Difficulty understanding how Babylonian reciprocals work












2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 11 '18 at 7:57









amWhy

1




1










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


















  • $begingroup$
    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










3 Answers
3






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















1












$begingroup$

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".






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















0












$begingroup$

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 $[72, 84. 96]_{60} = [72, 85. 36]_{60}
=[73, 25. 36]_{60}==[1,13, 25. 36]_{60}$






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
















    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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














    2












    2








    2





    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • $begingroup$
      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











    1












    $begingroup$

    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".






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
















    1












    $begingroup$

    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".






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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














    1












    1








    1





    $begingroup$

    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".






    share|cite|improve this answer









    $endgroup$



    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".







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • $begingroup$
      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











    0












    $begingroup$

    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 $[72, 84. 96]_{60} = [72, 85. 36]_{60}
    =[73, 25. 36]_{60}==[1,13, 25. 36]_{60}$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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 $[72, 84. 96]_{60} = [72, 85. 36]_{60}
      =[73, 25. 36]_{60}==[1,13, 25. 36]_{60}$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $[72, 84. 96]_{60} = [72, 85. 36]_{60}
        =[73, 25. 36]_{60}==[1,13, 25. 36]_{60}$






        share|cite|improve this answer









        $endgroup$



        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 $[72, 84. 96]_{60} = [72, 85. 36]_{60}
        =[73, 25. 36]_{60}==[1,13, 25. 36]_{60}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 3:58









        steven gregorysteven gregory

        17.9k32257




        17.9k32257






























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