CBSE Class 10 Maths Advanced Real Numbers Notes Set 01

Real Numbers

A. Introduction and Euclid's Division Lemma

Classification of Numbers

Earlier classes introduced natural numbers, whole numbers, integers, rational numbers and irrational numbers. Taken together they make up the real numbers. This chapter looks at some deeper properties of numbers, mainly of integers.

1. Real numbers. A real number is any number that can measure an actual quantity, and each one has its own point on the number line (a straight line with a fixed point chosen as zero). The set of real numbers contains every rational and every irrational number.

2. Prime numbers. A natural number is prime if the only natural numbers that divide it exactly are 1 and the number itself. The first few are 2, 3, 5, 7, 11, 13, 17, 19, 23, ...

3. Composite numbers. A natural number with more than two different factors is composite.

Note. The number 1 is neither prime nor composite, because it has just one factor.

4. Co-prime numbers. Two numbers (they need not be prime themselves) are co-prime when their H.C.F. is 1. For example 4 and 9 are co-prime since H.C.F.(4, 9) = 1.

Note. Two consecutive numbers are always co-prime, and so are any two different primes.

Divisibility

A non-zero integer b divides an integer a when a can be written as a = bq for some integer q. Here a is the dividend, b the divisor and q the quotient. For instance 5 divides 35 because 35 = 5 × 7. The symbol b | a is read "b divides a", and b ∤ a means that b does not divide a.

Lemma

A lemma is a statement that has already been proved and is used as a stepping stone for proving a larger result.

Euclid's Division Lemma

Take any two positive integers a and b. Then one can find exactly one pair of integers q and r with

a = bq + r, where 0 ≤ r < b.

When b divides a exactly, r = 0. The idea is just long division: the layout below shows the quotient q above the line, the dividend a under the bracket and the remainder r at the bottom.

In words: Dividend = Divisor × Quotient + Remainder.

Note. Either q or r can be 0 in the lemma, but the remainder is always strictly smaller than the divisor b.

Solved Examples

Example 1. Prove that any positive odd integer can be written in the form 4q + 1 or 4q + 3 where q is an integer.
Answer:
Let a be a positive odd integer. Divide it by 4: by the division lemma a = 4q + r with r = 0, 1, 2 or 3.
So a is one of 4q, 4q + 1, 4q + 2, 4q + 3.
4q and 4q + 2 = 2(2q + 1) are both even. An odd a therefore has to be 4q + 1 or 4q + 3.

Example 2. Prove that the square of any positive integer of the form 5q + 1 is of the same form.
Answer:
Let x = 5q + 1. Then x² = 25q² + 10q + 1 = 5(5q² + 2q) + 1.
Put m = 5q² + 2q = q(5q + 2). Then x² = 5m + 1, so x² has the same form 5m + 1.

Example 3. Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer.
Answer:
Write n = 3q + r with r = 0, 1 or 2, and look at the three numbers in each case.
r = 0: n = 3q is divisible by 3. Then n + 2 = 3q + 2 leaves remainder 2, and n + 4 = 3(q + 1) + 1 leaves remainder 1, so neither of them is divisible by 3.
r = 1: n = 3q + 1 is not divisible, n + 2 = 3q + 3 = 3(q + 1) is divisible, and n + 4 = 3q + 5 = 3(q + 1) + 2 is not.
r = 2: n = 3q + 2 is not divisible, n + 2 = 3q + 4 = 3(q + 1) + 1 is not, and n + 4 = 3q + 6 = 3(q + 2) is divisible.
In every case exactly one of the three numbers is a multiple of 3.

Example 4. Use Euclid's Division Lemma to show that the cube of any positive integer is of the form 9 m, 9 m + 1 or 9 m + 8, for some integer m.
Answer:
Any positive integer x is 3q, 3q + 1 or 3q + 2.
x = 3q: x³ = 27q³ = 9(3q³) = 9m with m = 3q³.
x = 3q + 1: x³ = 27q³ + 27q² + 9q + 1 = 9q(3q² + 3q + 1) + 1 = 9m + 1 with m = q(3q² + 3q + 1).
x = 3q + 2: x³ = 27q³ + 54q² + 36q + 8 = 9q(3q² + 6q + 4) + 8 = 9m + 8 with m = q(3q² + 6q + 4).
So every cube is of the form 9m, 9m + 1 or 9m + 8.

B. Euclid's Division Algorithm

Algorithm

An algorithm is a fixed list of steps, carried out in a definite order, that solves a given kind of problem. Computer programs are built from algorithms.

Euclid's Division Algorithm

If a = bq + r, then any number that divides both a and b also divides both b and r, and the other way round. In other words H.C.F.(a, b) = H.C.F.(b, r). This is why the following procedure finds the H.C.F. of two positive integers a and b, with a > b.

Step 1. Apply the division lemma to a and b: a = bq + r with 0 ≤ r < b.

Step 2. If r = 0, the H.C.F. is b. If r ≠ 0, apply the lemma again to the pair b and r.

Step 3. Keep repeating until a remainder of 0 appears. The divisor used in that last division is H.C.F.(a, b).

Solved Examples

Example 5. Use Euclid's division algorithm to find the H.C.F. of 196 and 38318.
Answer:
38318 = 196 × 195 + 98
196 = 98 × 2 + 0
The remainder becomes 0 in the second line, so the H.C.F. is the last divisor, 98.

Example 6. Use Euclid's division algorithm to find the HCF of (i) 56 and 814   (ii) 6265 and 76254
Answer:
(i) 814 = 56 × 14 + 30
56 = 30 × 1 + 26
30 = 26 × 1 + 4
26 = 4 × 6 + 2
4 = 2 × 2 + 0
The last divisor is 2, so HCF(56, 814) = 2.
(ii) 76254 = 6265 × 12 + 1074
6265 = 1074 × 5 + 895
1074 = 895 × 1 + 179
895 = 179 × 5 + 0
So HCF(6265, 76254) = 179.

Example 7. If the H.C.F of 657 and 963 is expressible in the form 657 x + 963 × (– 15), find x.
Answer:
963 = 657 × 1 + 306
657 = 306 × 2 + 45
306 = 45 × 6 + 36
45 = 36 × 1 + 9
36 = 9 × 4 + 0
So the H.C.F. is 9.
Set 657x + 963 × (−15) = 9. Then 657x = 9 + 14445 = 14454, so x = \( \frac{14454}{657} \) = 22.

Example 8. If d is the HCF of 468 and 222, find x, y satisfying d = 468x + 222y. Also, show that x and y are not unique.
Answer:
468 = 222 × 2 + 24 ... (i)
222 = 24 × 9 + 6 ... (ii)
24 = 6 × 4 + 0 ... (iii)
The last non-zero remainder is 6, so d = 6.
Work backwards. From (ii), 6 = 222 − 24 × 9. From (i), 24 = 468 − 222 × 2. Substituting, 6 = 222 − (468 − 222 × 2) × 9 = 222 × 19 − 468 × 9.
So one solution is x = −9, y = 19.
Now add and subtract the product 222 × 468 on the right: 6 = 222 × 19 + 222 × 468 − 468 × 9 − 222 × 468 = 222 × (19 + 468) − 468 × (9 + 222) = 222 × 487 − 468 × 231.
That gives a second solution, x = −231, y = 487. Two different pairs work, so x and y are not unique.

Example 9. 144 cartons of coke cans and 90 cartons of pepsi cans are to be stacked in a canteen. If each stack is of same height and is to contain cartons of the same drink, what would be the greatest number of cartons each stack would have ?
Answer:
Each stack must hold a number of cartons that divides both 144 and 90 exactly, and we want the largest such number, i.e. the H.C.F.
144 = 90 × 1 + 54
90 = 54 × 1 + 36
54 = 36 × 1 + 18
36 = 18 × 2 + 0
H.C.F.(144, 90) = 18, so each stack can have at most 18 cartons.

C. Fundamental Theorem of Arithmetic

Fundamental Theorem of Arithmetic

Every composite number can be written as a product of prime numbers, and this product is unique apart from the order in which the primes are listed.

HCF and LCM

The theorem gives a neat way to get the H.C.F. and L.C.M. of numbers: write each as a product of primes.

H.C.F. = product of the primes that are common to all the numbers, each taken with its smallest exponent.
L.C.M. = product of all the primes that occur in any of the numbers, each taken with its largest exponent.

For two natural numbers a and b: H.C.F. × L.C.M. = a × b.

Note. The L.C.M. of numbers is always divisible by their H.C.F.

Solved Examples

Example 10. Determine the prime factors of 45470971.
Answer:
Divide repeatedly by primes, as in the factor tree. At each stage the prime on the left is taken out and the quotient continues on the main branch.

So 45470971 = 7² × 13² × 17² × 19.

Example 11. Check whether 6ⁿ can end with the digit 0 for any natural number n.
Answer:
A number ending in 0 is a multiple of 10, so 5 must appear among its prime factors.
But 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ, whose only prime factors are 2 and 3.
By the uniqueness in the theorem, 5 is not a factor of 6ⁿ for any n, so 6ⁿ never ends with 0.

Example 13. In a morning walk three persons step off together, their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that they can cover the distance in complete steps ?
Answer:
The required distance is the smallest number that is a multiple of all three step lengths, i.e. their L.C.M.
80 = 2⁴ × 5, 85 = 5 × 17, 90 = 2 × 3² × 5.
L.C.M. = 2⁴ × 3² × 5 × 17 = 16 × 9 × 5 × 17 = 12240 cm = 122 m 40 cm.

D. Proof of Irrationality and Decimal Representation

Some Important Results

1. If a prime number p divides a², where a is a positive integer, then p divides a as well.

2. Suppose a rational number x has a terminating decimal expansion. Then x = \( \frac{p}{q} \) with p, q co-prime, and the prime factorisation of q has the form \( 2^m \times 5^n \), where m and n are non-negative integers.

3. Suppose x = \( \frac{p}{q} \) is rational (p, q co-prime) and the prime factorisation of q is not of the form \( 2^m \times 5^n \). Then the decimal expansion of x is non-terminating and repeating.

Results 2 and 3 together give a quick test: reduce the fraction to lowest terms and look at the denominator. If it contains only the primes 2 and 5 the decimal stops; if any other prime appears the decimal never stops but repeats a block of digits.

Solved Examples

Example 14. Prove that \( \sqrt{2} \) is an irrational number.
Answer:
Suppose, to the contrary, that \( \sqrt{2} \) is rational. Then \( \sqrt{2} = \frac{a}{b} \) for positive integers a and b with no common factor (H.C.F. 1).
Squaring both sides: \( 2 = \frac{a^2}{b^2} \), so \( a^2 = 2b^2 \). Hence a² is a multiple of 2, and by result 1, a is a multiple of 2. Write a = 2c for some integer c.
Then 4c² = 2b², i.e. b² = 2c². So b² is a multiple of 2 and therefore b is also a multiple of 2.
Now both a and b are divisible by 2, which contradicts the fact that they are co-prime. So the assumption was wrong, and \( \sqrt{2} \) is irrational.

Example 15. Prove that \( 5\sqrt{3} \) is not rational
Answer:
Assume it is rational: \( 5\sqrt3 = \frac{a}{b} \) with a, b co-prime integers (b ≠ 0).
Dividing by 5 gives \( \sqrt3 = \frac{a}{5b} \), a ratio of two integers.
That would make \( \sqrt{3} \) rational, but \( \sqrt{3} \) is known to be irrational. This contradiction shows that \( 5\sqrt3 \) cannot be rational.

Example 16. Prove that \( 3-\sqrt{5} \) is an irrational number.
Answer:
Suppose \( 3-\sqrt5 \) is rational, equal to \( \frac{a}{b} \) where a and b are co-prime positive integers.
Rearranging, \( \sqrt5 = 3-\frac{a}{b} = \frac{3b-a}{b} \). Since a and b are integers, \( \frac{3b-a}{b} \) is a rational number, which would make \( \sqrt{5} \) rational.
This contradicts the known fact that \( \sqrt{5} \) is irrational. Hence \( 3-\sqrt5 \) is irrational.

Example 17. Prove that \( \sqrt{2}+\sqrt{5} \) is irrational.
Answer:
Assume \( \sqrt2+\sqrt5 = \frac{a}{b} \) with a, b co-prime positive integers. Then \( \sqrt5 = \frac{a}{b}-\sqrt2 \).
Square both sides: \( 5 = \frac{a^2}{b^2}-\frac{2a}{b}\sqrt2+2 \), which rearranges to \( \frac{a^2}{b^2}-3 = \frac{2a}{b}\sqrt2 \).
Solving for the root: \( \sqrt2 = \frac{a^2-3b^2}{2ab} \). The right side is a ratio of integers, so it is rational, which would make \( \sqrt{2} \) rational.
But \( \sqrt{2} \) is irrational (Example 14). The assumption fails, so \( \sqrt2+\sqrt5 \) is irrational.

Example 18. Without actually performing the long division, state whether \( \frac{13}{3125} \) has terminating decimal expansion or not.
Answer:
3125 = 5⁵, so \( \frac{13}{3125} = \frac{13}{2^0\times5^5} \).
The denominator has the form 2^m × 5ⁿ (with m = 0, n = 5), so the decimal expansion is terminating.

Example 19. What can you say about the prime factorization of the denominators of the following rationals :
(i) 43.123456789    (ii) \( 43.\overline{123456789} \)
Answer:
(i) This decimal terminates, so, when the number is written as a fraction in lowest terms, the denominator has the form 2^m × 5ⁿ with m, n non-negative integers.
(ii) This decimal is non-terminating and repeating, so the denominator must contain at least one prime other than 2 and 5.