real number system proof

An algebraic expression of the form a 0 +a 1 x+a 2 x 2 +.+a n x n where a 0 , a 1 , a 2 ,.a n are real numbers, n is a positive integer is called a polynomial in x...

A - 1 does not exist But if (adj A) B = 0, then the system is consistent with infinite number of solutions or has no solution. the system is inconsistent i.e., it has no solutio..

forms the set of real numbers (ii..

If x and y are any real numbers, then x+iy is called a complex number. In the complex number x+iy, the real numbers x and y are respectively called the real part and imaginary part of x+iy. Two complex numbers are s..

= (14 - 12) - (7 - 3) + (4 - 2) = 2 - 4 + 2 = 0 The system may have infinite number of solutions or no solution. Put x = k in (1) and (2) and solve y + z = 6 - k 2y + 3z = 14 - k. Solving the above two equations, we have z = k + 2 and y = 4 - 2k When x = k, substituting t..

Some important results - Given a, b, c and d are non-zero real numbers, we can deduce other proportions by simple Algebra. These results are often referred by the names mentioned along each of the properties obtained. (1) If then bc = ad This property is known as INVERTENDO. (2)..

The given equations are 2x - y + z = -3 3x - 0.y - z = - 8 2x + 6y + 0.z= 2 = 2(6) +1(2) + 1(18) = 12 +2 + 18 = 32 The system has a unique solutions. A 1 1 = (0 + 6) = 6, A 1 2 = -(0 + 2) = -2, A 1 3 = 18 A 2 1 = 6, A 2 2 = -2, A 2 3 = -14 A 3 1 = 1, A 3 2 = 5, A ..

. To find integral of the type proceeding as in (7), we obtain the integral, using standard formulae. 09. To find the integral of the type where p, q, a, b, c are constants, we are to find real numbers A, B such that To determine A and B, we equate from both sides the coefficien..

Draw AL, BM and CN perpendicular to x-axis. LM = x 2 -x 1 MN = x 3 -x 2 LN = x 3 -x 1 Area of D ABC = area of trap {ALMB + BMNC - ALNC} This can be expressed in the form of a determinant Another form which is convenient to use for the area of triangles but which is very much useful wh..

Let A (x 1 , y 1 ) and B (x 2 , y 2 ) be two points in the plane. Let d = distance between the points A and B. Draw AL and BM perpendicular to x-axis (parallel to y-axis). Draw AC perpendicular to BM to cut BM at C. In the figure, OL = x 1 , OM = x 2 [AC = LM = OM - OL = x 2 - x 1 ] MB = y 2 , MC =..