Difficulty understanding how Babylonian reciprocals work
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According to the reciprocal tables for Babylonian's base $60$ system, dividing by $2$ is like multiplying like $30$. Dividing by $3$ is like multiplying by $20$. Dividing by $4$ is like multiplying by $15$. Dividing by $k$ is like multiplying by $60/k$.
I don't understand why the reciprocal works like this and not like $1/k$. You might say "It multiplies by $60/k$ because it's base $60$" but this doesn't make sense to me, it's not like our base-$10$ reciprocals look like $10/k$.
What's going on?
math-history number-systems
edited Nov 11 '18 at 7:57
amWhy
1
asked Nov 11 '18 at 7:52
user614482user614482
112
$endgroup$
|
show 2 more comments
$begingroup$
According to the reciprocal tables for Babylonian's base $60$ system, dividing by $2$ is like multiplying like $30$. Dividing by $3$ is like multiplying by $20$. Dividing by $4$ is like multiplying by $15$. Dividing by $k$ is like multiplying by $60/k$.
I don't understand why the reciprocal works like this and not like $1/k$. You might say "It multiplies by $60/k$ because it's base $60$" but this doesn't make sense to me, it's not like our base-$10$ reciprocals look like $10/k$.
What's going on?
math-history number-systems
edited Nov 11 '18 at 7:57
amWhy
1
asked Nov 11 '18 at 7:52
user614482user614482
112
$endgroup$
$begingroup$
Note that $frac n3=frac {20n}{60}$, for example
$endgroup$
– Mark Bennet
Nov 11 '18 at 8:30
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@MarkBennet I don't understand how it would get used though. Say I wanted to divide $8$ in half. How does multiplying by $30$ help me get $4$?
$endgroup$
– user614482
Nov 11 '18 at 14:43
$begingroup$
$30times 8=240 =40_{60}$ (ie translated to base $60$) and there is your $4$
$endgroup$
– Mark Bennet
Nov 11 '18 at 14:58
$begingroup$
@MarkBennet Oh, wow! But then why wouldn't they say it's $40$ as opposed to $4$? They had symbols for both $4$ and $40$ separately. How would they do the multiplication itself? The symbol for $8$ times the symbol for $30$ (both had their own symbols in base $60$) somehow giving them just.... $4$
$endgroup$
– user614482
Nov 11 '18 at 15:42
$begingroup$
Well, when I multiply by $25$ by dividing by $4$ I do need to make sure I get the decimal point in the right place.
$endgroup$
– Mark Bennet
Nov 11 '18 at 16:33
|
show 2 more comments
$begingroup$
According to the reciprocal tables for Babylonian's base $60$ system, dividing by $2$ is like multiplying like $30$. Dividing by $3$ is like multiplying by $20$. Dividing by $4$ is like multiplying by $15$. Dividing by $k$ is like multiplying by $60/k$.
I don't understand why the reciprocal works like this and not like $1/k$. You might say "It multiplies by $60/k$ because it's base $60$" but this doesn't make sense to me, it's not like our base-$10$ reciprocals look like $10/k$.
What's going on?
math-history number-systems
edited Nov 11 '18 at 7:57
amWhy
1
asked Nov 11 '18 at 7:52
user614482user614482
112
$endgroup$
According to the reciprocal tables for Babylonian's base $60$ system, dividing by $2$ is like multiplying like $30$. Dividing by $3$ is like multiplying by $20$. Dividing by $4$ is like multiplying by $15$. Dividing by $k$ is like multiplying by $60/k$.
I don't understand why the reciprocal works like this and not like $1/k$. You might say "It multiplies by $60/k$ because it's base $60$" but this doesn't make sense to me, it's not like our base-$10$ reciprocals look like $10/k$.
What's going on?
math-history number-systems
math-history number-systems
edited Nov 11 '18 at 7:57
amWhy
1
asked Nov 11 '18 at 7:52
user614482user614482
112
edited Nov 11 '18 at 7:57
amWhy
1
asked Nov 11 '18 at 7:52
user614482user614482
112
edited Nov 11 '18 at 7:57
amWhy
1
edited Nov 11 '18 at 7:57
amWhy
1
edited Nov 11 '18 at 7:57
amWhy
1
1
asked Nov 11 '18 at 7:52
user614482user614482
112
asked Nov 11 '18 at 7:52
user614482user614482
112
asked Nov 11 '18 at 7:52
user614482user614482
112
112
$begingroup$
Note that $frac n3=frac {20n}{60}$, for example
$endgroup$
– Mark Bennet
Nov 11 '18 at 8:30
$begingroup$
@MarkBennet I don't understand how it would get used though. Say I wanted to divide $8$ in half. How does multiplying by $30$ help me get $4$?
$endgroup$
– user614482
Nov 11 '18 at 14:43
$begingroup$
$30times 8=240 =40_{60}$ (ie translated to base $60$) and there is your $4$
$endgroup$
– Mark Bennet
Nov 11 '18 at 14:58
$begingroup$
@MarkBennet Oh, wow! But then why wouldn't they say it's $40$ as opposed to $4$? They had symbols for both $4$ and $40$ separately. How would they do the multiplication itself? The symbol for $8$ times the symbol for $30$ (both had their own symbols in base $60$) somehow giving them just.... $4$
$endgroup$
– user614482
Nov 11 '18 at 15:42
$begingroup$
Well, when I multiply by $25$ by dividing by $4$ I do need to make sure I get the decimal point in the right place.
$endgroup$
– Mark Bennet
Nov 11 '18 at 16:33
|
show 2 more comments
$begingroup$
Note that $frac n3=frac {20n}{60}$, for example
$endgroup$
– Mark Bennet
Nov 11 '18 at 8:30
$begingroup$
@MarkBennet I don't understand how it would get used though. Say I wanted to divide $8$ in half. How does multiplying by $30$ help me get $4$?
$endgroup$
– user614482
Nov 11 '18 at 14:43
$begingroup$
$30times 8=240 =40_{60}$ (ie translated to base $60$) and there is your $4$
$endgroup$
– Mark Bennet
Nov 11 '18 at 14:58
$begingroup$
@MarkBennet Oh, wow! But then why wouldn't they say it's $40$ as opposed to $4$? They had symbols for both $4$ and $40$ separately. How would they do the multiplication itself? The symbol for $8$ times the symbol for $30$ (both had their own symbols in base $60$) somehow giving them just.... $4$
$endgroup$
– user614482
Nov 11 '18 at 15:42
$begingroup$
Well, when I multiply by $25$ by dividing by $4$ I do need to make sure I get the decimal point in the right place.
$endgroup$
– Mark Bennet
Nov 11 '18 at 16:33
$begingroup$
Note that $frac n3=frac {20n}{60}$, for example
$endgroup$
– Mark Bennet
Nov 11 '18 at 8:30
$begingroup$
Note that $frac n3=frac {20n}{60}$, for example
$endgroup$
– Mark Bennet
Nov 11 '18 at 8:30
$begingroup$
@MarkBennet I don't understand how it would get used though. Say I wanted to divide $8$ in half. How does multiplying by $30$ help me get $4$?
$endgroup$
– user614482
Nov 11 '18 at 14:43
$begingroup$
@MarkBennet I don't understand how it would get used though. Say I wanted to divide $8$ in half. How does multiplying by $30$ help me get $4$?
$endgroup$
– user614482
Nov 11 '18 at 14:43
$begingroup$
$30times 8=240 =40_{60}$ (ie translated to base $60$) and there is your $4$
$endgroup$
– Mark Bennet
Nov 11 '18 at 14:58
$begingroup$
$30times 8=240 =40_{60}$ (ie translated to base $60$) and there is your $4$
$endgroup$
– Mark Bennet
Nov 11 '18 at 14:58
$begingroup$
@MarkBennet Oh, wow! But then why wouldn't they say it's $40$ as opposed to $4$? They had symbols for both $4$ and $40$ separately. How would they do the multiplication itself? The symbol for $8$ times the symbol for $30$ (both had their own symbols in base $60$) somehow giving them just.... $4$
$endgroup$
– user614482
Nov 11 '18 at 15:42
$begingroup$
@MarkBennet Oh, wow! But then why wouldn't they say it's $40$ as opposed to $4$? They had symbols for both $4$ and $40$ separately. How would they do the multiplication itself? The symbol for $8$ times the symbol for $30$ (both had their own symbols in base $60$) somehow giving them just.... $4$
$endgroup$
– user614482
Nov 11 '18 at 15:42
$begingroup$
Well, when I multiply by $25$ by dividing by $4$ I do need to make sure I get the decimal point in the right place.
$endgroup$
– Mark Bennet
Nov 11 '18 at 16:33
$begingroup$
Well, when I multiply by $25$ by dividing by $4$ I do need to make sure I get the decimal point in the right place.
$endgroup$
– Mark Bennet
Nov 11 '18 at 16:33
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Our base-$10$ reciprocals do look like $10/k$! Dividing by $2$ is like multiplying by $5$ (and then shifting the decimal point) and vice versa. Indeed, I regularly divide by $5$ by doubling the number and shifting the decimal point.
$10$ has fewer divisors than $60$, so this "trick" (if you like) doesn't have as many applications—basically only this one.
answered Nov 11 '18 at 8:09
Greg MartinGreg Martin
34.9k23161
$endgroup$
$begingroup$
For example I don't see how dividing by $2$ is like multiplying by $30$. For example if I wanted to divide $8$ in half to get $4$ it's somehow the same as taking $8$ times $30$?
$endgroup$
– user614482
Nov 11 '18 at 14:33
$begingroup$
8 times 30 is 240, that is 4*60 + 0
$endgroup$
– mau
Nov 18 '18 at 19:15
add a comment |
$begingroup$
Babylonians - at least in the tablets we have found - do not have the concept of order of magnitude: probably they calculated it in other ways. This means that - as far as the tablets are concerned - 1, 60, 3600=60*60, 1/60, 1/3600 are all represented as 1.
Therefore, if you have a number $n$ and you want to divide it by 4 (say), if the number is greater than 4 there is no problem; 9 / 4 = 2;15 (the semicolon has the same use as the comma in our system). For 3/4, they would have 0;45; but since they could not accept a 0 by itself, they multiplied the result for 60 obtaining 45, which is the same as 3*15. In this way, "divide by 4" is the same as "multiply by 15".
answered Nov 18 '18 at 20:32
maumau
7,00523263
$endgroup$
$begingroup$
A note that the semicolon notation is not Babylonian, but, I think, from the 20th century by Neugebauer (e.g., see www-groups.dcs.st-and.ac.uk/history/HistTopics/…).
$endgroup$
– Rolazaro Azeveires
Nov 19 '18 at 2:11
1
$begingroup$
yes, this is only a practical way for us to separate numbers.
$endgroup$
– mau
Nov 20 '18 at 8:03
add a comment |
$begingroup$
In our number system, $dfrac 15 = dfrac{2}{10}$. So,
to divide $678$ by $5$,
first divide by $10$, getting $67.8$,
and then multiply by $2$, getting $135.6$.
In a base $60$ number system, $dfrac 15 = dfrac{12}{60}$. So
to divide $[6,7,8]_{60}$ by $5$,
first divide by $[1,0]_{60}$, getting $[6,7 . 8]_{60}$
then multiply by $12_{60}$, getting $[72, 84. 96]_{60} = [72, 85. 36]_{60}
=[73, 25. 36]_{60}==[1,13, 25. 36]_{60}$
answered Jan 1 at 3:58
steven gregorysteven gregory
17.9k32257
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Our base-$10$ reciprocals do look like $10/k$! Dividing by $2$ is like multiplying by $5$ (and then shifting the decimal point) and vice versa. Indeed, I regularly divide by $5$ by doubling the number and shifting the decimal point.
$10$ has fewer divisors than $60$, so this "trick" (if you like) doesn't have as many applications—basically only this one.
answered Nov 11 '18 at 8:09
Greg MartinGreg Martin
34.9k23161
$endgroup$
$begingroup$
For example I don't see how dividing by $2$ is like multiplying by $30$. For example if I wanted to divide $8$ in half to get $4$ it's somehow the same as taking $8$ times $30$?
$endgroup$
– user614482
Nov 11 '18 at 14:33
$begingroup$
8 times 30 is 240, that is 4*60 + 0
$endgroup$
– mau
Nov 18 '18 at 19:15
add a comment |
$begingroup$
Our base-$10$ reciprocals do look like $10/k$! Dividing by $2$ is like multiplying by $5$ (and then shifting the decimal point) and vice versa. Indeed, I regularly divide by $5$ by doubling the number and shifting the decimal point.
$10$ has fewer divisors than $60$, so this "trick" (if you like) doesn't have as many applications—basically only this one.
answered Nov 11 '18 at 8:09
Greg MartinGreg Martin
34.9k23161
$endgroup$
$begingroup$
For example I don't see how dividing by $2$ is like multiplying by $30$. For example if I wanted to divide $8$ in half to get $4$ it's somehow the same as taking $8$ times $30$?
$endgroup$
– user614482
Nov 11 '18 at 14:33
$begingroup$
8 times 30 is 240, that is 4*60 + 0
$endgroup$
– mau
Nov 18 '18 at 19:15
add a comment |
$begingroup$
Our base-$10$ reciprocals do look like $10/k$! Dividing by $2$ is like multiplying by $5$ (and then shifting the decimal point) and vice versa. Indeed, I regularly divide by $5$ by doubling the number and shifting the decimal point.
$10$ has fewer divisors than $60$, so this "trick" (if you like) doesn't have as many applications—basically only this one.
answered Nov 11 '18 at 8:09
Greg MartinGreg Martin
34.9k23161
$endgroup$
Our base-$10$ reciprocals do look like $10/k$! Dividing by $2$ is like multiplying by $5$ (and then shifting the decimal point) and vice versa. Indeed, I regularly divide by $5$ by doubling the number and shifting the decimal point.
$10$ has fewer divisors than $60$, so this "trick" (if you like) doesn't have as many applications—basically only this one.
answered Nov 11 '18 at 8:09
Greg MartinGreg Martin
34.9k23161
answered Nov 11 '18 at 8:09
Greg MartinGreg Martin
34.9k23161
answered Nov 11 '18 at 8:09
Greg MartinGreg Martin
34.9k23161
answered Nov 11 '18 at 8:09
Greg MartinGreg Martin
34.9k23161
34.9k23161
$begingroup$
For example I don't see how dividing by $2$ is like multiplying by $30$. For example if I wanted to divide $8$ in half to get $4$ it's somehow the same as taking $8$ times $30$?
$endgroup$
– user614482
Nov 11 '18 at 14:33
$begingroup$
8 times 30 is 240, that is 4*60 + 0
$endgroup$
– mau
Nov 18 '18 at 19:15
add a comment |
$begingroup$
For example I don't see how dividing by $2$ is like multiplying by $30$. For example if I wanted to divide $8$ in half to get $4$ it's somehow the same as taking $8$ times $30$?
$endgroup$
– user614482
Nov 11 '18 at 14:33
$begingroup$
8 times 30 is 240, that is 4*60 + 0
$endgroup$
– mau
Nov 18 '18 at 19:15
$begingroup$
For example I don't see how dividing by $2$ is like multiplying by $30$. For example if I wanted to divide $8$ in half to get $4$ it's somehow the same as taking $8$ times $30$?
$endgroup$
– user614482
Nov 11 '18 at 14:33
$begingroup$
For example I don't see how dividing by $2$ is like multiplying by $30$. For example if I wanted to divide $8$ in half to get $4$ it's somehow the same as taking $8$ times $30$?
$endgroup$
– user614482
Nov 11 '18 at 14:33
$begingroup$
8 times 30 is 240, that is 4*60 + 0
$endgroup$
– mau
Nov 18 '18 at 19:15
$begingroup$
8 times 30 is 240, that is 4*60 + 0
$endgroup$
– mau
Nov 18 '18 at 19:15
add a comment |
$begingroup$
Babylonians - at least in the tablets we have found - do not have the concept of order of magnitude: probably they calculated it in other ways. This means that - as far as the tablets are concerned - 1, 60, 3600=60*60, 1/60, 1/3600 are all represented as 1.
Therefore, if you have a number $n$ and you want to divide it by 4 (say), if the number is greater than 4 there is no problem; 9 / 4 = 2;15 (the semicolon has the same use as the comma in our system). For 3/4, they would have 0;45; but since they could not accept a 0 by itself, they multiplied the result for 60 obtaining 45, which is the same as 3*15. In this way, "divide by 4" is the same as "multiply by 15".
answered Nov 18 '18 at 20:32
maumau
7,00523263
$endgroup$
$begingroup$
A note that the semicolon notation is not Babylonian, but, I think, from the 20th century by Neugebauer (e.g., see www-groups.dcs.st-and.ac.uk/history/HistTopics/…).
$endgroup$
– Rolazaro Azeveires
Nov 19 '18 at 2:11
1
$begingroup$
yes, this is only a practical way for us to separate numbers.
$endgroup$
– mau
Nov 20 '18 at 8:03
add a comment |
$begingroup$
Babylonians - at least in the tablets we have found - do not have the concept of order of magnitude: probably they calculated it in other ways. This means that - as far as the tablets are concerned - 1, 60, 3600=60*60, 1/60, 1/3600 are all represented as 1.
Therefore, if you have a number $n$ and you want to divide it by 4 (say), if the number is greater than 4 there is no problem; 9 / 4 = 2;15 (the semicolon has the same use as the comma in our system). For 3/4, they would have 0;45; but since they could not accept a 0 by itself, they multiplied the result for 60 obtaining 45, which is the same as 3*15. In this way, "divide by 4" is the same as "multiply by 15".
answered Nov 18 '18 at 20:32
maumau
7,00523263
$endgroup$
$begingroup$
A note that the semicolon notation is not Babylonian, but, I think, from the 20th century by Neugebauer (e.g., see www-groups.dcs.st-and.ac.uk/history/HistTopics/…).
$endgroup$
– Rolazaro Azeveires
Nov 19 '18 at 2:11
1
$begingroup$
yes, this is only a practical way for us to separate numbers.
$endgroup$
– mau
Nov 20 '18 at 8:03
add a comment |
$begingroup$
Babylonians - at least in the tablets we have found - do not have the concept of order of magnitude: probably they calculated it in other ways. This means that - as far as the tablets are concerned - 1, 60, 3600=60*60, 1/60, 1/3600 are all represented as 1.
Therefore, if you have a number $n$ and you want to divide it by 4 (say), if the number is greater than 4 there is no problem; 9 / 4 = 2;15 (the semicolon has the same use as the comma in our system). For 3/4, they would have 0;45; but since they could not accept a 0 by itself, they multiplied the result for 60 obtaining 45, which is the same as 3*15. In this way, "divide by 4" is the same as "multiply by 15".
answered Nov 18 '18 at 20:32
maumau
7,00523263
$endgroup$
Babylonians - at least in the tablets we have found - do not have the concept of order of magnitude: probably they calculated it in other ways. This means that - as far as the tablets are concerned - 1, 60, 3600=60*60, 1/60, 1/3600 are all represented as 1.
Therefore, if you have a number $n$ and you want to divide it by 4 (say), if the number is greater than 4 there is no problem; 9 / 4 = 2;15 (the semicolon has the same use as the comma in our system). For 3/4, they would have 0;45; but since they could not accept a 0 by itself, they multiplied the result for 60 obtaining 45, which is the same as 3*15. In this way, "divide by 4" is the same as "multiply by 15".
answered Nov 18 '18 at 20:32
maumau
7,00523263
answered Nov 18 '18 at 20:32
maumau
7,00523263
answered Nov 18 '18 at 20:32
maumau
7,00523263
answered Nov 18 '18 at 20:32
maumau
7,00523263
7,00523263
$begingroup$
A note that the semicolon notation is not Babylonian, but, I think, from the 20th century by Neugebauer (e.g., see www-groups.dcs.st-and.ac.uk/history/HistTopics/…).
$endgroup$
– Rolazaro Azeveires
Nov 19 '18 at 2:11
1
$begingroup$
yes, this is only a practical way for us to separate numbers.
$endgroup$
– mau
Nov 20 '18 at 8:03
add a comment |
$begingroup$
A note that the semicolon notation is not Babylonian, but, I think, from the 20th century by Neugebauer (e.g., see www-groups.dcs.st-and.ac.uk/history/HistTopics/…).
$endgroup$
– Rolazaro Azeveires
Nov 19 '18 at 2:11
1
$begingroup$
yes, this is only a practical way for us to separate numbers.
$endgroup$
– mau
Nov 20 '18 at 8:03
$begingroup$
A note that the semicolon notation is not Babylonian, but, I think, from the 20th century by Neugebauer (e.g., see www-groups.dcs.st-and.ac.uk/history/HistTopics/…).
$endgroup$
– Rolazaro Azeveires
Nov 19 '18 at 2:11
$begingroup$
A note that the semicolon notation is not Babylonian, but, I think, from the 20th century by Neugebauer (e.g., see www-groups.dcs.st-and.ac.uk/history/HistTopics/…).
$endgroup$
– Rolazaro Azeveires
Nov 19 '18 at 2:11
1
1
$begingroup$
yes, this is only a practical way for us to separate numbers.
$endgroup$
– mau
Nov 20 '18 at 8:03
$begingroup$
yes, this is only a practical way for us to separate numbers.
$endgroup$
– mau
Nov 20 '18 at 8:03
add a comment |
$begingroup$
In our number system, $dfrac 15 = dfrac{2}{10}$. So,
to divide $678$ by $5$,
first divide by $10$, getting $67.8$,
and then multiply by $2$, getting $135.6$.
In a base $60$ number system, $dfrac 15 = dfrac{12}{60}$. So
to divide $[6,7,8]_{60}$ by $5$,
first divide by $[1,0]_{60}$, getting $[6,7 . 8]_{60}$
then multiply by $12_{60}$, getting $[72, 84. 96]_{60} = [72, 85. 36]_{60}
=[73, 25. 36]_{60}==[1,13, 25. 36]_{60}$
answered Jan 1 at 3:58
steven gregorysteven gregory
17.9k32257
$endgroup$
add a comment |
$begingroup$
In our number system, $dfrac 15 = dfrac{2}{10}$. So,
to divide $678$ by $5$,
first divide by $10$, getting $67.8$,
and then multiply by $2$, getting $135.6$.
In a base $60$ number system, $dfrac 15 = dfrac{12}{60}$. So
to divide $[6,7,8]_{60}$ by $5$,
first divide by $[1,0]_{60}$, getting $[6,7 . 8]_{60}$
then multiply by $12_{60}$, getting $[72, 84. 96]_{60} = [72, 85. 36]_{60}
=[73, 25. 36]_{60}==[1,13, 25. 36]_{60}$
answered Jan 1 at 3:58
steven gregorysteven gregory
17.9k32257
$endgroup$
add a comment |
$begingroup$
In our number system, $dfrac 15 = dfrac{2}{10}$. So,
to divide $678$ by $5$,
first divide by $10$, getting $67.8$,
and then multiply by $2$, getting $135.6$.
In a base $60$ number system, $dfrac 15 = dfrac{12}{60}$. So
to divide $[6,7,8]_{60}$ by $5$,
first divide by $[1,0]_{60}$, getting $[6,7 . 8]_{60}$
then multiply by $12_{60}$, getting $[72, 84. 96]_{60} = [72, 85. 36]_{60}
=[73, 25. 36]_{60}==[1,13, 25. 36]_{60}$
answered Jan 1 at 3:58
steven gregorysteven gregory
17.9k32257
$endgroup$
In our number system, $dfrac 15 = dfrac{2}{10}$. So,
to divide $678$ by $5$,
first divide by $10$, getting $67.8$,
and then multiply by $2$, getting $135.6$.
In a base $60$ number system, $dfrac 15 = dfrac{12}{60}$. So
to divide $[6,7,8]_{60}$ by $5$,
first divide by $[1,0]_{60}$, getting $[6,7 . 8]_{60}$
then multiply by $12_{60}$, getting $[72, 84. 96]_{60} = [72, 85. 36]_{60}
=[73, 25. 36]_{60}==[1,13, 25. 36]_{60}$
answered Jan 1 at 3:58
steven gregorysteven gregory
17.9k32257
answered Jan 1 at 3:58
steven gregorysteven gregory
17.9k32257
answered Jan 1 at 3:58
steven gregorysteven gregory
17.9k32257
answered Jan 1 at 3:58
steven gregorysteven gregory
17.9k32257
17.9k32257
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$begingroup$
Note that $frac n3=frac {20n}{60}$, for example
$endgroup$
– Mark Bennet
Nov 11 '18 at 8:30
$begingroup$
@MarkBennet I don't understand how it would get used though. Say I wanted to divide $8$ in half. How does multiplying by $30$ help me get $4$?
$endgroup$
– user614482
Nov 11 '18 at 14:43
$begingroup$
$30times 8=240 =40_{60}$ (ie translated to base $60$) and there is your $4$
$endgroup$
– Mark Bennet
Nov 11 '18 at 14:58
$begingroup$
@MarkBennet Oh, wow! But then why wouldn't they say it's $40$ as opposed to $4$? They had symbols for both $4$ and $40$ separately. How would they do the multiplication itself? The symbol for $8$ times the symbol for $30$ (both had their own symbols in base $60$) somehow giving them just.... $4$
$endgroup$
– user614482
Nov 11 '18 at 15:42
$begingroup$
Well, when I multiply by $25$ by dividing by $4$ I do need to make sure I get the decimal point in the right place.
$endgroup$
– Mark Bennet
Nov 11 '18 at 16:33