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)
irrational-numbers
asked Dec 2 '18 at 0:08
marthamartha
71
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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.
add a comment |
$begingroup$
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)
irrational-numbers
asked Dec 2 '18 at 0:08
marthamartha
71
$endgroup$
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
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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
1
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What are the restrictions on $a,b?$ By "any number" I take it you mean "any real number," right?
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– saulspatz
Dec 2 '18 at 0:18
add a comment |
$begingroup$
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)
irrational-numbers
asked Dec 2 '18 at 0:08
marthamartha
71
$endgroup$
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
irrational-numbers
asked Dec 2 '18 at 0:08
marthamartha
71
asked Dec 2 '18 at 0:08
marthamartha
71
edited Dec 2 '18 at 1:26
martha
asked Dec 2 '18 at 0:08
marthamartha
71
asked Dec 2 '18 at 0:08
marthamartha
71
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
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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.
edited Dec 2 '18 at 0:37
Théophile
19.5k12946
answered Dec 2 '18 at 0:26
mlerma54mlerma54
1,162138
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
edited Dec 2 '18 at 0:37
Théophile
19.5k12946
answered Dec 2 '18 at 0:26
mlerma54mlerma54
1,162138
$endgroup$
add a comment |
$begingroup$
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.
edited Dec 2 '18 at 0:37
Théophile
19.5k12946
answered Dec 2 '18 at 0:26
mlerma54mlerma54
1,162138
$endgroup$
add a comment |
$begingroup$
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.
edited Dec 2 '18 at 0:37
Théophile
19.5k12946
answered Dec 2 '18 at 0:26
mlerma54mlerma54
1,162138
$endgroup$
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.
edited Dec 2 '18 at 0:37
Théophile
19.5k12946
answered Dec 2 '18 at 0:26
mlerma54mlerma54
1,162138
edited Dec 2 '18 at 0:37
Théophile
19.5k12946
edited Dec 2 '18 at 0:37
Théophile
19.5k12946
edited Dec 2 '18 at 0:37
Théophile
19.5k12946
19.5k12946
answered Dec 2 '18 at 0:26
mlerma54mlerma54
1,162138
answered Dec 2 '18 at 0:26
mlerma54mlerma54
1,162138
answered Dec 2 '18 at 0:26
mlerma54mlerma54
1,162138
1,162138
add a comment |
add a comment |
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