Did Babylonians know the Pythagorean theorem before his time?












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On old tablets the Babylonians were able to work out the digits to the square root of two from the hypotenuse of a $45^circ-45^circ-90^circ$ triangle.



How could they have figured this out without the use of Pythagorean theorem unless they came up with it first?






geometry math-history irrational-numbers pythagorean-triples







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    This is a good question for the History of Science and Mathematics Stack Exchange.
    $endgroup$
    – Blue
    Nov 11 '18 at 17:08










  • $begingroup$
    Wikipedia says: “According to Joran Friberg, a historian of mathematics, evidence indicates that the Pythagorean theorem was well-known to the mathematicians of the First Babylonian Dynasty (20th to 16th centuries BC), which would have been over a thousand years before Pythagoras was born,”
    $endgroup$
    – MJD
    Nov 11 '18 at 18:05












  • $begingroup$
    I saw that @MJD but didn't see direct evidence to support it
    $endgroup$
    – user614647
    Nov 11 '18 at 18:17






  • 1




    $begingroup$
    Well, did you follow up the cited reference? Presumably the direct evidence will be explained in the paper.
    $endgroup$
    – MJD
    Nov 11 '18 at 19:31










  • $begingroup$
    @MJD Is Plimpton really a proof that they knew the theorem though? You can verify a lot of integer valued triplets by brute force experimentation of constructing triangles
    $endgroup$
    – user614647
    Nov 12 '18 at 3:29
















1












$begingroup$


On old tablets the Babylonians were able to work out the digits to the square root of two from the hypotenuse of a $45^circ-45^circ-90^circ$ triangle.



How could they have figured this out without the use of Pythagorean theorem unless they came up with it first?






geometry math-history irrational-numbers pythagorean-triples







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    This is a good question for the History of Science and Mathematics Stack Exchange.
    $endgroup$
    – Blue
    Nov 11 '18 at 17:08










  • $begingroup$
    Wikipedia says: “According to Joran Friberg, a historian of mathematics, evidence indicates that the Pythagorean theorem was well-known to the mathematicians of the First Babylonian Dynasty (20th to 16th centuries BC), which would have been over a thousand years before Pythagoras was born,”
    $endgroup$
    – MJD
    Nov 11 '18 at 18:05












  • $begingroup$
    I saw that @MJD but didn't see direct evidence to support it
    $endgroup$
    – user614647
    Nov 11 '18 at 18:17






  • 1




    $begingroup$
    Well, did you follow up the cited reference? Presumably the direct evidence will be explained in the paper.
    $endgroup$
    – MJD
    Nov 11 '18 at 19:31










  • $begingroup$
    @MJD Is Plimpton really a proof that they knew the theorem though? You can verify a lot of integer valued triplets by brute force experimentation of constructing triangles
    $endgroup$
    – user614647
    Nov 12 '18 at 3:29














1












1








1





$begingroup$


On old tablets the Babylonians were able to work out the digits to the square root of two from the hypotenuse of a $45^circ-45^circ-90^circ$ triangle.



How could they have figured this out without the use of Pythagorean theorem unless they came up with it first?






geometry math-history irrational-numbers pythagorean-triples







share|cite|improve this question











$endgroup$




On old tablets the Babylonians were able to work out the digits to the square root of two from the hypotenuse of a $45^circ-45^circ-90^circ$ triangle.



How could they have figured this out without the use of Pythagorean theorem unless they came up with it first?






geometry math-history irrational-numbers pythagorean-triples




geometry math-history irrational-numbers pythagorean-triples






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 17:04









Tianlalu

3,08621038




3,08621038










asked Nov 11 '18 at 16:58









user614647user614647

61




61








  • 3




    $begingroup$
    This is a good question for the History of Science and Mathematics Stack Exchange.
    $endgroup$
    – Blue
    Nov 11 '18 at 17:08










  • $begingroup$
    Wikipedia says: “According to Joran Friberg, a historian of mathematics, evidence indicates that the Pythagorean theorem was well-known to the mathematicians of the First Babylonian Dynasty (20th to 16th centuries BC), which would have been over a thousand years before Pythagoras was born,”
    $endgroup$
    – MJD
    Nov 11 '18 at 18:05












  • $begingroup$
    I saw that @MJD but didn't see direct evidence to support it
    $endgroup$
    – user614647
    Nov 11 '18 at 18:17






  • 1




    $begingroup$
    Well, did you follow up the cited reference? Presumably the direct evidence will be explained in the paper.
    $endgroup$
    – MJD
    Nov 11 '18 at 19:31










  • $begingroup$
    @MJD Is Plimpton really a proof that they knew the theorem though? You can verify a lot of integer valued triplets by brute force experimentation of constructing triangles
    $endgroup$
    – user614647
    Nov 12 '18 at 3:29














  • 3




    $begingroup$
    This is a good question for the History of Science and Mathematics Stack Exchange.
    $endgroup$
    – Blue
    Nov 11 '18 at 17:08










  • $begingroup$
    Wikipedia says: “According to Joran Friberg, a historian of mathematics, evidence indicates that the Pythagorean theorem was well-known to the mathematicians of the First Babylonian Dynasty (20th to 16th centuries BC), which would have been over a thousand years before Pythagoras was born,”
    $endgroup$
    – MJD
    Nov 11 '18 at 18:05












  • $begingroup$
    I saw that @MJD but didn't see direct evidence to support it
    $endgroup$
    – user614647
    Nov 11 '18 at 18:17






  • 1




    $begingroup$
    Well, did you follow up the cited reference? Presumably the direct evidence will be explained in the paper.
    $endgroup$
    – MJD
    Nov 11 '18 at 19:31










  • $begingroup$
    @MJD Is Plimpton really a proof that they knew the theorem though? You can verify a lot of integer valued triplets by brute force experimentation of constructing triangles
    $endgroup$
    – user614647
    Nov 12 '18 at 3:29








3




3




$begingroup$
This is a good question for the History of Science and Mathematics Stack Exchange.
$endgroup$
– Blue
Nov 11 '18 at 17:08




$begingroup$
This is a good question for the History of Science and Mathematics Stack Exchange.
$endgroup$
– Blue
Nov 11 '18 at 17:08












$begingroup$
Wikipedia says: “According to Joran Friberg, a historian of mathematics, evidence indicates that the Pythagorean theorem was well-known to the mathematicians of the First Babylonian Dynasty (20th to 16th centuries BC), which would have been over a thousand years before Pythagoras was born,”
$endgroup$
– MJD
Nov 11 '18 at 18:05






$begingroup$
Wikipedia says: “According to Joran Friberg, a historian of mathematics, evidence indicates that the Pythagorean theorem was well-known to the mathematicians of the First Babylonian Dynasty (20th to 16th centuries BC), which would have been over a thousand years before Pythagoras was born,”
$endgroup$
– MJD
Nov 11 '18 at 18:05














$begingroup$
I saw that @MJD but didn't see direct evidence to support it
$endgroup$
– user614647
Nov 11 '18 at 18:17




$begingroup$
I saw that @MJD but didn't see direct evidence to support it
$endgroup$
– user614647
Nov 11 '18 at 18:17




1




1




$begingroup$
Well, did you follow up the cited reference? Presumably the direct evidence will be explained in the paper.
$endgroup$
– MJD
Nov 11 '18 at 19:31




$begingroup$
Well, did you follow up the cited reference? Presumably the direct evidence will be explained in the paper.
$endgroup$
– MJD
Nov 11 '18 at 19:31












$begingroup$
@MJD Is Plimpton really a proof that they knew the theorem though? You can verify a lot of integer valued triplets by brute force experimentation of constructing triangles
$endgroup$
– user614647
Nov 12 '18 at 3:29




$begingroup$
@MJD Is Plimpton really a proof that they knew the theorem though? You can verify a lot of integer valued triplets by brute force experimentation of constructing triangles
$endgroup$
– user614647
Nov 12 '18 at 3:29










2 Answers
2






active

oldest

votes


















2












$begingroup$

There is a cuneiform tablet in the British Museum, BM 85196, which is dated to the Old Babylonian period (2000 BCE - 1600 BCE). Problem 9 on that tablet is understood as being about a pole, initially vertical, sliding out from a wall so that the pole forms the hypotenuse of a right triangle and the wall and the ground the legs. Note that the wall is not explicitly mentioned in the problem, although it is possible that the damaged part contained such a mention. The following description, with translation due to Jens Høyrup, can be found in the article



Duncan J. Melville, Poles and walls in Mesopotamia and Egypt, Historia Mathematica, Volume 31, Issue 2, 2004,



which discusses a number of similar problems found across a variety of cultures and time periods.




Small portions of Problem 9 are missing or damaged, including, unfortunately, a part of the first line, but the structure of the problem can be safely restored. The problem reads as follows:



A pole. 30 (a reed). From its …



Above, it has descended 6, below, how far has it moved away?



You. Square 30: 15 you see. Subtract 6 from 30: 24 you see.



Square 24: 9,36 you see. Subtract 9,36 from 15:



5,24 you see. What is the square root of 5,24? 18 is the square root. 18



on the ground it has moved away. If 18 on the ground



above, what did it descend? Square 18: 5,24 you see.



Subtract 5,24 from 15: 9,36 you see. Of 9,36,



what is the square root? 24 is the square root. Subtract 24 from 30:



6 you see descended. Such is the procedure.




In this passage, base-60 numbers are given with digits separated by commas. As in the original there is no explicit sexagesimal point, and the magnitude of the numbers must be inferred from context. So $30$ means $30/60=1/2$, $6$ means $6/60=1/10$, $15$ means $15/60=1/4$, $9,36$ means $9/60+36/60^2=576/3600=4/25$, and so on. Following the procedure, the length along the ground is computed, in modern terms, as
$$
sqrt{text{(length of pole)}^2-text{(height on wall)}^2}=sqrt{(1/2)^2-(1/2-1/10)^2}=3/10
$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Computing the hypotenuse of the unit right triangle (same as the diagonal of the unit square) is significantly easier than the general Pythagorean Theorem, see, e.g., this



    For that matter, the so-called Babylonian method for extracting square roots doesn't rely on the Pythagorean Theorem.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How would they have figured this out naturally?
      $endgroup$
      – user614647
      Nov 11 '18 at 17:27










    • $begingroup$
      Figured what? The hypotenuse? That's really clear in this case. The square root method (Newton's method narrowed down to square roots) is, to my mind, much more impressive.
      $endgroup$
      – lulu
      Nov 11 '18 at 17:30










    • $begingroup$
      Yes the square root method sorry
      $endgroup$
      – user614647
      Nov 11 '18 at 17:51











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    2 Answers
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    2 Answers
    2






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    active

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    active

    oldest

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    2












    $begingroup$

    There is a cuneiform tablet in the British Museum, BM 85196, which is dated to the Old Babylonian period (2000 BCE - 1600 BCE). Problem 9 on that tablet is understood as being about a pole, initially vertical, sliding out from a wall so that the pole forms the hypotenuse of a right triangle and the wall and the ground the legs. Note that the wall is not explicitly mentioned in the problem, although it is possible that the damaged part contained such a mention. The following description, with translation due to Jens Høyrup, can be found in the article



    Duncan J. Melville, Poles and walls in Mesopotamia and Egypt, Historia Mathematica, Volume 31, Issue 2, 2004,



    which discusses a number of similar problems found across a variety of cultures and time periods.




    Small portions of Problem 9 are missing or damaged, including, unfortunately, a part of the first line, but the structure of the problem can be safely restored. The problem reads as follows:



    A pole. 30 (a reed). From its …



    Above, it has descended 6, below, how far has it moved away?



    You. Square 30: 15 you see. Subtract 6 from 30: 24 you see.



    Square 24: 9,36 you see. Subtract 9,36 from 15:



    5,24 you see. What is the square root of 5,24? 18 is the square root. 18



    on the ground it has moved away. If 18 on the ground



    above, what did it descend? Square 18: 5,24 you see.



    Subtract 5,24 from 15: 9,36 you see. Of 9,36,



    what is the square root? 24 is the square root. Subtract 24 from 30:



    6 you see descended. Such is the procedure.




    In this passage, base-60 numbers are given with digits separated by commas. As in the original there is no explicit sexagesimal point, and the magnitude of the numbers must be inferred from context. So $30$ means $30/60=1/2$, $6$ means $6/60=1/10$, $15$ means $15/60=1/4$, $9,36$ means $9/60+36/60^2=576/3600=4/25$, and so on. Following the procedure, the length along the ground is computed, in modern terms, as
    $$
    sqrt{text{(length of pole)}^2-text{(height on wall)}^2}=sqrt{(1/2)^2-(1/2-1/10)^2}=3/10
    $$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      There is a cuneiform tablet in the British Museum, BM 85196, which is dated to the Old Babylonian period (2000 BCE - 1600 BCE). Problem 9 on that tablet is understood as being about a pole, initially vertical, sliding out from a wall so that the pole forms the hypotenuse of a right triangle and the wall and the ground the legs. Note that the wall is not explicitly mentioned in the problem, although it is possible that the damaged part contained such a mention. The following description, with translation due to Jens Høyrup, can be found in the article



      Duncan J. Melville, Poles and walls in Mesopotamia and Egypt, Historia Mathematica, Volume 31, Issue 2, 2004,



      which discusses a number of similar problems found across a variety of cultures and time periods.




      Small portions of Problem 9 are missing or damaged, including, unfortunately, a part of the first line, but the structure of the problem can be safely restored. The problem reads as follows:



      A pole. 30 (a reed). From its …



      Above, it has descended 6, below, how far has it moved away?



      You. Square 30: 15 you see. Subtract 6 from 30: 24 you see.



      Square 24: 9,36 you see. Subtract 9,36 from 15:



      5,24 you see. What is the square root of 5,24? 18 is the square root. 18



      on the ground it has moved away. If 18 on the ground



      above, what did it descend? Square 18: 5,24 you see.



      Subtract 5,24 from 15: 9,36 you see. Of 9,36,



      what is the square root? 24 is the square root. Subtract 24 from 30:



      6 you see descended. Such is the procedure.




      In this passage, base-60 numbers are given with digits separated by commas. As in the original there is no explicit sexagesimal point, and the magnitude of the numbers must be inferred from context. So $30$ means $30/60=1/2$, $6$ means $6/60=1/10$, $15$ means $15/60=1/4$, $9,36$ means $9/60+36/60^2=576/3600=4/25$, and so on. Following the procedure, the length along the ground is computed, in modern terms, as
      $$
      sqrt{text{(length of pole)}^2-text{(height on wall)}^2}=sqrt{(1/2)^2-(1/2-1/10)^2}=3/10
      $$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        There is a cuneiform tablet in the British Museum, BM 85196, which is dated to the Old Babylonian period (2000 BCE - 1600 BCE). Problem 9 on that tablet is understood as being about a pole, initially vertical, sliding out from a wall so that the pole forms the hypotenuse of a right triangle and the wall and the ground the legs. Note that the wall is not explicitly mentioned in the problem, although it is possible that the damaged part contained such a mention. The following description, with translation due to Jens Høyrup, can be found in the article



        Duncan J. Melville, Poles and walls in Mesopotamia and Egypt, Historia Mathematica, Volume 31, Issue 2, 2004,



        which discusses a number of similar problems found across a variety of cultures and time periods.




        Small portions of Problem 9 are missing or damaged, including, unfortunately, a part of the first line, but the structure of the problem can be safely restored. The problem reads as follows:



        A pole. 30 (a reed). From its …



        Above, it has descended 6, below, how far has it moved away?



        You. Square 30: 15 you see. Subtract 6 from 30: 24 you see.



        Square 24: 9,36 you see. Subtract 9,36 from 15:



        5,24 you see. What is the square root of 5,24? 18 is the square root. 18



        on the ground it has moved away. If 18 on the ground



        above, what did it descend? Square 18: 5,24 you see.



        Subtract 5,24 from 15: 9,36 you see. Of 9,36,



        what is the square root? 24 is the square root. Subtract 24 from 30:



        6 you see descended. Such is the procedure.




        In this passage, base-60 numbers are given with digits separated by commas. As in the original there is no explicit sexagesimal point, and the magnitude of the numbers must be inferred from context. So $30$ means $30/60=1/2$, $6$ means $6/60=1/10$, $15$ means $15/60=1/4$, $9,36$ means $9/60+36/60^2=576/3600=4/25$, and so on. Following the procedure, the length along the ground is computed, in modern terms, as
        $$
        sqrt{text{(length of pole)}^2-text{(height on wall)}^2}=sqrt{(1/2)^2-(1/2-1/10)^2}=3/10
        $$






        share|cite|improve this answer









        $endgroup$



        There is a cuneiform tablet in the British Museum, BM 85196, which is dated to the Old Babylonian period (2000 BCE - 1600 BCE). Problem 9 on that tablet is understood as being about a pole, initially vertical, sliding out from a wall so that the pole forms the hypotenuse of a right triangle and the wall and the ground the legs. Note that the wall is not explicitly mentioned in the problem, although it is possible that the damaged part contained such a mention. The following description, with translation due to Jens Høyrup, can be found in the article



        Duncan J. Melville, Poles and walls in Mesopotamia and Egypt, Historia Mathematica, Volume 31, Issue 2, 2004,



        which discusses a number of similar problems found across a variety of cultures and time periods.




        Small portions of Problem 9 are missing or damaged, including, unfortunately, a part of the first line, but the structure of the problem can be safely restored. The problem reads as follows:



        A pole. 30 (a reed). From its …



        Above, it has descended 6, below, how far has it moved away?



        You. Square 30: 15 you see. Subtract 6 from 30: 24 you see.



        Square 24: 9,36 you see. Subtract 9,36 from 15:



        5,24 you see. What is the square root of 5,24? 18 is the square root. 18



        on the ground it has moved away. If 18 on the ground



        above, what did it descend? Square 18: 5,24 you see.



        Subtract 5,24 from 15: 9,36 you see. Of 9,36,



        what is the square root? 24 is the square root. Subtract 24 from 30:



        6 you see descended. Such is the procedure.




        In this passage, base-60 numbers are given with digits separated by commas. As in the original there is no explicit sexagesimal point, and the magnitude of the numbers must be inferred from context. So $30$ means $30/60=1/2$, $6$ means $6/60=1/10$, $15$ means $15/60=1/4$, $9,36$ means $9/60+36/60^2=576/3600=4/25$, and so on. Following the procedure, the length along the ground is computed, in modern terms, as
        $$
        sqrt{text{(length of pole)}^2-text{(height on wall)}^2}=sqrt{(1/2)^2-(1/2-1/10)^2}=3/10
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 at 11:25









        Will OrrickWill Orrick

        13.6k13360




        13.6k13360























            0












            $begingroup$

            Computing the hypotenuse of the unit right triangle (same as the diagonal of the unit square) is significantly easier than the general Pythagorean Theorem, see, e.g., this



            For that matter, the so-called Babylonian method for extracting square roots doesn't rely on the Pythagorean Theorem.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How would they have figured this out naturally?
              $endgroup$
              – user614647
              Nov 11 '18 at 17:27










            • $begingroup$
              Figured what? The hypotenuse? That's really clear in this case. The square root method (Newton's method narrowed down to square roots) is, to my mind, much more impressive.
              $endgroup$
              – lulu
              Nov 11 '18 at 17:30










            • $begingroup$
              Yes the square root method sorry
              $endgroup$
              – user614647
              Nov 11 '18 at 17:51
















            0












            $begingroup$

            Computing the hypotenuse of the unit right triangle (same as the diagonal of the unit square) is significantly easier than the general Pythagorean Theorem, see, e.g., this



            For that matter, the so-called Babylonian method for extracting square roots doesn't rely on the Pythagorean Theorem.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How would they have figured this out naturally?
              $endgroup$
              – user614647
              Nov 11 '18 at 17:27










            • $begingroup$
              Figured what? The hypotenuse? That's really clear in this case. The square root method (Newton's method narrowed down to square roots) is, to my mind, much more impressive.
              $endgroup$
              – lulu
              Nov 11 '18 at 17:30










            • $begingroup$
              Yes the square root method sorry
              $endgroup$
              – user614647
              Nov 11 '18 at 17:51














            0












            0








            0





            $begingroup$

            Computing the hypotenuse of the unit right triangle (same as the diagonal of the unit square) is significantly easier than the general Pythagorean Theorem, see, e.g., this



            For that matter, the so-called Babylonian method for extracting square roots doesn't rely on the Pythagorean Theorem.






            share|cite|improve this answer









            $endgroup$



            Computing the hypotenuse of the unit right triangle (same as the diagonal of the unit square) is significantly easier than the general Pythagorean Theorem, see, e.g., this



            For that matter, the so-called Babylonian method for extracting square roots doesn't rely on the Pythagorean Theorem.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 11 '18 at 17:04









            lulululu

            40k24778




            40k24778












            • $begingroup$
              How would they have figured this out naturally?
              $endgroup$
              – user614647
              Nov 11 '18 at 17:27










            • $begingroup$
              Figured what? The hypotenuse? That's really clear in this case. The square root method (Newton's method narrowed down to square roots) is, to my mind, much more impressive.
              $endgroup$
              – lulu
              Nov 11 '18 at 17:30










            • $begingroup$
              Yes the square root method sorry
              $endgroup$
              – user614647
              Nov 11 '18 at 17:51


















            • $begingroup$
              How would they have figured this out naturally?
              $endgroup$
              – user614647
              Nov 11 '18 at 17:27










            • $begingroup$
              Figured what? The hypotenuse? That's really clear in this case. The square root method (Newton's method narrowed down to square roots) is, to my mind, much more impressive.
              $endgroup$
              – lulu
              Nov 11 '18 at 17:30










            • $begingroup$
              Yes the square root method sorry
              $endgroup$
              – user614647
              Nov 11 '18 at 17:51
















            $begingroup$
            How would they have figured this out naturally?
            $endgroup$
            – user614647
            Nov 11 '18 at 17:27




            $begingroup$
            How would they have figured this out naturally?
            $endgroup$
            – user614647
            Nov 11 '18 at 17:27












            $begingroup$
            Figured what? The hypotenuse? That's really clear in this case. The square root method (Newton's method narrowed down to square roots) is, to my mind, much more impressive.
            $endgroup$
            – lulu
            Nov 11 '18 at 17:30




            $begingroup$
            Figured what? The hypotenuse? That's really clear in this case. The square root method (Newton's method narrowed down to square roots) is, to my mind, much more impressive.
            $endgroup$
            – lulu
            Nov 11 '18 at 17:30












            $begingroup$
            Yes the square root method sorry
            $endgroup$
            – user614647
            Nov 11 '18 at 17:51




            $begingroup$
            Yes the square root method sorry
            $endgroup$
            – user614647
            Nov 11 '18 at 17:51


















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