SELECTED CHAPTERS FROM ALGEBRA I. R. Shafarevich Preface

4 I. R. Shafarevicha) b)Fig. 1Then, the unit lessens, breaking up E into m equal segments E 0 . If E 0 is containedin A exactly n times, then we say that the length of A is equal to n (relative tomthe unit E). Thousands of years, men, in dierent parts of the world, had appliedthis procedure in a variety of situations until the question: is this breaking reallypossible?, arose. This completely new setting of question already belongs to thehistorical epoch|Pythagoras' School, in the period of VI or V century B.C. Thesegments A and E are called commensurable if there exists a segment E 0 exactlym times contained in E and n times in A. Thus the above question modies tothe following: are each two segments commensurable? Or further: is the length ofeach segment (a unit being xed) equal to a rational numbern ? The answer ismnegative and an example of a pair of incommensurable segments is simple. Considera square, its side being E and its diagonal A.THEOREM 1. The side of the square isincommensurable with its diagonal.Before we proceed with the proof, we give this theorem another form. Accordingto the famous Pythagorean theorem, the area of the square over the hypotenuseof a right triangle is equal to the sum of areas of the squares over the other twosides. Or, in other words, the square of the length of the hypotenuse is equal tothe sum of squares of the lengths of the other two sides. However, the diagonalA of our square is the hypotenuse of the isosceles triangle whose other two sidescoincide with the sides E of the square (Fig. 2) and hence in our case A 2 =2E 2 , n 2 nand if A = nE 0 , E = mE 0 , then =2orm m = p 2. Therefore, Theorem 1 canbe reformulated asTHEOREM 2. p 2 is not a rational number.Fig. 2 Fig. 3We shall give a proof of this form of the theorem, but rst we make thefollowing remark. Although we have leaned on the Pythagorean theorem, we have