CBSE Class 10 Mathematics Real Numbers Notes Set 04

Number System

Natural numbers, whole numbers, integers, rational numbers and irrational numbers together make up the real numbers.

Real numbers are denoted by R and consist of all rational and irrational numbers. Real numbers (R) include rational numbers (Q), which include integers (Z), which include whole numbers (W), which in turn include natural numbers (N).

1. Natural numbers: numbers used for counting and ordering, N = {1, 2, 3, 4, …} (excluding 0).
2. Whole numbers: natural numbers together with zero, W = {0, 1, 2, 3, 4, …}.
3. Integers: natural numbers, their negatives and zero, Z = {…, −4, −3, −2, −1, 0, 1, 2, 3, 4, …}.
4. Prime numbers: a number that can be divided only by 1 and by itself, for example 2, 3, 5, 7. A prime number has exactly two factors.
5. Composite numbers: a number that has more than two factors, for example 4, 6, 8, 10. Note that 1 is neither prime nor composite.
6. Irrational numbers: real numbers that are not rational. A non-terminating, non-repeating decimal is an irrational number, for example \( \sqrt{2} \), \( \pi \).
7. Rational numbers: numbers that can be written as \( \frac{p}{q} \), where p and q are integers and q ≠ 0.

Mind it
1. Terminating decimals and non-terminating recurring (repeating) decimals are rational numbers.
2. The sum of a rational number and an irrational number is always irrational.
3. The sum of two irrational numbers is not always irrational.
4. The product of a non-zero rational number and an irrational number is always irrational.
5. The product of two irrational numbers is not always irrational.

Complex numbers: a complex number is written as a + ib, where a and b are real numbers and \( i = \sqrt{-1} \). Here a is the real part and b is the imaginary part. Complex numbers are studied in Class 11.

Euclid's Division Lemma

Theorem 1 (Euclid's division lemma): For two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. Here, when a is divided by b, q and r are the quotient and the remainder.

(i) Take 27 and 5. Then 27 = 5 × 5 + 2. Comparing with a = bq + r, a = 27, b = 5, q = 5, r = 2, and 0 ≤ r < b (as 0 ≤ 2 < 5).
(ii) Take 14 and 4. Then 14 = 4 × 3 + 2. Here a = 14, b = 4, q = 3, r = 2 and 0 ≤ r < b (as 0 ≤ 2 < 4).

The relation 0 ≤ r < b is just a statement of long division of a by b, in which q is the quotient and r is the remainder. Thus dividend = divisor × quotient + remainder, that is, a = bq + r.

Example 1. How will you prove that every odd integer is of the form 2q + 1, where q is an integer?
Answer:
Let a be any positive integer. Applying the division lemma to a and b = 2, a = 2q + r, where q is an integer ≥ 0 and 0 ≤ r < 2, so r can be 0 or 1.
For r = 0: a = 2q, an even integer. For r = 1: a = 2q + 1.
Since a is odd, a = 2q is not possible. Hence every odd positive integer is of the form 2q + 1.

Example 2. Prove that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.
Answer:
Any positive integer can be of the form 5m, 5m + 1, 5m + 2, 5m + 3 or 5m + 4, since 0 ≤ r < 5.
\( (5m)^2 = 25m^2 = 5(5m^2) = 5q \), where q = 5m².
\( (5m+1)^2 = 25m^2 + 10m + 1 = 5(5m^2 + 2m) + 1 = 5q + 1 \).
\( (5m+2)^2 = 25m^2 + 20m + 4 = 5(5m^2 + 4m) + 4 = 5q + 4 \).
\( (5m+3)^2 = 25m^2 + 30m + 9 = 5(5m^2 + 6m + 1) + 4 = 5q + 4 \).
\( (5m+4)^2 = 25m^2 + 40m + 16 = 5(5m^2 + 8m + 3) + 1 = 5q + 1 \).
So the square of a positive integer is of the form 5q, 5q + 1 or 5q + 4, and never 5q + 2 or 5q + 3.

Example 3. If x and y are both odd positive integers, then prove that x² + y² is even but not divisible by 4.
Answer:
Let x = 2m + 1 and y = 2n + 1 for some integers m and n.
\( x^2 + y^2 = (2m+1)^2 + (2n+1)^2 = 4m^2 + 4m + 4n^2 + 4n + 2 = 2[2(m^2 + n^2) + 2(m + n) + 1] \), which is of the form 2q, so it is even.
Also \( x^2 + y^2 = 4[(m^2 + n^2) + (m + n)] + 2 = 4q + 2 \), so it leaves remainder 2 on division by 4.
Therefore x² + y² is even but not divisible by 4.

Euclid's Division Algorithm

Euclid's algorithm is a technique to compute the highest common factor (HCF), also called the greatest common divisor (GCD). To obtain the HCF of two positive integers m and n, with m > n, follow these steps.

1. Apply Euclid's division lemma to m and n to find whole numbers q and r such that m = nq + r (0 ≤ r < n).
2. If r = 0, then n is the HCF of m and n. If r ≠ 0, apply the division lemma to n and r.
3. Continue the process till the remainder is zero. The divisor at that stage is the required HCF.

Example 1. Explain the principle behind Euclid's division algorithm.
Answer:
The principle is that if the smaller number is subtracted from the larger number, the greatest common divisor of the two numbers does not change.
Consider 252 and 105, whose HCF is 21. We have 252 = 21 × 12 and 105 = 21 × 5. Since 252 − 105 = 147, the HCF of 147 and 105 is also 21.
Since the larger number is reduced at each step, repeating this gives successively smaller numbers until zero is obtained. The remaining non-zero number is the HCF.
Step 1: \( 252 = 105 \times 2 + 42 \) (252 − 105 = 147, 147 − 105 = 42)
Step 2: \( 105 = 42 \times 2 + 21 \) (105 − 42 = 63, 63 − 42 = 21)
Step 3: \( 42 = 21 \times 2 + 0 \) (42 − 21 = 21, 21 − 21 = 0)
Therefore HCF (252, 105) = 21.

Example 2. Find the HCF of 420, 130 and 600 by using Euclid's division algorithm.
Answer:
First find the HCF of 420 and 130: 420 = 130 × 3 + 30, 130 = 30 × 4 + 10, 30 = 10 × 3 + 0. So HCF (420, 130) = 10.
Now find the HCF of 10 and 600: 600 = 10 × 60 + 0. So HCF (10, 600) = 10.
Hence HCF (420, 130, 600) = 10.

Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic (FTA), also called the unique prime factorisation theorem, gives the relationship between prime numbers and composite numbers.

Theorem 2: Every composite number can be expressed as a product of its prime factors, and this factorisation is unique, apart from the order in which the prime factors occur. For example, \( 12600 = 2^3 \cdot 3^2 \cdot 5^2 \cdot 7 \). The composite number 12600 is written as a product of powers of primes in ascending order, and this decomposition is unique.

H.C.F. and L.C.M. of Numbers

H.C.F. (Highest Common Factor): the H.C.F. of two or more numbers is the largest positive integer that divides each of the integers.
L.C.M. (Least Common Multiple): the L.C.M. of two or more numbers is the smallest positive integer that is exactly divisible by each of the given numbers.

To find the HCF and LCM by prime factorisation (using the FTA), first express each number as a product of prime factors. Then HCF = product of the smallest powers of the common prime factors, and LCM = product of the greatest powers of every prime factor involved in the numbers.

Important relations

1. L.C.M. × H.C.F. = product of the two numbers.
2. The L.C.M. of two or more prime numbers is equal to their product.
3. The H.C.F. of two or more prime numbers is always 1.

Mind it
1. For three positive integers a, b, c: HCF (a, b, c) × LCM (a, b, c) ≠ a × b × c.
2. \( \text{LCM}(a,b,c) = \dfrac{a \cdot b \cdot c \cdot \text{HCF}(a,b,c)}{\text{HCF}(a,b) \cdot \text{HCF}(b,c) \cdot \text{HCF}(c,a)} \)
3. \( \text{HCF}(a,b,c) = \dfrac{a \cdot b \cdot c \cdot \text{LCM}(a,b,c)}{\text{LCM}(a,b) \cdot \text{LCM}(b,c) \cdot \text{LCM}(c,a)} \)
4. For fractions \( \frac{p}{q} \) and \( \frac{a}{b} \): L.C.M. of fractions = \( \dfrac{\text{L.C.M. of numerators}}{\text{H.C.F. of denominators}} \) and H.C.F. of fractions = \( \dfrac{\text{H.C.F. of numerators}}{\text{L.C.M. of denominators}} \).

Example 1. On an evening walk, three people step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?
Answer:
\( 40 = 2^3 \times 5 \), \( 42 = 2 \times 3 \times 7 \), \( 45 = 3^2 \times 5 \).
The minimum distance that each can cover in complete steps is the LCM of 40, 42 and 45.
LCM = \( 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 35 = 2520 \) cm.

Example 2. 7 × 11 × 13 + 13 is a composite number. Explain.
Answer:
\( 7 \times 11 \times 13 + 13 = (7 \times 11 + 1) \times 13 = 78 \times 13 = 2 \times 3 \times 13 \times 13 = 2 \times 3 \times 13^2 \).
The number has more than two factors, so it is a composite number.

Example 3. By using the prime factorisation method, find the L.C.M. and H.C.F. of 1296 and 2520.
Answer:

\( 1296 = 2^4 \times 3^4 \) and \( 2520 = 2^3 \times 3^2 \times 5 \times 7 \).
LCM (1296, 2520) = \( 2^4 \times 3^4 \times 5 \times 7 = 45360 \).
HCF (1296, 2520) = \( 2^3 \times 3^2 = 72 \).

Example 4. Find the greatest number which can divide 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively.
Answer:
The numbers that the required number divides exactly are (1251 − 1), (9377 − 2) and (15628 − 3), that is, 1250, 9375 and 15625.
The required number is the HCF of 1250, 9375 and 15625.
\( 1250 = 2 \times 5^4 \), \( 9375 = 3 \times 5^5 \), \( 15625 = 5^6 \), so HCF = \( 5^4 = 625 \).
Hence 625 is the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively.

Irrational Numbers

An irrational number cannot be expressed as the ratio of two integers. Examples are \( \sqrt{2} \), \( \sqrt{3} \), \( \sqrt{5} \), \( -\sqrt{15} \), \( \pi \). A non-terminating and non-repeating decimal represents an irrational number, for example 0.101001000…, 93.47803945…, 2.395634532….

Theorem 3: Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.

Proof: Let the prime factorisation of a be \( a = p_1 p_2 \cdots p_n \), where \( p_1, p_2, \ldots, p_n \) are primes, not necessarily distinct. Then \( a^2 = (p_1 p_2 \cdots p_n)(p_1 p_2 \cdots p_n) = p_1^2 p_2^2 \cdots p_n^2 \). We are given that p divides a², so p is one of the prime factors of a². By the uniqueness part of the Fundamental Theorem of Arithmetic, the only prime factors of a² are \( p_1, p_2, \ldots, p_n \). So p is one of \( p_1, p_2, \ldots, p_n \). Since a = \( p_1 p_2 \cdots p_n \), we conclude that p divides a.

Mind it
1. If \( \sqrt{pq} \) is irrational, then \( \sqrt{p} + \sqrt{q} \) is definitely irrational.
2. If a is a positive prime number, then \( \sqrt{a} \) is an irrational number.

Revisiting Rational Numbers and Their Decimal Expansions

A rational number is any number that can be written as \( \frac{a}{b} \), where a and b are integers and b ≠ 0. Its decimal representation is terminating or non-terminating but repeating, for example \( \frac{4}{7} \), 1, 0, −2, 4.1, 3.777…, 2.689689….

Theorem 4: Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form \( \frac{a}{b} \), where a and b are coprime and the prime factorisation of b is of the form \( 2^m 5^n \), where m and n are non-negative integers.

Theorem 5: Let \( x = \frac{a}{b} \) be a rational number in which the prime factorisation of b is of the form \( 2^m 5^n \), where m and n are non-negative integers. Then x has a decimal expansion which terminates.

For example, \( \frac{64}{125} = \frac{2^6}{5^3} = \frac{2^9}{(2 \times 5)^3} = \frac{512}{10^3} = 0.512 \) and \( \frac{9}{25} = \frac{9 \times 2^2}{5^2 \times 2^2} = \frac{36}{(2 \times 5)^2} = \frac{36}{10^2} = 0.36 \). So a rational number \( \frac{a}{b} \) in which b is of the form \( 2^m 5^n \) can be converted to an equivalent rational number \( \frac{p}{q} \) in which q is a power of 10, and its decimal expansion terminates.

Theorem 6: Let \( x = \frac{a}{b} \) be a rational number in which the prime factorisation of b is not of the form \( 2^m 5^n \), where m and n are non-negative integers. Then x has a decimal expansion which is non-terminating repeating (recurring). For example, \( \frac{1}{7} = 0.142857142857\ldots \); since the denominator 7 is not of the form \( 2^m 5^n \), this rational number does not terminate.

Hence the decimal expansion of every rational number is either terminating or non-terminating repeating.

Example 1. How will you prove that \( \sqrt{2} \) is irrational?
Answer:
Assume, to the contrary, that \( \sqrt{2} \) is rational. Then \( \sqrt{2} = \frac{a}{b} \), where a and b are integers having no common factor other than 1 (coprime).
So \( \sqrt{2}\,b = a \). Squaring both sides gives \( a^2 = 2b^2 \), so 2 divides a². By Theorem 3, 2 divides a. Write a = 2c for some integer c.
Then \( a^2 = 4c^2 \), so \( 2b^2 = 4c^2 \) and \( b^2 = 2c^2 \). So 2 divides b², and by Theorem 3, 2 divides b.
Hence a and b have at least 2 as a common factor, which contradicts the fact that a and b have no common factor other than 1.
This contradiction arose from our wrong assumption that \( \sqrt{2} \) is rational. So \( \sqrt{2} \) is irrational.

Example 2. Prove that \( 2 + \sqrt{3} \) is irrational.
Answer:
Assume, to the contrary, that \( 2 + \sqrt{3} \) is a rational number r. Then \( \sqrt{3} = r - 2 \).
The left side \( \sqrt{3} \) is irrational, while the right side r − 2 is rational, so L.H.S. ≠ R.H.S. This contradicts our assumption.
Hence \( 2 + \sqrt{3} \) is irrational.

Example 3. Prove that \( \sqrt{2} + \sqrt{3} \) is irrational.
Answer:
Assume, to the contrary, that \( \sqrt{2} + \sqrt{3} = x \) is rational. Squaring, \( x^2 = 2 + 3 + 2\sqrt{3}\cdot\sqrt{2} = 5 + 2\sqrt{6} \).
So \( \sqrt{6} = \dfrac{x^2 - 5}{2} \). Since x, 5 and 2 are rational, \( \dfrac{x^2 - 5}{2} \) is rational, so \( \sqrt{6} \) would be rational. But \( \sqrt{6} \) is irrational, a contradiction.
Hence \( \sqrt{2} + \sqrt{3} \) is irrational.

Example 4. Prove that \( \sqrt{a} + \sqrt{b} \) is irrational, where a and b are primes.
Answer:
Suppose, to the contrary, that \( \sqrt{a} + \sqrt{b} = r \), where r is rational. Then \( \sqrt{a} = r - \sqrt{b} \).
Squaring both sides, \( a = r^2 + b - 2r\sqrt{b} \), so \( \sqrt{b} = \dfrac{r^2 + b - a}{2r} \).
Now r², b, a and r are all rational numbers, and rational numbers are closed under addition, subtraction, multiplication and division (division by zero excluded). So \( \dfrac{r^2 + b - a}{2r} \) is rational.
But b is a prime number, so b is not a perfect square and \( \sqrt{b} \) is irrational. An irrational number cannot equal a rational number, which is a contradiction.
Hence \( \sqrt{a} + \sqrt{b} \) is irrational.

Example 5. Find which of the following rational numbers have terminating decimals and which have non-terminating recurring decimals: (i) \( \frac{17}{8} \) (ii) \( \frac{13}{3125} \) (iii) \( \frac{64}{455} \) (iv) \( \frac{29}{343} \) (v) \( \frac{15}{1600} \)
Answer:
(i) \( \frac{17}{8} = \frac{17}{2^3} = \frac{17 \times 5^3}{(2 \times 5)^3} = \frac{17 \times 125}{10^3} \). Terminating.
(ii) \( \frac{13}{3125} = \frac{13}{5^5} = \frac{13 \times 2^5}{(2 \times 5)^5} = \frac{13 \times 32}{10^5} \). Terminating.
(iii) \( \frac{64}{455} = \frac{2^6}{5 \times 7 \times 13} \). The factors 7 and 13 cannot be removed from the denominator, so it is non-terminating recurring.
(iv) \( \frac{29}{343} = \frac{29}{7^3} \). Non-terminating recurring.
(v) \( \frac{15}{1600} = \frac{3 \times 5}{2^6 \times 5^2} = \frac{3}{2^6 \times 5} = \frac{3 \times 5^5}{(2 \times 5)^6} = \frac{3 \times 5^5}{10^6} \). Terminating.

Important Concepts Regarding H.C.F. and L.C.M. of Numbers

1. Remainder when divided by a product. A number, when divided by a₁ and a₂, leaves remainders r₁ and r₂ respectively. If the number is divided by (a₁ × a₂), the remainder is (a₁ × r₂ + r₁).

Example 1. A number when divided by 5 and 7 leaves the remainders 3 and 4, respectively. Find the remainder when the same number is divided by 35.
Answer:
Required remainder = a₁ × r₂ + r₁ = 5 × 4 + 3 = 23.

2. How many numbers are divisible by a certain integer. Divide the limit by the integer; the quotient is the count.

Example 1. (i) Find how many numbers up to 594 are divisible by 15. (ii) Find how many numbers up to 600 are divisible by 7 and 5 together.
Answer:
(i) Divide 594 by 15: 594 = 39 × 15 + 9. The quotient 39 is the required number, so there are 39 such numbers.
(ii) The L.C.M. of 7 and 5 is 35. Divide 600 by 35: 600 = 17 × 35 + 5. So there are 17 such numbers.

3. Two numbers leave remainders r₁ and r₂ and their sum leaves remainder r₃. The divisor is given by (r₁ + r₂ − r₃).

Example 1. Two numbers when divided by a certain divisor leave remainders 473 and 298, respectively. When their sum is divided by the same divisor, the remainder is 236. Find the divisor.
Answer:
The required divisor = 473 + 298 − 236 = 535.

4. Largest number which divides a, b and c leaving remainders x, y and z. Required number = H.C.F. of (a − x), (b − y) and (c − z).

Example 1. Find the largest number which will divide 148, 246 and 623 leaving remainders 4, 6 and 11 respectively.
Answer:
Required number = H.C.F. of (148 − 4), (246 − 6) and (623 − 11) = H.C.F. of 144, 240 and 612 = 12.
Hence the largest number is 12.

5. Smallest number which when divided by a, b and c leaves remainders x, y and z, where (a − x) = (b − y) = (c − z) = k. Required number = (L.C.M. of a, b and c) − k.

Example 1. Find the smallest number which when divided by 12, 15 and 21 leaves the remainders 7, 10 and 16 respectively.
Answer:
Since (12 − 7) = (15 − 10) = (21 − 16) = 5, the required number = (L.C.M. of 12, 15 and 21) − 5 = 420 − 5 = 415.

6. Smallest number which when divided by a, b and c leaves the same remainder r in each case. Required number = (L.C.M. of a, b and c) + r.

Example 1. Find the smallest number which when divided by 16, 20 and 24 will leave a remainder 5 in each case.
Answer:
Required number = (L.C.M. of 16, 20 and 24) + 5 = 240 + 5 = 245.

7. Largest number which divides a, b and c leaving the same remainder in each case. (a) When the remainder is not given: required number = H.C.F. of |a − b|, |b − c| and |c − a|. (b) When the remainder r is given: required number = H.C.F. of (a − r), (b − r) and (c − r).

Example 1. (i) Find the largest number which on dividing 34, 90 and 104 leaves the same remainder in each case. (ii) Find the largest number which will divide 155 and 192 and leaves the remainder 7 in each case.
Answer:
(i) Required number = H.C.F. of |34 − 90|, |90 − 104| and |104 − 34| = H.C.F. of 56, 14 and 70 = 14.
(ii) Required number = H.C.F. of (155 − 7) and (192 − 7) = H.C.F. of 148 and 185 = 37.

Number of Factors of a Given Number

Let N be a composite number whose prime factors are a, b, c, d, … with powers k, l, m, n, …, that is, \( N = a^k \cdot b^l \cdot c^m \cdot d^n \cdots \). Then the number of total factors (divisors) of N is \( (k+1)(l+1)(m+1)(n+1)\cdots \).

Example 1. Find the total number of factors of 1080.
Answer:
\( 1080 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 = 2^3 \times 3^3 \times 5^1 \).
Total number of factors = (3 + 1)(3 + 1)(1 + 1) = 32.

Sum of Factors of a Given Number

If \( N = a^k \cdot b^l \cdot c^m \cdot d^n \cdots \), then the sum of all the factors of N is \( \dfrac{(a^{k+1}-1)(b^{l+1}-1)(c^{m+1}-1)(d^{n+1}-1)\cdots}{(a-1)(b-1)(c-1)(d-1)\cdots} \).

Example 1. Find the sum of the factors of 180.
Answer:
\( 180 = 2^2 \times 3^2 \times 5 \).
Sum of the factors = \( \dfrac{(2^{3}-1)(3^{3}-1)(5^{2}-1)}{(2-1)(3-1)(5-1)} = \dfrac{7 \times 26 \times 24}{1 \times 2 \times 4} = 546 \).

Product of Factors

The factors of 24 are 1 × 24 = 2 × 12 = 3 × 8 = 4 × 6, so the product of the factors of 24 = (1 × 24)(2 × 12)(3 × 8)(4 × 6) = \( 24^4 \). Hence the product of the factors of a composite number N is \( N^{n/2} \), where n is the total number of factors of N.

Number of Odd Factors of a Given Number

Express N as \( N = (a_1^p \times a_2^q \times a_3^r \times \cdots) \times e^x \), where \( a_1, a_2, a_3, \ldots \) are the odd prime factors and e is the even prime factor (2). Then the total number of odd factors is (p + 1)(q + 1)(r + 1)….

Example 1. Find the number of odd factors of 72.
Answer:
\( 72 = 2^3 \times 3^2 \). Number of odd factors = (2 + 1) = 3.

Number of Even Factors of a Composite Number

Number of even factors of a number = (total number of factors of the number) − (total number of odd factors of the number).

Example 1. Find the number of even factors (divisors) of 72.
Answer:
\( 72 = 2^3 \times 3^2 \). Total number of factors = (3 + 1)(2 + 1) = 12.
Number of odd factors = (2 + 1) = 3.
Number of even factors = 12 − 3 = 9.

Summary

1. Euclid's division lemma: for any two positive integers a and b with a > b, there exist unique integers q and r such that a = bq + r (0 ≤ r < b).
2. Euclid's division algorithm: gives a step-wise procedure for computing the H.C.F. of two natural numbers.
3. Fundamental Theorem of Arithmetic: every composite number can be expressed as a product of primes, and this factorisation is unique.
4. H.C.F.: the H.C.F. of two or more positive integers is the largest positive integer that divides each of them exactly. The H.C.F. of two or more prime numbers is always 1, the L.C.M. of two or more prime numbers is their product, and HCF (a, b) × LCM (a, b) = a × b.
5. An irrational number is a non-terminating, non-recurring decimal and cannot be expressed in the form p/q.
6. Rational numbers can be expressed as p/q, where p and q are integers and q ≠ 0. If q has a prime factorisation of the form \( 2^m 5^n \), the decimal expansion terminates; otherwise it is non-terminating repeating.
7. If p is a positive prime, then \( \sqrt{p} \) is irrational. The sum of a rational and an irrational number is irrational, and the product of a non-zero rational and an irrational number is irrational. If \( \sqrt{ab} \) is irrational, then \( \sqrt{a} + \sqrt{b} \) is also irrational.