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  1. Mar 03,  · a) v1 = k[A][B]^2 >1. if A is halved and B is tripled then v2 = k[A/2][3B]^2 >2. v1 = k[A][B]^2 and v2 = k[A/2][3B]^2 => v1 / v2 =[A / A/2 ][ B / 3B ]^2.
  2. In , k/a is an integer, and so is by definition (or by casting out the modulus): b≡c (mod m) Hence when a and m are relatively prime, we can divide as normal.
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  4. 2 if a+ ib=0 wherei= p −1, then a= b=0 if a+ ib= x+ iy,wherei= p −1, then a= xand b= y The roots of the quadratic equationax2+bx+c=0;a6= 0 are −b p b2 −4ac 2a The solution set of the equation is (−b+ p 2a −b− p 2a where = discriminant = b2 −4ac
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