Math’s ProjectWorkReal

numbers and

polynomials

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Real Numbers

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Theorem 1.1(Euclid’s Division Lemma) :Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r,0 ≤ r < b.Euclid’s division algorithm is a technique to compute the Highest Common Factor(HCF) of two given positive integers. Recall that the HCF of two positive integers a and b is the largest positive integer d that divides both a and b.Let us see how the algorithm works, through an example first. Suppose we need to find the HCF of the integers 455 and 42. We start with the larger integer, that is,455. Then we use Euclid’s lemma to get455 =42 × 10 + 35Now consider the divisor 42 and the remainder 35, and apply the division lemma to get42 =35 × 1 + 7Now consider the divisor 35 and the remainder 7, and apply the division lemmato get 35 =7 × 5 + 0 3

Notice that the remainder has become zero, and we cannot proceed any further. We claim that the HCF of 455 and 42 is the divisor at this stage, i.e., 7. You can easily verify this by listing all the factors of 455 and 42. Why does this method work? It works because of the following result. So, let us stateEuclid’s division algorithm clearly.To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below:Step 1 : Apply Euclid’s division lemma, to c and d . So, we find whole numbers, q and r such that c = dq + r , 0 ≤ r < d .Step 2 :If r = 0, d is the HCF of c and d . If r ≠0, apply the division lemma to d and r .Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.This algorithm works because HCF (c, d ) = HCF (d ,r ) where the symbolHCF (c, d ) denotes the HCF of c and d ,etc. 4

Example 1 : Use Euclid’s algorithm to find the HCF of 4052 and 12576.Solution :Step 1 : Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get 12576 =4052 × 3 + 420Step 2 :Since the remainder 420 ≠ 0, we apply the division lemma to 4052 and 420, to get 4052 =420 × 9 + 272Step 3 : We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get 420 =272 × 1 + 148We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get 272 =148 × 1 + 124We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get 5

148 =124 × 1 + 24We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get124 =24 × 5 + 4We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get 24 =4 × 6 + 0The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.

Theorem 1.2(Fundamental Theorem of Arithmetic) : Every composite number can be expressed (factorised )as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

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Note that HCF(6, 20) = 21=Product of the smallest power of each Common prime factor in the numbers.LCM (6, 20) = 22× 31× 51=Product of the greatest power of each prime factor, involved in the numbers.

 Theorem1.3-Let p be a prime number. If p divides a , then p divides a, where a is a positive integer.

Theorem1.4-Let x be rational number whose decimal expansion termin-ates. Then x can be expressed in the form p/q, where p and q are coprime,And the prime factorisation of q is of the form 2n x 5m, where n, m arenon negative numbers.

Theorem1.5-Let x = p, q be a rational number, such that the prime Factorisation of q is of the form 2n x 5m , where n, m are non-negative integers. Then x has a decimal expansion which terminates. 7

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Polynomials

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Polynomial: In mathematics, a polynomial is an expression of finite length constructed from variables (also called indeterminates) and constants, using only the operations of addition, subtraction, multiplication, and non-negative integer exponents. For example, x2 − x/4 + 7 is a polynomial, but x2 − 4/x + 7x3/2 is not, because its second term involves division by the variable x (4/x), and also because its third term contains an exponent that is not an integer (3/2).

Degree:  if p(x) is a polynomial in x, the highest power of  x in p(x) is called the degree of the polynomial p(x). For example, 4 x+ 2 A polynomial of degree 1 is called a linear polynomial. For example,2 x– 3A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’ has been derived from the word ‘quadrate’, which means ‘square’. 2x+ 3x + 4

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A polynomial of degree 3 is called a cubic polynomial. Example: 4x + 5

If  p(x) is a polynomial in x, and if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of  p(x) at x =k , and is denoted by p(k).

Zero: a real number k is said to be a zero of a polynomial p(x), if p(k ) = 0.

If k is a zero of p( x) =ax+ b, then p(k ) =ak +b= 0, i.e., k= -b/a So, the zero of the linear polynomial ax+ b is –b/a: -(Constant term)/ Coefficient of x 

A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can have at most 3 zeroes.

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If α and β are the zeroes of the quadratic polynomial ax2+bx+c, then α +β = -b/a , αβ = c/aIf α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d = 0, then α +β + γ = -b/a, αβ+ βγ+ γα = c/a, αβγ = -d/a

The division algorithm states that given any polynomial p(x) and any non-zero polynomial g(x), there are polynomials q(x) and r( x) such that  p( x) = g(x) q( x) + r(x), where r(x) = 0 or degree r( x) < degree g( x).Example: Divide 3x – x – 3 x+ 5 by x– 1 –x2, and verify the division algorithm.Solution : Note that the given polynomials are not in standard form. To carry out division, we first write both the dividend and divisor in decreasing orders of their degrees.

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So, dividend = – x3+ 3 x2– 3 x+ 5 and divisor = – x2+ x– 1.Division process is shown on the right side.We stop here since degree (3) = 0 < 2 = degree (–x2+x– 1).So, quotient = x - 2, remainder = 3.

Graph for linear polynomial:

y= 2x + 3

X-axis -2 2 -3/2

Y-axis -1 7 0

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Graph for quadratic polynomial: x2 – 3x - 4

X-axis -2 -1 0 1 2 3 4 5

Y-axis 6 0 -4 -6 -6 -4 0 6

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Graph of cubic polynomial: x3 – 4x

X-axis -2 -1 0 1 2

Y-axis 0 3 0 -3 0