Simplifying radicals without using prime factorization

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Is there an easy way to simplify radicals?



For example, take the case of $sqrt{252}$. We can the find prime factorization of $252$ as $252=2times 2times 3 times 3times 7$ and thus we get $sqrt{252}=sqrt{2times 2times 3 times 3times 7}=6sqrt{7}.$



This method takes more time for large numbers. Without doing these calculations, i.e., without finding out prime factorization, is there any approach available to simplify radicals?



Please help. thanks.










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  • $begingroup$
    For this kind of simplification, not really, I think. If all you're after is $15.87$, then there are numerous methods (not that I personally know very many, but I know they are there).
    $endgroup$
    – Arthur
    Nov 26 '18 at 17:28












  • $begingroup$
    thanks for the info
    $endgroup$
    – Kiran
    Nov 26 '18 at 17:34






  • 1




    $begingroup$
    If the number if definable as a multiple of some perfect square factors, that would mean the radical is reducible.For example:$252=16^2-2^2=(16-2)(16+2)=2^2.3^2.7$. I think this takes shorter time than factorizing.
    $endgroup$
    – sirous
    Nov 27 '18 at 3:21










  • $begingroup$
    If the number is too large to complete the factorization, you cannot do any better than factor the number as far as possible. If you continue upto, lets say, $10^6$, chances that the cofactor is squarefree, are very good. In most cases, you will not miss a better solution, but only the factorization makes all doubts vanish.
    $endgroup$
    – Peter
    Jan 19 at 15:39










  • $begingroup$
    @sirous But in general, deciding whether a number is squarefree is not significantly easier than factoring.
    $endgroup$
    – Peter
    Jan 20 at 12:50
















2












$begingroup$


Is there an easy way to simplify radicals?



For example, take the case of $sqrt{252}$. We can the find prime factorization of $252$ as $252=2times 2times 3 times 3times 7$ and thus we get $sqrt{252}=sqrt{2times 2times 3 times 3times 7}=6sqrt{7}.$



This method takes more time for large numbers. Without doing these calculations, i.e., without finding out prime factorization, is there any approach available to simplify radicals?



Please help. thanks.










share|cite|improve this question











$endgroup$












  • $begingroup$
    For this kind of simplification, not really, I think. If all you're after is $15.87$, then there are numerous methods (not that I personally know very many, but I know they are there).
    $endgroup$
    – Arthur
    Nov 26 '18 at 17:28












  • $begingroup$
    thanks for the info
    $endgroup$
    – Kiran
    Nov 26 '18 at 17:34






  • 1




    $begingroup$
    If the number if definable as a multiple of some perfect square factors, that would mean the radical is reducible.For example:$252=16^2-2^2=(16-2)(16+2)=2^2.3^2.7$. I think this takes shorter time than factorizing.
    $endgroup$
    – sirous
    Nov 27 '18 at 3:21










  • $begingroup$
    If the number is too large to complete the factorization, you cannot do any better than factor the number as far as possible. If you continue upto, lets say, $10^6$, chances that the cofactor is squarefree, are very good. In most cases, you will not miss a better solution, but only the factorization makes all doubts vanish.
    $endgroup$
    – Peter
    Jan 19 at 15:39










  • $begingroup$
    @sirous But in general, deciding whether a number is squarefree is not significantly easier than factoring.
    $endgroup$
    – Peter
    Jan 20 at 12:50














2












2








2





$begingroup$


Is there an easy way to simplify radicals?



For example, take the case of $sqrt{252}$. We can the find prime factorization of $252$ as $252=2times 2times 3 times 3times 7$ and thus we get $sqrt{252}=sqrt{2times 2times 3 times 3times 7}=6sqrt{7}.$



This method takes more time for large numbers. Without doing these calculations, i.e., without finding out prime factorization, is there any approach available to simplify radicals?



Please help. thanks.










share|cite|improve this question











$endgroup$




Is there an easy way to simplify radicals?



For example, take the case of $sqrt{252}$. We can the find prime factorization of $252$ as $252=2times 2times 3 times 3times 7$ and thus we get $sqrt{252}=sqrt{2times 2times 3 times 3times 7}=6sqrt{7}.$



This method takes more time for large numbers. Without doing these calculations, i.e., without finding out prime factorization, is there any approach available to simplify radicals?



Please help. thanks.







elementary-number-theory prime-numbers prime-factorization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 10:01









daniel

6,22022157




6,22022157










asked Nov 26 '18 at 17:19









KiranKiran

3,21411634




3,21411634












  • $begingroup$
    For this kind of simplification, not really, I think. If all you're after is $15.87$, then there are numerous methods (not that I personally know very many, but I know they are there).
    $endgroup$
    – Arthur
    Nov 26 '18 at 17:28












  • $begingroup$
    thanks for the info
    $endgroup$
    – Kiran
    Nov 26 '18 at 17:34






  • 1




    $begingroup$
    If the number if definable as a multiple of some perfect square factors, that would mean the radical is reducible.For example:$252=16^2-2^2=(16-2)(16+2)=2^2.3^2.7$. I think this takes shorter time than factorizing.
    $endgroup$
    – sirous
    Nov 27 '18 at 3:21










  • $begingroup$
    If the number is too large to complete the factorization, you cannot do any better than factor the number as far as possible. If you continue upto, lets say, $10^6$, chances that the cofactor is squarefree, are very good. In most cases, you will not miss a better solution, but only the factorization makes all doubts vanish.
    $endgroup$
    – Peter
    Jan 19 at 15:39










  • $begingroup$
    @sirous But in general, deciding whether a number is squarefree is not significantly easier than factoring.
    $endgroup$
    – Peter
    Jan 20 at 12:50


















  • $begingroup$
    For this kind of simplification, not really, I think. If all you're after is $15.87$, then there are numerous methods (not that I personally know very many, but I know they are there).
    $endgroup$
    – Arthur
    Nov 26 '18 at 17:28












  • $begingroup$
    thanks for the info
    $endgroup$
    – Kiran
    Nov 26 '18 at 17:34






  • 1




    $begingroup$
    If the number if definable as a multiple of some perfect square factors, that would mean the radical is reducible.For example:$252=16^2-2^2=(16-2)(16+2)=2^2.3^2.7$. I think this takes shorter time than factorizing.
    $endgroup$
    – sirous
    Nov 27 '18 at 3:21










  • $begingroup$
    If the number is too large to complete the factorization, you cannot do any better than factor the number as far as possible. If you continue upto, lets say, $10^6$, chances that the cofactor is squarefree, are very good. In most cases, you will not miss a better solution, but only the factorization makes all doubts vanish.
    $endgroup$
    – Peter
    Jan 19 at 15:39










  • $begingroup$
    @sirous But in general, deciding whether a number is squarefree is not significantly easier than factoring.
    $endgroup$
    – Peter
    Jan 20 at 12:50
















$begingroup$
For this kind of simplification, not really, I think. If all you're after is $15.87$, then there are numerous methods (not that I personally know very many, but I know they are there).
$endgroup$
– Arthur
Nov 26 '18 at 17:28






$begingroup$
For this kind of simplification, not really, I think. If all you're after is $15.87$, then there are numerous methods (not that I personally know very many, but I know they are there).
$endgroup$
– Arthur
Nov 26 '18 at 17:28














$begingroup$
thanks for the info
$endgroup$
– Kiran
Nov 26 '18 at 17:34




$begingroup$
thanks for the info
$endgroup$
– Kiran
Nov 26 '18 at 17:34




1




1




$begingroup$
If the number if definable as a multiple of some perfect square factors, that would mean the radical is reducible.For example:$252=16^2-2^2=(16-2)(16+2)=2^2.3^2.7$. I think this takes shorter time than factorizing.
$endgroup$
– sirous
Nov 27 '18 at 3:21




$begingroup$
If the number if definable as a multiple of some perfect square factors, that would mean the radical is reducible.For example:$252=16^2-2^2=(16-2)(16+2)=2^2.3^2.7$. I think this takes shorter time than factorizing.
$endgroup$
– sirous
Nov 27 '18 at 3:21












$begingroup$
If the number is too large to complete the factorization, you cannot do any better than factor the number as far as possible. If you continue upto, lets say, $10^6$, chances that the cofactor is squarefree, are very good. In most cases, you will not miss a better solution, but only the factorization makes all doubts vanish.
$endgroup$
– Peter
Jan 19 at 15:39




$begingroup$
If the number is too large to complete the factorization, you cannot do any better than factor the number as far as possible. If you continue upto, lets say, $10^6$, chances that the cofactor is squarefree, are very good. In most cases, you will not miss a better solution, but only the factorization makes all doubts vanish.
$endgroup$
– Peter
Jan 19 at 15:39












$begingroup$
@sirous But in general, deciding whether a number is squarefree is not significantly easier than factoring.
$endgroup$
– Peter
Jan 20 at 12:50




$begingroup$
@sirous But in general, deciding whether a number is squarefree is not significantly easier than factoring.
$endgroup$
– Peter
Jan 20 at 12:50










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