you're getting expressions of the form $a+bsqrt{c}$, so you might as well just write that.






          share|cite|improve this answer









          $endgroup$



          They're valid or invalid depending on the definition you use. What do you take these expressions to mean? If you divide each digit (treating surds as honorary digits) by successively larger powers of $10$, then sum, you'll have a definition that generalises the usual no-surds-allowed one. I sincerely doubt it'll catch on, though; you're getting expressions of the form $a+bsqrt{c}$, so you might as well just write that.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 18:41









          J.G.J.G.

          24.5k22539




          24.5k22539























              0












              $begingroup$

              I asked myself a similar question a while ago about a supposed function $f(x) = 0.x$. It turns out you can express this function in terms of a more widely accepted notation as



              $$f(x) = 10^{-lfloor log_{10}(x)+1 rfloor}x.$$



              Using this definition, then indeed it turns out



              $$f(sqrt{9}) = 0.3.$$



              $65 sqrt 2$ is commonly interpreted as $65 cdot sqrt 2 = 91.9239...$ as opposed to the concatenation of the two, $65 || sqrt 2 = 651.41421...$. In either case that you mean, we have
              $$f(65 cdot sqrt 2) = frac{13}{10sqrt 2}, f(65||sqrt2) = frac{650+sqrt 2}{1000}$$



              and so on.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I asked myself a similar question a while ago about a supposed function $f(x) = 0.x$. It turns out you can express this function in terms of a more widely accepted notation as



                $$f(x) = 10^{-lfloor log_{10}(x)+1 rfloor}x.$$



                Using this definition, then indeed it turns out



                $$f(sqrt{9}) = 0.3.$$



                $65 sqrt 2$ is commonly interpreted as $65 cdot sqrt 2 = 91.9239...$ as opposed to the concatenation of the two, $65 || sqrt 2 = 651.41421...$. In either case that you mean, we have
                $$f(65 cdot sqrt 2) = frac{13}{10sqrt 2}, f(65||sqrt2) = frac{650+sqrt 2}{1000}$$



                and so on.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I asked myself a similar question a while ago about a supposed function $f(x) = 0.x$. It turns out you can express this function in terms of a more widely accepted notation as



                  $$f(x) = 10^{-lfloor log_{10}(x)+1 rfloor}x.$$



                  Using this definition, then indeed it turns out



                  $$f(sqrt{9}) = 0.3.$$



                  $65 sqrt 2$ is commonly interpreted as $65 cdot sqrt 2 = 91.9239...$ as opposed to the concatenation of the two, $65 || sqrt 2 = 651.41421...$. In either case that you mean, we have
                  $$f(65 cdot sqrt 2) = frac{13}{10sqrt 2}, f(65||sqrt2) = frac{650+sqrt 2}{1000}$$



                  and so on.






                  share|cite|improve this answer









                  $endgroup$



                  I asked myself a similar question a while ago about a supposed function $f(x) = 0.x$. It turns out you can express this function in terms of a more widely accepted notation as



                  $$f(x) = 10^{-lfloor log_{10}(x)+1 rfloor}x.$$



                  Using this definition, then indeed it turns out



                  $$f(sqrt{9}) = 0.3.$$



                  $65 sqrt 2$ is commonly interpreted as $65 cdot sqrt 2 = 91.9239...$ as opposed to the concatenation of the two, $65 || sqrt 2 = 651.41421...$. In either case that you mean, we have
                  $$f(65 cdot sqrt 2) = frac{13}{10sqrt 2}, f(65||sqrt2) = frac{650+sqrt 2}{1000}$$



                  and so on.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 19:46









                  SkipSkip

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