CBSE Class 10 Mathematics Real Numbers Notes Set 03

Euclid's Division Lemma

Dividend = Divisor × Quotient + Remainder. Can you divide 14 by 6? After division we get 2 as the quotient and 2 as the remainder, so 14 = 6 × 2 + 2. For other sets of dividends and divisors:
(1) Divide 100 by 20: 100 = 20 × 5 + 0
(2) Divide 117 by 15: 117 = 15 × 7 + 12
(3) Divide 67 by 17: 67 = 17 × 3 + 16

If we have a dividend and a divisor, there is a unique pair of a quotient and a remainder that fits the equation above. This is Euclid's division lemma.

This lemma is very useful for finding the H.C.F. of large numbers where breaking them into factors is difficult. The method is called Euclid's division algorithm. Note that Euclid's division algorithm can be applied to polynomials also.

Example 1: Find the H.C.F. of 4032 and 262 using Euclid's division algorithm.

Solution:
Step 1: Apply the lemma to 4032 and 262: 4032 = 262 × 15 + 102
Step 2: The remainder is non-zero, so apply it to 262 and 102: 262 = 102 × 2 + 58
Step 3: Apply it to 102 and 58: 102 = 58 × 1 + 44
Step 4: Apply it to 58 and 44: 58 = 44 × 1 + 14
Step 5: Apply it to 44 and 14: 44 = 14 × 3 + 2
Step 6: Apply it to 14 and 2: 14 = 2 × 7 + 0
The remainder is 0 when the divisor is 2. Hence, 2 is the H.C.F. of 4032 and 262.

Example 2: A rectangular garden of dimensions 190 m × 60 m is to be divided in square blocks to plant different flowers in each block. Into how many blocks can this garden be divided so that no land is wasted?

Solution:
To waste no land, we need the largest number that divides both 190 and 60, i.e. the H.C.F. of (190, 60).
190 = 60 × 3 + 10
60 = 10 × 6 + 0
So the H.C.F. of 190 and 60 is 10.
Number of square blocks along the length = \( \frac{190}{10} \) = 19 and along the breadth = \( \frac{60}{10} \) = 6.
Total number of blocks = 19 × 6 = 114.

Example 3: Find the H.C.F. of 336 and 90 using Euclid's division algorithm.

Solution:
As 336 > 90, we apply the division lemma to 336 and 90.
336 = 90 × 3 + 66
90 = 66 × 1 + 24
66 = 24 × 2 + 18
24 = 18 × 1 + 6
18 = 6 × 3 + 0
The remainder is zero, so the divisor 6 is the required H.C.F.

Example 4: Find the H.C.F. of 45, 81 and 117 using Euclid's division algorithm.

Solution:
Take any two of the numbers, say 45 and 81. As 81 > 45:
81 = 45 × 1 + 36
45 = 36 × 1 + 9
36 = 9 × 4 + 0
So the H.C.F. of 45 and 81 is 9.
Now apply the algorithm to this H.C.F. (9) and the remaining number 117:
117 = 9 × 13 + 0
So the H.C.F. of 9 and 117 is 9. Thus, the H.C.F. of 45, 81 and 117 is 9.

Example 5: In an inter-school essay writing competition, the numbers of participants from schools A, B and C are 20, 16 and 28 respectively. If the participants in each room are from the same school, find the minimum number of rooms required such that each room has the same number of participants.

Solution:
To have the minimum number of rooms, each room must hold the maximum possible number of participants, i.e. the largest number that divides 20, 16 and 28 (their H.C.F.).
20 = 16 × 1 + 4 and 16 = 4 × 4 + 0, so the H.C.F. of 20 and 16 is 4.
28 = 4 × 7 + 0, so the H.C.F. of 28 and 4 is 4.
H.C.F. of 20, 16, 28 = 4, so each room has 4 participants.
Rooms needed: \( \frac{20}{4} \) = 5 for school A, \( \frac{16}{4} \) = 4 for school B and \( \frac{28}{4} \) = 7 for school C.
Total rooms = 5 + 4 + 7 = 16.

Using Euclid's Division Lemma to Prove Results

Whenever we divide 32 by 5, we get 6 as the quotient and 2 as the remainder, so 32 = 6 × 5 + 2. This expression is unique: when we divide 54 by 11 we always get the quotient 4 and the remainder 10, never any other pair. Hence for positive integers a and b there exist unique integers q and r with a = bq + r, 0 ≤ r < b. One application of this lemma is to prove relationships among numbers.

Example 1: Prove that every positive integer is of the form 3p, 3p + 1 or 3p + 2, where p is an integer.

Solution:
Let a be any positive integer and b = 3. By Euclid's division lemma, a = 3p + r for some integer p and 0 ≤ r < 3.
So a can be 3p, 3p + 1 or 3p + 2. Hence every positive integer is of one of these forms.

Example 2: Prove that every positive even integer is of the form 2m and every positive odd integer is of the form 2m + 1, where m is an integer.

Solution:
Let a be any positive integer and b = 2. Then a = 2m + r, where 0 ≤ r < 2, so r = 0 or 1.
If r = 0, a = 2m is divisible by 2, so it is even.
If r = 1, a = 2m + 1 is not divisible by 2, so it is odd.
Hence every positive even integer is of the form 2m and every positive odd integer is of the form 2m + 1.

Example 3: Prove that the expression y(y + 1) always represents an even number, where y is a positive integer.

Solution:
y is either odd or even.
When y is odd, y = 2p + 1 for an integer p: y(y + 1) = (2p + 1)(2p + 2) = 2(2p + 1)(p + 1), a multiple of 2.
When y is even, y = 2q for an integer q: y(y + 1) = 2q(2q + 1), a multiple of 2.
In both cases y(y + 1) is even.

Example 4: Show that (a) the sum, the difference and the product of two even numbers are always even; (b) the sum and the difference of two odd numbers are always even, whereas their product is always odd.

Solution:
(a) Let x = 2m and y = 2n, where m and n are integers.
x + y = 2(m + n) = 2p, where p = m + n is an integer, so x + y is even.
x − y = 2(m − n) = 2q, where q = m − n is an integer, so x − y is even.
xy = 4mn = 2(2mn) = 2t, where t = 2mn is an integer, so xy is even.
(b) Let x = 2m + 1 and y = 2n + 1, where m and n are integers.
x + y = 2(m + n + 1) = 2p, where p = m + n + 1, so x + y is even.
x − y = 2(m − n) = 2q, where q = m − n, so x − y is even.
xy = (2m + 1)(2n + 1) = 2(2mn + m + n) + 1 = 2t + 1, where t = 2mn + m + n, so xy is odd.

Fundamental Theorem of Arithmetic and Prime Factorisation

All composite numbers can be written as a product of two or more prime numbers. Whichever way we factorise 36, we end up with the same prime factors; only the order differs. For a larger number, 21560 can be uniquely broken down as \( 2^3 \times 5 \times 7^2 \times 11 \). The product 2 × 3 × 7 × 11 is the same as 7 × 11 × 2 × 3.

Thus, this theorem can be used to write the prime factorisation of any number.

Example 1: Write the prime factorisation of 31250. What are its prime factors?

Solution:
31250 = 2 × 15625 = 2 × 5 × 3125 = 2 × 5 × 5 × 625 = 2 × 5 × 5 × 5 × 125 = 2 × 5 × 5 × 5 × 5 × 25 = 2 × 5 × 5 × 5 × 5 × 5 × 5 = \( 2 \times 5^6 \)
Hence \( 2 \times 5^6 \) is the prime factorisation of 31250, and its prime factors are 2 and 5.

Example 2: If 13125 = \( 2^a \times 3^b \times 5^c \times 7^d \), find the value of a + 2b + 7c + 11d.

Solution:
13125 = 3 × 4375 = 3 × 5 × 875 = 3 × 5 × 5 × 175 = 3 × 5 × 5 × 5 × 35 = 3 × 5 × 5 × 5 × 5 × 7 = \( 3^1 \times 5^4 \times 7^1 \)
So \( 2^a \times 3^b \times 5^c \times 7^d = 2^0 \times 3^1 \times 5^4 \times 7^1 \).
Comparing exponents: a = 0, b = 1, c = 4, d = 1.
a + 2b + 7c + 11d = 0 + 2 × 1 + 7 × 4 + 11 × 1 = 0 + 2 + 28 + 11 = 41.

Example 3: Show that the following expressions are composite numbers: (a) 3 × 5 × 7 × 23 + 2 × 7 × 11 × 13 (b) 29 × 35 + 14 (c) \( 3^4 + 6^3 \).

Solution:
(a) 3 × 5 × 7 × 23 + 2 × 7 × 11 × 13 = 7(3 × 5 × 23 + 2 × 11 × 13) = 7(345 + 286) = 7 × 631.
Both 7 and 631 are primes, so the number is a product of two primes and hence composite.
(b) 29 × 35 + 14 = 29 × 5 × 7 + 2 × 7 = 7(29 × 5 + 2) = 7 × 147 = 7 × 3 × 7 × 7 = \( 3 \times 7^3 \).
It is a product of primes, so it is a composite number.
(c) \( 3^4 + 6^3 = 3^4 + (2 \times 3)^3 = 3^4 + 2^3 \times 3^3 = 3^3(3 + 2^3) = 3^3 \times 11 \).
It is a product of primes, so it is a composite number.

HCF and LCM by the Prime Factorisation Method

For example, 20 = \( 2^2 \times 5 \) and 54 = \( 2 \times 3^3 \). Apart from the order, every number can be prime factorised in only one way. The prime factorisation of 980 is 980 = 2 × 490 = 2 × 2 × 245 = 2 × 2 × 5 × 49 = 2 × 2 × 5 × 7 × 7 = \( 2^2 \times 5^1 \times 7^2 \). So \( 2^2 \times 5^1 \times 7^2 \) is the prime factorisation of 980, and 2, 5 and 7 are its prime factors.

Example 1: Find the LCM and the HCF of 432 and 676 using the prime factorisation method.

Solution:
432 = \( 2^4 \times 3^3 \) and 676 = \( 2^2 \times 13^2 \).
HCF: the only common prime factor is 2 and its smallest power is 2. HCF (432, 676) = \( 2^2 \) = 4.
LCM: the prime factors are 2, 3 and 13 with greatest powers 4, 3 and 2. LCM (432, 676) = \( 2^4 \times 3^3 \times 13^2 \) = 73008.

Example 2: Find the HCF and the LCM of 28, 42 and 64 using the prime factorisation method.

Solution:
28 = \( 2^2 \times 7^1 \), 42 = \( 2 \times 3^1 \times 7^1 \), 64 = \( 2^6 \).
The only common prime factor is 2 and its smallest power is 1. HCF (28, 42, 64) = \( 2^1 \) = 2.
LCM (28, 42, 64) = \( 2^6 \times 3^1 \times 7^1 \) = 1344.

Example 3: Find the HCF and the LCM of 1080 and 900 using prime factorisation and show that HCF × LCM = product of the two numbers.

Solution:
1080 = \( 2^3 \times 3^3 \times 5 \) and 900 = \( 2^2 \times 3^2 \times 5^2 \).
HCF (1080, 900) = \( 2^2 \times 3^2 \times 5 \) = 180.
LCM (1080, 900) = \( 2^3 \times 3^3 \times 5^2 \) = 5400.
HCF × LCM = 180 × 5400 = 972000 and the product of the numbers = 1080 × 900 = 972000.
Hence HCF × LCM = product of the two numbers.

Example 4: The HCF of 273 and another number is 7, while their LCM is 3003. Find the other number.

Solution:
Let the first number a = 273 and the second number be b. HCF (a, b) = 7 and LCM (a, b) = 3003.
HCF × LCM = product of the two numbers, so 7 × 3003 = 273 × b, which gives b = 77.
Hence the other number is 77.

Example 5: Anurag takes 6 minutes to complete one round of jogging around the circular track of a park, while Twinkle takes 8 minutes. If both start jogging at the same time from the same point, after how much time will they meet at the starting point?

Solution:
The time after which they meet at the starting point is the lowest common multiple of 6 and 8.
6 = 2 × 3 and 8 = \( 2^3 \), so LCM (6, 8) = \( 2^3 \times 3 \) = 24.
They will meet at the starting point after 24 minutes.

Example 6: There are 120 students in a class. Arranged by roll numbers, every second student got distinction in Mathematics, every third in Science and every fifth in English. How many students got distinction in all three subjects?

Solution:
The roll numbers of students with distinction in all three subjects are the multiples of the LCM of 2, 3 and 5.
LCM (2, 3, 5) = 2 × 3 × 5 = 30, so every 30th student got distinction in all three subjects.
Number of such students = \( \frac{120}{30} \) = 4.

Properties of Prime Numbers

Consider the number \( 8^n \), where n is a natural number. Is there any value of n for which \( 8^n \) ends with zero? This is difficult to answer directly, but the fundamental theorem of arithmetic helps. It states that every composite number can be factorised as a product of primes and this factorisation is unique, apart from the order of the prime factors. For example, 255 = 3 × 5 × 17, and it can also be written as 3 × 17 × 5, 5 × 3 × 17, 5 × 17 × 3, 17 × 3 × 5 or 17 × 5 × 3. So 255 is a product of the unique primes 3, 5 and 17, only the order may differ.

Suppose \( 8^n \) ends with zero for some n. Then it is divisible by 10 = 2 × 5, so it must be divisible by both 2 and 5, and its prime factorisation must contain both 2 and 5. But \( 8^n = (2^3)^n = 2^{3n} \), whose only prime factor is 2. By the fundamental theorem of arithmetic there is no other prime in the factorisation of \( 8^n \). Hence there is no natural number n for which \( 8^n \) ends with the digit zero.

Example 1: Prove that the number \( 9^n \), where n is a natural number, cannot end with a zero.

Solution:
Suppose \( 9^n \) ends with a zero for some n. Then it is divisible by 10 = 2 × 5, so its prime factorisation must contain both 2 and 5.
But \( 9^n = (3^2)^n = 3^{2n} \), so the only prime in its factorisation is 3. By the fundamental theorem of arithmetic there is no other prime.
Hence there is no natural number n for which \( 9^n \) ends with the digit zero.

Example 2: Check whether the numbers \( 49^n \), where n is a natural number, can end with a zero.

Solution:
Suppose \( 49^n \) ends with a zero for some n. Then it is divisible by 10 = 2 × 5, so its prime factorisation must contain both 2 and 5.
But \( 49^n = (7^2)^n = 7^{2n} \), so the only prime in its factorisation is 7. By the fundamental theorem of arithmetic there is no other prime.
Hence there is no natural number n for which \( 49^n \) ends with the digit zero.

Irrational Numbers

A number which cannot be written in the form \( \frac{p}{q} \), where p and q are integers and q ≠ 0, is called an irrational number. For example, all numbers of the form \( \sqrt{p} \), where p is a prime number, such as \( \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7} \), are irrational.

Proof of the theorem: Let a = \( p_1 \times p_2 \times \ldots \times p_n \), where \( p_1, p_2, \ldots, p_n \) are the prime factors of a. Squaring, \( a^2 = (p_1 \times p_2 \times \ldots \times p_n)^2 = p_1^2 \times p_2^2 \times \ldots \times p_n^2 \). Since p divides \( a^2 \), p is a factor of \( p_1^2 \times p_2^2 \times \ldots \times p_n^2 \). By the fundamental theorem of arithmetic, the primes in the factorisation of \( a^2 \) are unique, so p is one of the primes \( p_1, p_2, \ldots, p_n \). If p = \( p_k \) for some k from 1 to n, then \( p_k \) divides \( p_1 \times p_2 \times \ldots \times p_n \), that is, p divides a.

Proof that \( \sqrt{2} \) is irrational: Assume \( \sqrt{2} \) is rational. Then \( \sqrt{2} = \frac{a}{b} \), where a and b are co-prime integers and b ≠ 0. Squaring, \( 2b^2 = a^2 \), so 2 divides \( a^2 \) and hence 2 divides a. Let a = 2c. Then \( 2b^2 = 4c^2 \), so \( b^2 = 2c^2 \), and 2 divides b. So 2 divides both a and b, which contradicts the fact that a and b are co-prime. Hence \( \sqrt{2} \) is irrational. In the same way we can prove that the square roots of other primes such as \( \sqrt{3}, \sqrt{5}, \sqrt{11} \) are irrational.

Besides these, there are other irrational numbers like \( 9\sqrt{2}, 5\sqrt{7} \). Before proving that such numbers are irrational, let us see what happens when we apply mathematical operations to irrational numbers.

\( 9\sqrt{2} \) is irrational: We know that \( \sqrt{2} \) is irrational (proved above). The product of a non-zero rational number and an irrational number is irrational. Therefore \( 9\sqrt{2} \) is an irrational number.

Example 1: Prove that \( 5 + 2\sqrt{7} \) is irrational.

Solution:
Let us assume that \( 5 + 2\sqrt{7} \) is rational. Then \( 5 + 2\sqrt{7} = \frac{a}{b} \), where a and b are integers and b ≠ 0. Dividing by any common factor, we obtain \( 5 + 2\sqrt{7} = \frac{c}{d} \), where c and d are co-prime.
\( \Rightarrow 5 - \frac{c}{d} = -2\sqrt{7} \)
\( \Rightarrow 2\sqrt{7} = \frac{c}{d} - 5 \)
\( \Rightarrow \sqrt{7} = \frac{c}{2d} - \frac{5}{2} \)
As c, d and 2 are integers, \( \frac{c}{2d} \) and \( \frac{5}{2} \) are rational numbers, so \( \frac{c}{2d} - \frac{5}{2} \) is rational. So \( \sqrt{7} \) is rational, which contradicts the fact that \( \sqrt{7} \) is irrational.
Therefore our assumption is wrong. Hence \( 5 + 2\sqrt{7} \) is irrational.

Example 2: Prove that \( 3 - \sqrt{5} \) is irrational.

Solution:
Let us assume that \( 3 - \sqrt{5} \) is rational. Then \( 3 - \sqrt{5} = \frac{a}{b} \), where a and b are co-prime and b ≠ 0.
\( \Rightarrow \sqrt{5} = 3 - \frac{a}{b} \)
As a and b are integers, \( \frac{a}{b} \) is rational, so \( 3 - \frac{a}{b} \) is rational. This means \( \sqrt{5} \) is rational, which contradicts the fact that \( \sqrt{5} \) is irrational.
Therefore our assumption is wrong. Hence \( 3 - \sqrt{5} \) is irrational.

Decimal Expansions of Rational Numbers

A carpenter wants to mark a point at 95 mm from the end of a wooden plank, but he has a centimetre tape. He converts 95 mm into 9.5 cm and measures that length. Situations like this need numbers to be converted into decimals. Here we convert rational numbers into decimals, observe the types of decimal expansions and solve a few examples.

Know more: Two rational numbers \( \frac{a}{b} \) and \( \frac{c}{d} \) are equal if and only if ad = bc. For example, for \( \frac{2}{4} \) and \( \frac{3}{6} \): a = 2, b = 4, c = 3, d = 6, so ad = 2 × 6 = 12 and bc = 4 × 3 = 12. Since ad = bc, \( \frac{2}{4} = \frac{3}{6} \).

The form \( \frac{p}{q} \) represents the division of the integer p by the integer q. By solving this division we find the decimal equivalent of the rational number \( \frac{p}{q} \). Let us convert \( \frac{5}{8} \), \( \frac{4}{3} \) and \( \frac{2}{7} \) into decimals by long division.

      0.625              1.33...            0.285714...
    -------            -------            ----------
 8 ) 5.000          3 ) 4.00           7 ) 2.000000
     48                 3                  14
     --                 -                  --
      20               10                   60
      16                9                   56
      --               --                   --
       40               10                   40
       40                9                   35
       --               --                   --
        0                1                    50
                                              49
                                              --
                                              10
                                               7
                                              --
                                              30
                                              28
                                              --
                                               2

While the remainder is zero in the division of 5 by 8, it is not so in the other two divisions. Thus we get two different cases for the decimal expansion of a rational number.

Case I: When the remainder is zero. The quotient (decimal expansion) terminates after a finite number of digits after the decimal point. For example, in the decimal expansion of \( \frac{5}{8} \) the remainder is zero and the quotient is 0.625.

Case II: When the remainder is never zero. The remainder never becomes zero and the decimal expansion is non-terminating. For example, for \( \frac{4}{3} \) and \( \frac{2}{7} \) the quotients are 1.3333… and 0.285714285714…, which are non-terminating decimals. In these, the digit 3 and the group of digits 285714 get repeated. Therefore we write \( \frac{4}{3} = 1.3333\ldots = 1.\overline{3} \) and \( \frac{2}{7} = 0.285714285714\ldots = 0.\overline{285714} \). The bar over the digits indicates the digit or group of digits that gets repeated.

Example 1: Write the decimal expansion of \( \frac{1237}{25} \) and find if it is terminating or non-terminating and repeating.

Solution:
Long division of \( \frac{1237}{25} \):

        49.48
      -------
 25 ) 1237.00
      100
      ---
       237
       225
       ---
        120
        100
        ---
        200
        200
        ---
          0

Hence the decimal expansion of \( \frac{1237}{25} \) is 49.48. Since the remainder is zero, the decimal number is terminating.

Example 2: Write the decimal expansion of \( \frac{2358}{27} \) and find if it is terminating or non-terminating and repeating.

Solution:
Long division of \( \frac{2358}{27} \):

        87.33...
      ----------
 27 ) 2358.00
      216
      ---
      198
      189
      ---
       90
       81
       --
       90
       81
       --
        9...

Hence the decimal expansion of \( \frac{2358}{27} \) is 87.33…. Since the remainder 9 is obtained again and again, the decimal number is non-terminating and repeating. It can also be written as \( 87.\overline{3} \).

Example 3: Find the decimal expansion of each of the following rational numbers and write the nature of the same: (1) \( \frac{65}{101} \) (2) \( \frac{923}{400} \) (3) \( \frac{37}{99} \) (4) \( \frac{67}{100} \).

Solution:
(i) Dividing 65 by 101 gives 0.643564… The remainders repeat, so \( \frac{65}{101} = 0.64356435\ldots = 0.\overline{6435} \). The group of digits 6435 repeats after the decimal point. Hence the decimal expansion is non-terminating and repeating.
(ii) Dividing 923 by 400 gives 2.3075 and the remainder becomes zero. So \( \frac{923}{400} = 2.3075 \). Hence the given rational number has a terminating decimal expansion.
(iii) Dividing 37 by 99 gives 0.3737… So \( \frac{37}{99} = 0.3737\ldots = 0.\overline{37} \). The pair of digits 37 repeats after the decimal point. Hence the decimal expansion is non-terminating and repeating.
(iv) Dividing 67 by 100 gives 0.67 and the remainder becomes zero. So \( \frac{67}{100} = 0.67 \). Hence the given rational number has a terminating decimal expansion.

Terminating and Non-terminating Repeating Decimal Expansions

We can find the decimal expansion of a rational number by long division. However, it is possible to check whether the decimal expansion is terminating or non-terminating without actually carrying out the long division. Let us start by taking a few terminating decimals and writing them as rational numbers.

(a) \( 0.5632 = \frac{5632}{10000} \). On prime factorising the numerator and the denominator, \( \frac{5632}{10000} = \frac{2^9 \times 11}{2^4 \times 5^4} = \frac{2^5 \times 11}{5^4} \).
(b) \( 0.275 = \frac{275}{1000} \). On prime factorising, \( \frac{275}{1000} = \frac{5^2 \times 11}{2^3 \times 5^3} = \frac{11}{2^3 \times 5} \).

Can you see a pattern in the two examples? Both are rational numbers with terminating decimal expansions. When they are written in the \( \frac{p}{q} \) form, where p and q are co-prime (the HCF of p and q is 1), the denominator, written in terms of prime factors, has 2 or 5 or both.

Let us see a few examples that verify this theorem.

(a) \( \frac{7}{12} = \frac{7}{2^2 \times 3} = 0.58333\ldots \)
(b) \( \frac{15}{16} = \frac{3 \times 5}{2^4} = \frac{3 \times 5 \times 5^4}{2^4 \times 5^4} = \frac{3 \times 5^5}{(10)^4} = 0.9375 \)
(c) \( \frac{1}{14} = \frac{1}{7 \times 2} = 0.0714285714\ldots \)
(d) \( \frac{125}{16} = \frac{5^3}{2^4} = \frac{5^3 \times 5^4}{2^4 \times 5^4} = \frac{5^7}{(10)^4} = \frac{78125}{10^4} = 7.8125 \)

In examples (b) and (d), the denominators are composed only of the prime factors 2 and 5, so the decimal expansion is terminating. In examples (a) and (c), each denominator has at least one prime factor other than 2 and 5, so the decimal expansion is non-terminating and repeating.

Example 1: Without carrying out the actual division, find if the following rational numbers have a terminating or a non-terminating decimal expansion: (a) \( \frac{17}{1600} \) (b) \( \frac{723}{392} \).

Solution:
(a) \( \frac{17}{1600} = \frac{17}{2^6 \times 5^2} \). The denominator is of the form \( 2^n 5^m \) with n = 6 and m = 2, which are non-negative integers. So the rational number has a terminating decimal expansion.
(b) \( \frac{723}{392} = \frac{3 \times 241}{2^3 \times 7^2} \). The denominator cannot be written in the form \( 2^n 5^m \). So the rational number has a non-terminating (repeating) decimal expansion.

Example 2: Without carrying out the actual division, find if the expression \( \frac{715}{128} \) has a terminating or a non-terminating decimal expansion. If it terminates, find the decimal expansion.

Solution:
\( \frac{715}{128} = \frac{715}{2^7} \). The denominator is of the form \( 2^n 5^m \) with n = 7 and m = 0, so the rational number has a terminating decimal expansion.
\( \frac{715}{128} = \frac{715}{2^7} = \frac{715 \times 5^7}{2^7 \times 5^7} = \frac{715 \times 5^7}{10^7} = \frac{715 \times 78125}{10^7} = \frac{55859375}{10^7} = 5.5859375 \)
Hence 5.5859375 is the decimal expansion of the given rational number.