CBSE Class 10 Mathematics Real Numbers Notes Set 09

Fundamental Theorem of Arithmetic

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

H.C.F. and L.C.M. by Prime Factorisation

H.C.F. and L.C.M. of numbers can be found by prime factorisation, which is an application of the fundamental theorem of arithmetic.

H.C.F. = product of the smallest power of each common prime factor.
L.C.M. = product of the greatest power of each prime factor involved in the numbers.

For two natural numbers a and b: \( \text{H.C.F.} \times \text{L.C.M.} = a \times b \).
Note: L.C.M. is always divisible by H.C.F.

Solved Examples on the Fundamental Theorem of Arithmetic

Example 10. Determine the prime factors of 45470971.
Answer:
Divide successively by prime factors: \( 45470971 = 7 \times 6495853 \), \( 6495853 = 7 \times 927979 \), \( 927979 = 13 \times 71383 \), \( 71383 = 13 \times 5491 \), \( 5491 = 17 \times 323 \), \( 323 = 17 \times 19 \).
Therefore \( 45470971 = 7^2 \times 13^2 \times 17^2 \times 19 \).

Example 11. Check whether \( 6^n \) can end with the digit 0 for any natural number n.
Answer:
Any positive integer ending with the digit zero is divisible by 5, so its prime factorisation must contain the prime 5.
\( 6^n = (2 \times 3)^n = 2^n \times 3^n \), so the only primes in the factorisation of \( 6^n \) are 2 and 3.
5 does not occur in the prime factorisation of \( 6^n \) for any n, so \( 6^n \) does not end with the digit zero for any natural number n.

Example 13. In a morning walk three persons step off together and 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 minimum distance is the L.C.M. of 80 cm, 85 cm and 90 cm.
\( 80 = 2^4 \times 5 \), \( 85 = 5 \times 17 \), \( 90 = 2 \times 3^2 \times 5 \).
L.C.M. = \( 2^4 \times 3^2 \times 5 \times 17 = 16 \times 9 \times 5 \times 17 = 12240 \) cm = 122 m 40 cm.

Check Your Progress: Fundamental Theorem of Arithmetic

Check Your Progress 1. Express the following numbers as a product of prime factors: (a) 1771 (b) 8232 (c) 10584.
Answer:
(a) \( 1771 = 7 \times 11 \times 23 \)
(b) \( 8232 = 2^3 \times 3 \times 7^3 \)
(c) \( 10584 = 2^3 \times 3^3 \times 7^2 \)

Check Your Progress 2. Find the H.C.F. and L.C.M. of the following pairs of numbers by prime factorisation and verify that L.C.M. × H.C.F. = product of the numbers: (a) 1080 and 252 (b) 252 and 294.
Answer:
(a) \( 1080 = 2^3 \times 3^3 \times 5 \), \( 252 = 2^2 \times 3^2 \times 7 \). H.C.F. = 36, L.C.M. = 7560. Check: \( 36 \times 7560 = 272160 = 1080 \times 252 \).
(b) \( 252 = 2^2 \times 3^2 \times 7 \), \( 294 = 2 \times 3 \times 7^2 \). H.C.F. = 42, L.C.M. = 1764. Check: \( 42 \times 1764 = 74088 = 252 \times 294 \).

Check Your Progress 3. If the L.C.M. of two numbers is 252, the H.C.F. is 2 and one of the numbers is 28, find the other number.
Answer:
Other number = \( \dfrac{252 \times 2}{28} = \) 18.

Check Your Progress 4. Check whether \( 4^n \) can end with the digit 0 for any natural number n.
Answer:
\( 4^n = 2^{2n} \) has only the prime factor 2, so 5 is not a factor and \( 4^n \) never ends with the digit 0.

Check Your Progress 5. Leena has music class on alternate days, dance class once in 3 days and yoga once in 5 days. On the 1st of January she had all the three classes. When will she have all three classes again?
Answer:
The classes repeat every 2, 3 and 5 days. L.C.M.(2, 3, 5) = 30, so the three classes fall together again after 30 days, that is, on 31st January.

Check Your Progress 6. There is an oval-shaped park with a pathway running round it. Babu and Raju start jogging at P at the same time and in the same direction. If Babu can complete a full round in 16 minutes while Raju can do it in 20 minutes, after how many minutes will they meet at P again?
Answer:
They meet again at P after L.C.M.(16, 20) = 80 minutes.

Check Your Progress 7. A rope of length 140 cm has to be cut into 2 pieces in the ratio 3 : 4. What is the maximum length of the measuring stick which can be used to measure both the lengths?
Answer:
The pieces are \( \tfrac{3}{7} \times 140 = 60 \) cm and \( \tfrac{4}{7} \times 140 = 80 \) cm. The longest stick that measures both exactly has length H.C.F.(60, 80) = 20 cm.

Proof of Irrationality and Decimal Representation

Some important results
1. Let p be a prime number and a a positive integer. If p divides \( a^2 \), then p divides a.
2. Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form \( \tfrac{p}{q} \), where p and q are co-prime and the prime factorisation of q is of the form \( 2^m \times 5^n \), where m and n are non-negative integers.
3. Let \( x = \tfrac{p}{q} \) be a rational number such that the prime factorisation of q is not of the form \( 2^m \times 5^n \), where m and n are non-negative integers. Then x has a decimal expansion which is non-terminating repeating.

Solved Examples on Irrational Numbers

Example 14. Prove that \( \sqrt{2} \) is an irrational number.
Answer:
Let us assume, on the contrary, that \( \sqrt{2} \) is rational. Then there exist co-prime positive integers a and b such that \( \sqrt{2} = \dfrac{a}{b} \).
Squaring: \( 2 = \dfrac{a^2}{b^2} \), so \( a^2 = 2b^2 \). Hence \( a^2 \) is a multiple of 2, so a is a multiple of 2 ... (i)
Let a = 2c for some integer c. Then \( 4c^2 = 2b^2 \), so \( b^2 = 2c^2 \). Hence \( b^2 \) is a multiple of 2, so b is a multiple of 2 ... (ii)
From (i) and (ii), a and b have at least 2 as a common factor. This contradicts the fact that a and b are co-prime. Hence \( \sqrt{2} \) is an irrational number.

Example 15. Prove that \( 5\sqrt{3} \) is not rational.
Answer:
If possible, let \( 5\sqrt{3} \) be rational. Then \( 5\sqrt{3} = \dfrac{a}{b} \), where a and b are co-prime integers, so \( \sqrt{3} = \dfrac{a}{5b} \).
This means \( \sqrt{3} \), which is irrational, is equal to the rational number \( \dfrac{a}{5b} \), which is a contradiction. Hence \( 5\sqrt{3} \) cannot be rational.

Example 16. Prove that \( 3 - \sqrt{5} \) is an irrational number.
Answer:
Assume, on the contrary, that \( 3 - \sqrt{5} \) is rational. Then there exist co-prime positive integers a and b such that \( 3 - \sqrt{5} = \dfrac{a}{b} \).
So \( \sqrt{5} = 3 - \dfrac{a}{b} = \dfrac{3b - a}{b} \), which is rational because a and b are integers.
This contradicts the fact that \( \sqrt{5} \) is irrational. Hence \( 3 - \sqrt{5} \) is an irrational number.

Example 17. Prove that \( \sqrt{2} + \sqrt{5} \) is irrational.
Answer:
Assume, on the contrary, that \( \sqrt{2} + \sqrt{5} = \dfrac{a}{b} \) is rational, with a and b co-prime positive integers. Then \( \sqrt{5} = \dfrac{a}{b} - \sqrt{2} \).
Squaring both sides: \( 5 = \dfrac{a^2}{b^2} - \dfrac{2a}{b}\sqrt{2} + 2 \), so \( \dfrac{a^2}{b^2} - 3 = \dfrac{2a}{b}\sqrt{2} \).
Hence \( \sqrt{2} = \dfrac{a^2 - 3b^2}{2ab} \), which is a rational number. This contradicts the fact that \( \sqrt{2} \) is irrational.
So our assumption is wrong. Hence \( \sqrt{2} + \sqrt{5} \) is irrational.

Example 18. Without actually performing the long division, state whether \( \dfrac{13}{3125} \) has a terminating decimal expansion or not.
Answer:
\( \dfrac{13}{3125} = \dfrac{13}{5^5} = \dfrac{13}{2^0 \times 5^5} \).
The prime factorisation of the denominator is of the form \( 2^m \times 5^n \), so the decimal expansion is terminating.

Example 19. What can you say about the prime factorisations of the denominators of the following rationals: (i) 43.123456789 (ii) \( 43.\overline{123456789} \)?
Answer:
(i) 43.123456789 has a terminating decimal expansion, so the prime factorisation of the denominator is of the form \( 2^m \times 5^n \), where m and n are non-negative integers.
(ii) \( 43.\overline{123456789} \) has a non-terminating repeating decimal expansion, so the denominator has prime factors other than 2 and 5.

Check Your Progress: Irrational Numbers and Decimal Expansions

Check Your Progress 1. Show that \( \sqrt{11} \) is an irrational number.
Answer:
11 is prime. If \( \sqrt{11} = \tfrac{a}{b} \) with a, b co-prime, then \( a^2 = 11b^2 \), so 11 divides a; put a = 11c to get \( b^2 = 11c^2 \), so 11 divides b. This contradicts co-primality, so \( \sqrt{11} \) is irrational.

Check Your Progress 2. Show that \( 4 - \sqrt{3} \) is an irrational number.
Answer:
If \( 4 - \sqrt{3} = \tfrac{a}{b} \) is rational, then \( \sqrt{3} = \tfrac{4b - a}{b} \) is rational, contradicting the irrationality of \( \sqrt{3} \).

Check Your Progress 3. Show that \( 2\sqrt{5} \) is an irrational number.
Answer:
If \( 2\sqrt{5} = \tfrac{a}{b} \) is rational, then \( \sqrt{5} = \tfrac{a}{2b} \) is rational, a contradiction.

Check Your Progress 4. Show that \( \sqrt{8} \) is an irrational number.
Answer:
\( \sqrt{8} = 2\sqrt{2} \). If it were rational, \( \sqrt{2} = \tfrac{\sqrt{8}}{2} \) would be rational, but \( \sqrt{2} \) is irrational.

Check Your Progress 5. Show that \( \sqrt{5} - \sqrt{3} \) is an irrational number.
Answer:
Suppose \( \sqrt{5} - \sqrt{3} = r \) is rational. Squaring, \( 8 - 2\sqrt{15} = r^2 \), so \( \sqrt{15} = \tfrac{8 - r^2}{2} \) is rational. But 15 is not a perfect square, so \( \sqrt{15} \) is irrational, a contradiction.

Check Your Progress 6. Without actually performing the long division, state whether \( \dfrac{343}{875} \) has a terminating decimal expansion or not.
Answer:
\( \dfrac{343}{875} = \dfrac{7^3}{5^3 \times 7} = \dfrac{49}{5^3} \). The denominator is of the form \( 2^0 \times 5^3 \), so the expansion is terminating.