
01. Real Numbers
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02. Polynomials
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03. Linear Equation
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04. Quadratic Equation
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05. Arithmetic Progressions
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06. Some Applications of Trigonometry
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07. Coordinate Geometry
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08. Triangles
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09. Circles
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10. Areas Related to Circles
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11. Introduction to Trigonometry
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12. Surface Areas and Volumes
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13. Statistics
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14. Probability
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15. Construction
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Chapter Notes – Real Numbers
(1) Euclid’s Division Lemma:
Theorem: Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.
(2) Euclid’s division algorithm: 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.
For Example: Use Euclid’s division algorithm to find the HCF of 135 and 225.
Step 1: Here 225 > 135, on applying the division lemma to 225 and 135, we get 225 = 135 x 1 + 90
Step 2: Since, remainder ≠ 0, we again apply division lemma to 135 and 90, we get 135 = 90 x 1 + 45
Step 3: Again, applying division lemma to 90 and 45, we get 90 = 45 x 2 + 0
The remainder has become zero. And since the divisor at this step is 45, the HCF of 135 and 225 is 45.
For Example: Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Let a be any positive odd integer. And we apply division algorithm with a and b = 4.
As 0 ≤ r < 4, the possible remainders could be 0, 1, 2 and 3.
So, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.
Now, since a is odd, so a cannot be 4q or 4q + 2 (as both are divisible by 2).
Hence, any odd integer is of the form 4q + 1 or 4q + 3.
(3) The Fundamental Theorem of Arithmetic:
Theorem: 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. In general, given a composite number x, we factorise it as x = p_{1} p_{2} … p_{n} , where p_{1}, p_{2} ,…, p_{n} are primes and written in ascending order, i.e., p_{1} ≤ p_{2} ≤ . . . ≤ p_{n }. If we combine the same primes, we will get powers of primes.
For Example: The prime factors of 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 2^{3} × 3^{2} × 5 × 7 × 13.
For Example: Find the HCF and LCM of 96 and 404 by prime factorisation method.
The prime factorisation of 96 is 2^{5} x 3. And that of 404 is 2^{2} x 101.
Hence, HCF of 96 and 404 will be 2^{2} = 4.
Now, LCM (96, 404) = (96 x 404)/(HCF (96, 404)) = (96 x 404)/4 = 9696.
For Example: Check whether 6^{n} can end with the digit 0 for any natural number n.
If a number ends with digit 0, then, it must be divisible by 10 or in other words, it will be divisible by 2 and 5 as 10 = 2 x 5.
Now, prime factorisation of 6^{n} = (2 x 3)^{n}.
Here, 5 is not in the prime factorisation of 6^{n}. Hence, for any value of n, 6^{n} will not be divisible by 5.
Thus, 6^{n} cannot end with the digit 0 for any natural number n.
(4) Revisiting Irrational Numbers:
Irrational Number : are the numbers which cannot be written in p/q form, where p and q are integers and q ≠ 0.
Theorem 1: Let p be a prime number. If p divides a^{2}, then p divides a, where a is a positive integer.
Proof: Suppose the prime factorisation of a is as follows:
(i) a = p_{1}p_{2}….p_{n}, where p_{1},p_{2},….pn are primes.
(ii) On squaring both the sides, we get,
(iii) a^{2} = (p_{1}p_{2}….p_{n}) ( p_{1}p_{2}….p_{n}) = p_{1}^{2}p_{2}^{2}….p_{n}^{2}.
(iv) It is given that p divides a^{2}. Hence, we can say that p is one of the prime factors of a^{2} as per the Fundamental Theorem of Arithmetic.
(v) However, as per the uniqueness part of the Fundamental Theorem of Arithmetic, we can deduce that the only prime factors of a^{2 }are p_{1}p_{2}….p_{n}. Thus, p is one of p_{1}p_{2}….p_{n}.
Since, a = p_{1}p_{2}….p_{n},p divides a.
Theorem 2: √2 is irrational.
Proof: We shall start by assuming √2 as rational. In other words, we need to find integers x and y such that √2 = x/y.
(i) Let x and y have a common factor other than 1, and so we can divide by that common factor and assume that x and y are coprime. So, y√2 = x.
(ii) Squaring both side, we get, 2y^{2} = x^{2}.
(iii) Thus, 2 divides x^{2}. and by theorem we can say that 2 divides x.
(iv) Hence, x = 2z for some integer z.
(v) Substituting x, we get, 2x^{2} = 4z^{2}e. y^{2} = 4z^{2}; which means y^{2 }is divisible by 2, and so y will also be divisible by 2.
(vi) Now, from theorem, x and y will have 2 as a common factor. But, it is opposite to fact that x and y are coprime.
(vii) Hence, we can conclude √2 is irrational.
For Example: Prove that √3 is irrational.
We shall start by assuming √3 as rational. In other words, we need to find integers x and y such that √3 = x/y.
Let x and y have a common factor other than 1, and so we can divide by that common factor and assume that x and y are coprime. So, y√3 = x.
Squaring both side, we get, 3y^{2} = x^{2}.
Thus, x^{2} is divisible by 3, and by theorem we can say that x is also divisible by 3.
Hence, x = 3z for some integer z.
Substituting a, we get, 3x^{2} = 9z^{2} i.e. y^{2} = 3z^{2}; which means y^{2 }is divisible by 3, and so y will also be divisible by 3.
Now, from theorem, x and y will have 3 as a common factor. But, it is opposite to fact that x and y are coprime.
Hence, we can conclude √3 is irrational.
For Example: Prove that 6 + √2 is irrational.
Let us assume 6 + √2 to be rational.
Therefore, we must find two integers a, b (b ≠ 0) such that
6 + √2 = a/b i.e. √2 = a/b – 6.
Since, a and b are integers, a/b – 6 is also rational and hence √2 must be rational.
Now, this contradicts the fact that √2 is irrational.
Hence, 6 + √2 is irrational.
(5) Revisiting Rational Numbers and Their Decimal Expansions:
Theorem 1: Let x be a rational number whose decimal expansion terminates. 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 2^{n} 5^{m}, where n, m are nonnegative integers.
For Example: 13/125 = 13/5^{3} = (13 x 2^{3})/(2^{3 }x 5^{3}) = 104/10^{3} = 0.104
Theorem 2: Let x = p/q be a rational number, such that the prime factorisation of q is of the form 2^{n }5^{m}, where n, m are nonnegative integers. Then x has a decimal expansion which terminates.
Theorem 3: Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2^{n} 5^{m}, where n, m are nonnegative integers. Then, x has a decimal expansion which is nonterminating repeating (recurring).
For Example: Without actually performing the long division, state whether 6/15 will have a terminating decimal expansion or a nonterminating repeating decimal expansion.
The prime factorisation of 6/15 can be written as
6/15 = (2 x 3)/(3 x 5) = 2/5
Here, the denominator is of the form 5^{n}.
Hence, decimal expansion of 6/15 is terminating.
For Example: Write down the decimal expansions of 17/8.
The decimal expansion of 17/8 is
For Example: The following real number has decimal expansions as given below. Decide whether it is rational or not. If it is rational, and of the form, p/q what can you say about the prime factors of q?
Here, as the decimal expansion is nonterminating recurring, the given number is a rational number of the form p/q.
Moreover, q is not of the form 2^{n} 5^{m}, hence, prime factors of q will also have factors other than 2 or 5.
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