Egyptian Fraction when numerator is greater than denominator
$begingroup$
I am doing an assignment about Egyptian fractions and I am a bit confused about what to do when the given fraction's numerator is greater than denominator. My initial idea was to subtract the fraction by 1, 1/2, 1/3 etc and when the numerator becomes less than denominator I would apply the proper algorithm. And I saw that this way does not work out. Do you have any suggestions?
Thank you.
EDIT: I cannot subtract by 1 because it will not be an Egyptian fraction.
number-theory elementary-number-theory egyptian-fractions
asked Dec 24 '18 at 15:01
ManosLManosL
385
$endgroup$
add a comment |
$begingroup$
I am doing an assignment about Egyptian fractions and I am a bit confused about what to do when the given fraction's numerator is greater than denominator. My initial idea was to subtract the fraction by 1, 1/2, 1/3 etc and when the numerator becomes less than denominator I would apply the proper algorithm. And I saw that this way does not work out. Do you have any suggestions?
Thank you.
EDIT: I cannot subtract by 1 because it will not be an Egyptian fraction.
number-theory elementary-number-theory egyptian-fractions
asked Dec 24 '18 at 15:01
ManosLManosL
385
$endgroup$
$begingroup$
Convert to a mixed number first, I think.
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:03
$begingroup$
To be honest I don't want my number to have the form 1 + 1+1+ 1/n1. In the final representation each 1/n for n belongs to natural numbers, should be appeared one time.
$endgroup$
– ManosL
Dec 24 '18 at 15:08
$begingroup$
Right, it'd be, for instance, 17/5 = 3 + 1/3 + 1/15
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:10
$begingroup$
I don't want that too
$endgroup$
– ManosL
Dec 24 '18 at 15:11
$begingroup$
Norm is to let the integer part be a single term, with denominator $1$ as it were. Thus if you ask Wolfram alpha for the egyptian fraction representation of $frac 72$ it will return $3+frac 12$.
$endgroup$
– lulu
Dec 24 '18 at 15:26
add a comment |
$begingroup$
I am doing an assignment about Egyptian fractions and I am a bit confused about what to do when the given fraction's numerator is greater than denominator. My initial idea was to subtract the fraction by 1, 1/2, 1/3 etc and when the numerator becomes less than denominator I would apply the proper algorithm. And I saw that this way does not work out. Do you have any suggestions?
Thank you.
EDIT: I cannot subtract by 1 because it will not be an Egyptian fraction.
number-theory elementary-number-theory egyptian-fractions
asked Dec 24 '18 at 15:01
ManosLManosL
385
$endgroup$
I am doing an assignment about Egyptian fractions and I am a bit confused about what to do when the given fraction's numerator is greater than denominator. My initial idea was to subtract the fraction by 1, 1/2, 1/3 etc and when the numerator becomes less than denominator I would apply the proper algorithm. And I saw that this way does not work out. Do you have any suggestions?
Thank you.
EDIT: I cannot subtract by 1 because it will not be an Egyptian fraction.
number-theory elementary-number-theory egyptian-fractions
number-theory elementary-number-theory egyptian-fractions
asked Dec 24 '18 at 15:01
ManosLManosL
385
asked Dec 24 '18 at 15:01
ManosLManosL
385
edited Dec 24 '18 at 15:03
ManosL
asked Dec 24 '18 at 15:01
ManosLManosL
385
asked Dec 24 '18 at 15:01
ManosLManosL
385
asked Dec 24 '18 at 15:01
ManosLManosL
385
385
$begingroup$
Convert to a mixed number first, I think.
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:03
$begingroup$
To be honest I don't want my number to have the form 1 + 1+1+ 1/n1. In the final representation each 1/n for n belongs to natural numbers, should be appeared one time.
$endgroup$
– ManosL
Dec 24 '18 at 15:08
$begingroup$
Right, it'd be, for instance, 17/5 = 3 + 1/3 + 1/15
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:10
$begingroup$
I don't want that too
$endgroup$
– ManosL
Dec 24 '18 at 15:11
$begingroup$
Norm is to let the integer part be a single term, with denominator $1$ as it were. Thus if you ask Wolfram alpha for the egyptian fraction representation of $frac 72$ it will return $3+frac 12$.
$endgroup$
– lulu
Dec 24 '18 at 15:26
add a comment |
$begingroup$
Convert to a mixed number first, I think.
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:03
$begingroup$
To be honest I don't want my number to have the form 1 + 1+1+ 1/n1. In the final representation each 1/n for n belongs to natural numbers, should be appeared one time.
$endgroup$
– ManosL
Dec 24 '18 at 15:08
$begingroup$
Right, it'd be, for instance, 17/5 = 3 + 1/3 + 1/15
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:10
$begingroup$
I don't want that too
$endgroup$
– ManosL
Dec 24 '18 at 15:11
$begingroup$
Norm is to let the integer part be a single term, with denominator $1$ as it were. Thus if you ask Wolfram alpha for the egyptian fraction representation of $frac 72$ it will return $3+frac 12$.
$endgroup$
– lulu
Dec 24 '18 at 15:26
$begingroup$
Convert to a mixed number first, I think.
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:03
$begingroup$
Convert to a mixed number first, I think.
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:03
$begingroup$
To be honest I don't want my number to have the form 1 + 1+1+ 1/n1. In the final representation each 1/n for n belongs to natural numbers, should be appeared one time.
$endgroup$
– ManosL
Dec 24 '18 at 15:08
$begingroup$
To be honest I don't want my number to have the form 1 + 1+1+ 1/n1. In the final representation each 1/n for n belongs to natural numbers, should be appeared one time.
$endgroup$
– ManosL
Dec 24 '18 at 15:08
$begingroup$
Right, it'd be, for instance, 17/5 = 3 + 1/3 + 1/15
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:10
$begingroup$
Right, it'd be, for instance, 17/5 = 3 + 1/3 + 1/15
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:10
$begingroup$
I don't want that too
$endgroup$
– ManosL
Dec 24 '18 at 15:11
$begingroup$
I don't want that too
$endgroup$
– ManosL
Dec 24 '18 at 15:11
$begingroup$
Norm is to let the integer part be a single term, with denominator $1$ as it were. Thus if you ask Wolfram alpha for the egyptian fraction representation of $frac 72$ it will return $3+frac 12$.
$endgroup$
– lulu
Dec 24 '18 at 15:26
$begingroup$
Norm is to let the integer part be a single term, with denominator $1$ as it were. Thus if you ask Wolfram alpha for the egyptian fraction representation of $frac 72$ it will return $3+frac 12$.
$endgroup$
– lulu
Dec 24 '18 at 15:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your trick is pretty close to getting what you want, but you stop too early. As you probably know, the series
$$
1 + frac12 + frac13 + frac14 + cdots
$$
diverges, which means that for any initial value $frac pq$ you started with, there is an $n$ such that
$$
H_n = 1 + frac12 + cdots + frac1n leq frac pq < 1 + frac12 + cdots + frac1{n+1} = H_{n+1}.
$$
Now, you can look at the rational $frac pq - H_n$, which must be less than $frac1{n+1}$. Thus, if we apply any ordinary Egyptian fraction algorithm that works for fractions less than 1, you get a representation of $frac pq - H_n$, and any term in this Egyptian fraction decomposition will have denominator greater than $n + 1$. Now just add $H_n$ back to this representation, and you have a decomposition of $frac pq$ of the form that you are looking for.
answered Dec 24 '18 at 15:41
Mees de VriesMees de Vries
17.6k13060
$endgroup$
1
$begingroup$
As $H_n approx ln n+0.577$ if the integer part of the number to be expressed is at all large it will take a lot of terms. If we want $H_n gt 10$ we need $n$ about $12370$
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:46
1
$begingroup$
True, but also note that this is basically optimal; within the constraints given by OP, $H_n$ is the "fastest" way to get the value down to below 1.
$endgroup$
– Mees de Vries
Dec 24 '18 at 15:49
$begingroup$
I agree. I just thought it good to show how slow it is.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:51
$begingroup$
In the particular case of $frac{17}{15}$ I used above, this gives $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{13 889}{720 720}$, which expands finally to $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{1}{52} + frac{1}{24 853} + frac{1}{1 053 650 245} + frac{1}{3 774 608 050 827 453 840}$
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:56
$begingroup$
Wow! Thank you very much this looks good.
$endgroup$
– ManosL
Dec 24 '18 at 15:59
add a comment |
Your Answer
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1 Answer
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$begingroup$
Your trick is pretty close to getting what you want, but you stop too early. As you probably know, the series
$$
1 + frac12 + frac13 + frac14 + cdots
$$
diverges, which means that for any initial value $frac pq$ you started with, there is an $n$ such that
$$
H_n = 1 + frac12 + cdots + frac1n leq frac pq < 1 + frac12 + cdots + frac1{n+1} = H_{n+1}.
$$
Now, you can look at the rational $frac pq - H_n$, which must be less than $frac1{n+1}$. Thus, if we apply any ordinary Egyptian fraction algorithm that works for fractions less than 1, you get a representation of $frac pq - H_n$, and any term in this Egyptian fraction decomposition will have denominator greater than $n + 1$. Now just add $H_n$ back to this representation, and you have a decomposition of $frac pq$ of the form that you are looking for.
answered Dec 24 '18 at 15:41
Mees de VriesMees de Vries
17.6k13060
$endgroup$
1
$begingroup$
As $H_n approx ln n+0.577$ if the integer part of the number to be expressed is at all large it will take a lot of terms. If we want $H_n gt 10$ we need $n$ about $12370$
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:46
1
$begingroup$
True, but also note that this is basically optimal; within the constraints given by OP, $H_n$ is the "fastest" way to get the value down to below 1.
$endgroup$
– Mees de Vries
Dec 24 '18 at 15:49
$begingroup$
I agree. I just thought it good to show how slow it is.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:51
$begingroup$
In the particular case of $frac{17}{15}$ I used above, this gives $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{13 889}{720 720}$, which expands finally to $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{1}{52} + frac{1}{24 853} + frac{1}{1 053 650 245} + frac{1}{3 774 608 050 827 453 840}$
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:56
$begingroup$
Wow! Thank you very much this looks good.
$endgroup$
– ManosL
Dec 24 '18 at 15:59
add a comment |
$begingroup$
Your trick is pretty close to getting what you want, but you stop too early. As you probably know, the series
$$
1 + frac12 + frac13 + frac14 + cdots
$$
diverges, which means that for any initial value $frac pq$ you started with, there is an $n$ such that
$$
H_n = 1 + frac12 + cdots + frac1n leq frac pq < 1 + frac12 + cdots + frac1{n+1} = H_{n+1}.
$$
Now, you can look at the rational $frac pq - H_n$, which must be less than $frac1{n+1}$. Thus, if we apply any ordinary Egyptian fraction algorithm that works for fractions less than 1, you get a representation of $frac pq - H_n$, and any term in this Egyptian fraction decomposition will have denominator greater than $n + 1$. Now just add $H_n$ back to this representation, and you have a decomposition of $frac pq$ of the form that you are looking for.
answered Dec 24 '18 at 15:41
Mees de VriesMees de Vries
17.6k13060
$endgroup$
1
$begingroup$
As $H_n approx ln n+0.577$ if the integer part of the number to be expressed is at all large it will take a lot of terms. If we want $H_n gt 10$ we need $n$ about $12370$
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:46
1
$begingroup$
True, but also note that this is basically optimal; within the constraints given by OP, $H_n$ is the "fastest" way to get the value down to below 1.
$endgroup$
– Mees de Vries
Dec 24 '18 at 15:49
$begingroup$
I agree. I just thought it good to show how slow it is.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:51
$begingroup$
In the particular case of $frac{17}{15}$ I used above, this gives $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{13 889}{720 720}$, which expands finally to $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{1}{52} + frac{1}{24 853} + frac{1}{1 053 650 245} + frac{1}{3 774 608 050 827 453 840}$
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:56
$begingroup$
Wow! Thank you very much this looks good.
$endgroup$
– ManosL
Dec 24 '18 at 15:59
add a comment |
$begingroup$
Your trick is pretty close to getting what you want, but you stop too early. As you probably know, the series
$$
1 + frac12 + frac13 + frac14 + cdots
$$
diverges, which means that for any initial value $frac pq$ you started with, there is an $n$ such that
$$
H_n = 1 + frac12 + cdots + frac1n leq frac pq < 1 + frac12 + cdots + frac1{n+1} = H_{n+1}.
$$
Now, you can look at the rational $frac pq - H_n$, which must be less than $frac1{n+1}$. Thus, if we apply any ordinary Egyptian fraction algorithm that works for fractions less than 1, you get a representation of $frac pq - H_n$, and any term in this Egyptian fraction decomposition will have denominator greater than $n + 1$. Now just add $H_n$ back to this representation, and you have a decomposition of $frac pq$ of the form that you are looking for.
answered Dec 24 '18 at 15:41
Mees de VriesMees de Vries
17.6k13060
$endgroup$
Your trick is pretty close to getting what you want, but you stop too early. As you probably know, the series
$$
1 + frac12 + frac13 + frac14 + cdots
$$
diverges, which means that for any initial value $frac pq$ you started with, there is an $n$ such that
$$
H_n = 1 + frac12 + cdots + frac1n leq frac pq < 1 + frac12 + cdots + frac1{n+1} = H_{n+1}.
$$
Now, you can look at the rational $frac pq - H_n$, which must be less than $frac1{n+1}$. Thus, if we apply any ordinary Egyptian fraction algorithm that works for fractions less than 1, you get a representation of $frac pq - H_n$, and any term in this Egyptian fraction decomposition will have denominator greater than $n + 1$. Now just add $H_n$ back to this representation, and you have a decomposition of $frac pq$ of the form that you are looking for.
answered Dec 24 '18 at 15:41
Mees de VriesMees de Vries
17.6k13060
answered Dec 24 '18 at 15:41
Mees de VriesMees de Vries
17.6k13060
answered Dec 24 '18 at 15:41
Mees de VriesMees de Vries
17.6k13060
answered Dec 24 '18 at 15:41
Mees de VriesMees de Vries
17.6k13060
17.6k13060
1
$begingroup$
As $H_n approx ln n+0.577$ if the integer part of the number to be expressed is at all large it will take a lot of terms. If we want $H_n gt 10$ we need $n$ about $12370$
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:46
1
$begingroup$
True, but also note that this is basically optimal; within the constraints given by OP, $H_n$ is the "fastest" way to get the value down to below 1.
$endgroup$
– Mees de Vries
Dec 24 '18 at 15:49
$begingroup$
I agree. I just thought it good to show how slow it is.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:51
$begingroup$
In the particular case of $frac{17}{15}$ I used above, this gives $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{13 889}{720 720}$, which expands finally to $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{1}{52} + frac{1}{24 853} + frac{1}{1 053 650 245} + frac{1}{3 774 608 050 827 453 840}$
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:56
$begingroup$
Wow! Thank you very much this looks good.
$endgroup$
– ManosL
Dec 24 '18 at 15:59
add a comment |
1
$begingroup$
As $H_n approx ln n+0.577$ if the integer part of the number to be expressed is at all large it will take a lot of terms. If we want $H_n gt 10$ we need $n$ about $12370$
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:46
1
$begingroup$
True, but also note that this is basically optimal; within the constraints given by OP, $H_n$ is the "fastest" way to get the value down to below 1.
$endgroup$
– Mees de Vries
Dec 24 '18 at 15:49
$begingroup$
I agree. I just thought it good to show how slow it is.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:51
$begingroup$
In the particular case of $frac{17}{15}$ I used above, this gives $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{13 889}{720 720}$, which expands finally to $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{1}{52} + frac{1}{24 853} + frac{1}{1 053 650 245} + frac{1}{3 774 608 050 827 453 840}$
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:56
$begingroup$
Wow! Thank you very much this looks good.
$endgroup$
– ManosL
Dec 24 '18 at 15:59
1
1
$begingroup$
As $H_n approx ln n+0.577$ if the integer part of the number to be expressed is at all large it will take a lot of terms. If we want $H_n gt 10$ we need $n$ about $12370$
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:46
$begingroup$
As $H_n approx ln n+0.577$ if the integer part of the number to be expressed is at all large it will take a lot of terms. If we want $H_n gt 10$ we need $n$ about $12370$
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:46
1
1
$begingroup$
True, but also note that this is basically optimal; within the constraints given by OP, $H_n$ is the "fastest" way to get the value down to below 1.
$endgroup$
– Mees de Vries
Dec 24 '18 at 15:49
$begingroup$
True, but also note that this is basically optimal; within the constraints given by OP, $H_n$ is the "fastest" way to get the value down to below 1.
$endgroup$
– Mees de Vries
Dec 24 '18 at 15:49
$begingroup$
I agree. I just thought it good to show how slow it is.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:51
$begingroup$
I agree. I just thought it good to show how slow it is.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:51
$begingroup$
In the particular case of $frac{17}{15}$ I used above, this gives $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{13 889}{720 720}$, which expands finally to $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{1}{52} + frac{1}{24 853} + frac{1}{1 053 650 245} + frac{1}{3 774 608 050 827 453 840}$
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:56
$begingroup$
In the particular case of $frac{17}{15}$ I used above, this gives $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{13 889}{720 720}$, which expands finally to $1 + frac{1}{2} + frac{1}{3} + cdots + frac{1}{16} + frac{1}{52} + frac{1}{24 853} + frac{1}{1 053 650 245} + frac{1}{3 774 608 050 827 453 840}$
$endgroup$
– Dan Uznanski
Dec 24 '18 at 15:56
$begingroup$
Wow! Thank you very much this looks good.
$endgroup$
– ManosL
Dec 24 '18 at 15:59
$begingroup$
Wow! Thank you very much this looks good.
$endgroup$
– ManosL
Dec 24 '18 at 15:59
add a comment |
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Convert to a mixed number first, I think.
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– Dan Uznanski
Dec 24 '18 at 15:03
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To be honest I don't want my number to have the form 1 + 1+1+ 1/n1. In the final representation each 1/n for n belongs to natural numbers, should be appeared one time.
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– ManosL
Dec 24 '18 at 15:08
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Right, it'd be, for instance, 17/5 = 3 + 1/3 + 1/15
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– Dan Uznanski
Dec 24 '18 at 15:10
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I don't want that too
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– ManosL
Dec 24 '18 at 15:11
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Norm is to let the integer part be a single term, with denominator $1$ as it were. Thus if you ask Wolfram alpha for the egyptian fraction representation of $frac 72$ it will return $3+frac 12$.
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– lulu
Dec 24 '18 at 15:26