Algebra in Ancient and Modern Times by V. S. Varadarajan

By V. S. Varadarajan

This article deals a distinct account of Indian paintings in diophantine equations in the course of the sixth via twelfth centuries and Italian paintings on options of cubic and biquadratic equations from the eleventh via sixteenth centuries. the quantity lines the ancient improvement of algebra and the speculation of equations from precedent days to the start of contemporary algebra, outlining a few sleek issues corresponding to the elemental theorem of algebra, Clifford algebras, and quarternions. it truly is aimed toward undergraduates who've no heritage in calculus.

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We turn t o hi m next . FIBONACCI ( ^ L E O N A R D O O F P I S A ) Fibonacci (1 1 80—1 240 ) i s shor t fo r Filio Bonacci whic h wa s ho w Leonardo of Pisa calle d himself , a member of the Bonacci Family. Hi s mos t famou s work s are 1. Libe r Abbac i (1 202 , revised 1 208 ) 2. Flo s (1 225 ) 3. Libe r Quadratoru m About 1 22 5 Frederic k I I hel d cour t a t Pisa , an d o n tha t occasio n Fibonacc i wa s presented t o him , an d i t appear s tha t Flos wa s presented t o th e Empero r then .

I t follow s a t onc e from th e abov e identit y tha t i f (p , q) i s a solutio n t o (P m ) an d (r , s) i s a solution t o (P n ), the n (pr ± Nqs,ps ± qr) i s a solutio n t o ( P m n ) . ps + qr) 20 V. S . VARADARAJA N We thin k o f thi s a s th e composition o f (p , q) an d (r , s). Thi s proces s wa s calle d the bhdvand b y Brahmagupta , th e wor d meanin g "production " i n Sanskrit . I t i s easy t o verif y tha t i f p, q, r, s ar e al l positiv e integers , the n (pr + Nqs,ps - h qr) i s different fro m th e previou s tw o (se e exercis e 2 below).

Suppos e first tha t bot h a and b are even. The n c 2 = a 2 + b 2 is even, whic h make s c even. S o 2 divide s al l thre e o f them , violatin g ou r assumptio n o f primitivity . O n th e othe r hand, suppos e tha t a and b are bot h odd . The n a 2 an d b 2 ar e bot h odd , s o that c 2 is even, showin g that c must b e even. Bu t a 2 an d b 2 must bot h leav e the remainde r 1 whe n divide d b y 4 s o tha t c 2 mus t leav e th e remainde r 2 whe n divide d b y 4 , a contradiction, becaus e a s c i s even , c 2 i s divisibl e b y 4 !

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