BEGINNINGS IN GREECE 53 



tecture of the fifth century, with fine effect, and must have been 

 systematically employed. 



As to the celebrated theorem which bears the name of Pythag- 

 oras, he may well have learned from the Egyptian rope-stretchers 

 that a right angle is formed by taking sides of lengths 3 and 4 and 

 separating the other ends a distance 5, while his study of numbers 

 would easily have led to the discovery that in the series of squares 

 the adjacent 9 and 16 make 25. It would naturally be investi- 

 gated whether a similar relation could be verified for other right 

 triangles. In the most familiar case of the isosceles right tri- 

 angle it soon appears that the length of the equal sides being taken 

 as 1, the length of the hypotenuse could be only approximately 

 expressed. It cannot indeed be exactly expressed by any whole 

 number, or fraction; it is irrational. 



If it is true as Whewell says, that the essence of the triumphs of 

 science and its progress consists in that it enables us to consider evident 

 and necessary, views which our ancestors held to be unintelligible and 

 were unable to comprehend, then the extension of the number concept 

 to include the irrational, and we will at once add, the imaginary, 

 is the greatest forward step which pure mathematics has ever taken. 



Hankel 



In this case the proof of the Pythagorean theorem is easily 

 effected by a simple graphical construction, involving merely the 

 drawing of diagonals of squares. The smaller 

 triangles in the figure are evidently all equal. 

 The larger square contains four of them, the 

 smaller squares, two each. It seems possible 

 that this was the Pythagorean method, but as 

 to how the proof was accomplished in other cases 

 we have no information, the simpler proof of Euclid having com- 

 pletely superseded the earlier. On the other hand, for the cor- 

 responding arithmetical problem of finding three whole numbers 

 which can be the sides of a right triangle, Pythagoras is said to 

 have given a correct solution, equivalent in our notation to 



(2 a + I) 2 + (2a 2 + 2a) 2 = (2a 2 + 2a + I) 2 , 



