How to convert any decimal number in form of a√b [closed]












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Is there any method to directly convert any decimal number in form of a√b, given a and b are integers?
like 8.66=5√3(approx)










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closed as unclear what you're asking by Shailesh, Toby Mak, Brahadeesh, user10354138, KReiser Dec 2 '18 at 6:07


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











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    How about $2 = 1sqrt{2^2}$, $pi = 1sqrt{pi^2}$?
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    – Toby Mak
    Dec 2 '18 at 0:09






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    What are the restrictions on $a,b?$ By "any number" I take it you mean "any real number," right?
    $endgroup$
    – saulspatz
    Dec 2 '18 at 0:18


















0












$begingroup$


Is there any method to directly convert any decimal number in form of a√b, given a and b are integers?
like 8.66=5√3(approx)










share|cite|improve this question











$endgroup$



closed as unclear what you're asking by Shailesh, Toby Mak, Brahadeesh, user10354138, KReiser Dec 2 '18 at 6:07


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.











  • 1




    $begingroup$
    How about $2 = 1sqrt{2^2}$, $pi = 1sqrt{pi^2}$?
    $endgroup$
    – Toby Mak
    Dec 2 '18 at 0:09






  • 1




    $begingroup$
    What are the restrictions on $a,b?$ By "any number" I take it you mean "any real number," right?
    $endgroup$
    – saulspatz
    Dec 2 '18 at 0:18
















0












0








0





$begingroup$


Is there any method to directly convert any decimal number in form of a√b, given a and b are integers?
like 8.66=5√3(approx)










share|cite|improve this question











$endgroup$




Is there any method to directly convert any decimal number in form of a√b, given a and b are integers?
like 8.66=5√3(approx)







irrational-numbers






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edited Dec 2 '18 at 1:26







martha

















asked Dec 2 '18 at 0:08









marthamartha

71




71




closed as unclear what you're asking by Shailesh, Toby Mak, Brahadeesh, user10354138, KReiser Dec 2 '18 at 6:07


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






closed as unclear what you're asking by Shailesh, Toby Mak, Brahadeesh, user10354138, KReiser Dec 2 '18 at 6:07


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 1




    $begingroup$
    How about $2 = 1sqrt{2^2}$, $pi = 1sqrt{pi^2}$?
    $endgroup$
    – Toby Mak
    Dec 2 '18 at 0:09






  • 1




    $begingroup$
    What are the restrictions on $a,b?$ By "any number" I take it you mean "any real number," right?
    $endgroup$
    – saulspatz
    Dec 2 '18 at 0:18
















  • 1




    $begingroup$
    How about $2 = 1sqrt{2^2}$, $pi = 1sqrt{pi^2}$?
    $endgroup$
    – Toby Mak
    Dec 2 '18 at 0:09






  • 1




    $begingroup$
    What are the restrictions on $a,b?$ By "any number" I take it you mean "any real number," right?
    $endgroup$
    – saulspatz
    Dec 2 '18 at 0:18










1




1




$begingroup$
How about $2 = 1sqrt{2^2}$, $pi = 1sqrt{pi^2}$?
$endgroup$
– Toby Mak
Dec 2 '18 at 0:09




$begingroup$
How about $2 = 1sqrt{2^2}$, $pi = 1sqrt{pi^2}$?
$endgroup$
– Toby Mak
Dec 2 '18 at 0:09




1




1




$begingroup$
What are the restrictions on $a,b?$ By "any number" I take it you mean "any real number," right?
$endgroup$
– saulspatz
Dec 2 '18 at 0:18






$begingroup$
What are the restrictions on $a,b?$ By "any number" I take it you mean "any real number," right?
$endgroup$
– saulspatz
Dec 2 '18 at 0:18












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I presume $a$ and $b$ must be integers. In that case there is a very simple (almost trivial) solution: just take your number $x$, square it to get $x^2$, round it to the closest integer $n = lfloor x^2+1/2 rfloor$, and then you can express it as $x approx sqrt{n}$. Factoring $n$ you may be able to simplify the radical.






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    1 Answer
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    $begingroup$

    I presume $a$ and $b$ must be integers. In that case there is a very simple (almost trivial) solution: just take your number $x$, square it to get $x^2$, round it to the closest integer $n = lfloor x^2+1/2 rfloor$, and then you can express it as $x approx sqrt{n}$. Factoring $n$ you may be able to simplify the radical.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      I presume $a$ and $b$ must be integers. In that case there is a very simple (almost trivial) solution: just take your number $x$, square it to get $x^2$, round it to the closest integer $n = lfloor x^2+1/2 rfloor$, and then you can express it as $x approx sqrt{n}$. Factoring $n$ you may be able to simplify the radical.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        I presume $a$ and $b$ must be integers. In that case there is a very simple (almost trivial) solution: just take your number $x$, square it to get $x^2$, round it to the closest integer $n = lfloor x^2+1/2 rfloor$, and then you can express it as $x approx sqrt{n}$. Factoring $n$ you may be able to simplify the radical.






        share|cite|improve this answer











        $endgroup$



        I presume $a$ and $b$ must be integers. In that case there is a very simple (almost trivial) solution: just take your number $x$, square it to get $x^2$, round it to the closest integer $n = lfloor x^2+1/2 rfloor$, and then you can express it as $x approx sqrt{n}$. Factoring $n$ you may be able to simplify the radical.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 2 '18 at 0:37









        Théophile

        19.5k12946




        19.5k12946










        answered Dec 2 '18 at 0:26









        mlerma54mlerma54

        1,162138




        1,162138















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