Simplifying radicals without using prime factorization
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.
elementary-number-theory prime-numbers prime-factorization
edited Nov 28 '18 at 10:01
daniel
6,22022157
asked Nov 26 '18 at 17:19
KiranKiran
3,21411634
$endgroup$
add a comment |
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.
elementary-number-theory prime-numbers prime-factorization
edited Nov 28 '18 at 10:01
daniel
6,22022157
asked Nov 26 '18 at 17:19
KiranKiran
3,21411634
$endgroup$
$begingroup$
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
add a comment |
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.
elementary-number-theory prime-numbers prime-factorization
edited Nov 28 '18 at 10:01
daniel
6,22022157
asked Nov 26 '18 at 17:19
KiranKiran
3,21411634
$endgroup$
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
elementary-number-theory prime-numbers prime-factorization
edited Nov 28 '18 at 10:01
daniel
6,22022157
asked Nov 26 '18 at 17:19
KiranKiran
3,21411634
edited Nov 28 '18 at 10:01
daniel
6,22022157
asked Nov 26 '18 at 17:19
KiranKiran
3,21411634
edited Nov 28 '18 at 10:01
daniel
6,22022157
edited Nov 28 '18 at 10:01
daniel
6,22022157
edited Nov 28 '18 at 10:01
daniel
6,22022157
6,22022157
asked Nov 26 '18 at 17:19
KiranKiran
3,21411634
asked Nov 26 '18 at 17:19
KiranKiran
3,21411634
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
add a comment |
$begingroup$
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
add a comment |
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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