GSEB Class 9 Maths Solutions Chapter 1 Number Systems Exercise 1.3

Question 1. Write the following in decimal form and say what kind of decimal expansion each has
(i) \( \frac { 36 }{ 100 } \)
(ii) \( \frac {1}{11} \)
(iii) \( 4\frac {1}{8} \)
(iv) \( \frac { 3 }{ 13 } \)
(v) \( \frac { 2 }{11} \)
(vi) \( \frac { 329 }{ 400 } \)
Answer:
(i) We have
\( \frac { 36 }{ 100 } = 0.36 \)
The decimal expansion is terminating, which means it ends after a specific number of digits.
(ii) We have
11) 1.00 (0.090909... -0.99 ----- 0.0100 -0.0099 ------ 0.0001
\( \frac { 1 }{ 11 } = 0.090909 ... = 0.\overline{09} \)
The decimal expansion is non-terminating repeating, showing a pattern that goes on forever.
(iii) We have \( 4\frac {1}{8} = \frac { 33 }{8} \)
8) 33 (4.125 -32 --- 10 -8 -- 20 -16 --- 40 -40 --- 0
\( 4\frac {1}{8} = 4.125 \)
The decimal expansion is terminating because it ends exactly at 4.125.
(iv) We have
13) 3.00 (0.230769... -2.6 ---- 40 -39 --- 100 -91 ---- 90 -78 --- 120 -117 ---- 3
\( \frac { 3 }{ 13 } = 0.230769230769... = 0.\overline{230769} \)
The decimal expansion is non-terminating repeating, as the sequence '230769' repeats endlessly.
(v) We have
11) 2.00 (0.1818... -1.1 ---- 90 -88 --- 20 -11 ---- 90 -88 ---- 2
\( \frac { 2 }{ 11 } = 0.1818 ... = 0.\overline{18} \)
The decimal expansion is non-terminating repeating, with '18' repeating over and over.
(vi) We have
400)329.00 (0.8225 -3200 ----- 900 -800 ----- 1000 -800 ----- 2000 -2000 ----- 0
\( \frac { 329 }{ 400 } = 0.8225 \)
The decimal expansion is terminating since it stops at 0.8225.
In simple words: When a fraction is written as a decimal, it either ends (terminating) or some digits repeat forever (non-terminating repeating). We perform division to see which type it is and identify the repeating part with a bar.

Exam Tip: To identify the type of decimal expansion, always perform the division. If the remainder becomes 0, it's terminating. If a remainder repeats, the decimal is non-terminating and repeating.

Question 2. You know that \( \frac {1}{7} = 0.\overline { 142857 } \). Can you predict what the decimal expansions of \( \frac { 2 }{ 7 } \), \( \frac { 3 }{ 7 } \), \( \frac { 4 }{ 7 } \), \( \frac { 5 }{ 7 } \), \( \frac { 6 }{ 7 } \) are, without actually doing the long division? If so, how?
(Hint. Study the remainders while finding the value of \( \frac {1}{7} \) carefully.)
Answer: Yes, we can predict the decimal expansions of \( \frac { 2 }{ 7 } \), \( \frac { 3 }{ 7 } \), \( \frac { 4 }{ 7 } \), \( \frac { 5 }{ 7 } \), \( \frac {6}{7} \) without actually doing the long division.
We know \( \frac { 1 }{ 7 } \):
7) 1.000000 (0.1428571... -0.7 ---- 30 -28 ---- 20 -14 ---- 60 -56 ---- 40 -35 ---- 50 -49 ---- 10 -7 -- 3
\( \frac {1}{7} = 0.142857142857... \)
So, \( \frac { 1 }{7} = 0.\overline { 142857 } \)
Now we can find the others by multiplying:
\( \frac { 2 }{7} = 2 \times \frac {1}{7} = 2 \times 0.\overline { 142857 } \)
\( \implies \frac { 2 }{7} = 0.\overline { 285714 } \)
Similarly,
\( \frac { 3 }{ 7 } = 3 \times \frac { 1 }{ 7 } = 3 \times 0.\overline { 142857 } \)
\( \implies \frac { 3 }{ 7 } = 0.\overline { 428571 } \)
\( \frac { 4 }{7} = 4 \times \frac {1}{ 7 } = 4 \times 0.\overline { 142857 } \)
\( \implies \frac { 4 }{ 7 } = 0.\overline { 571428 } \)
\( \frac {5}{7} = 5 \times \frac { 1 }{ 7 } = 5 \times 0.\overline { 142857 } \)
\( \implies \frac { 5}{ 7 } = 0.\overline { 714285 } \)
\( \frac { 6 }{ 7 } = 6 \times \frac { 1 }{ 7 } = 6 \times 0.\overline { 142857 } \)
\( \implies \frac { 6}{ 7 } = 0.\overline { 857142 } \)
In simple words: Since we know the repeating decimal for \( \frac{1}{7} \), we can find the decimals for \( \frac{2}{7}, \frac{3}{7} \) and so on by simply multiplying the decimal of \( \frac{1}{7} \) by 2, 3, or whatever the numerator is. The repeating pattern stays the same, just the starting point shifts.

Exam Tip: For fractions with the same denominator, if you know the repeating block of one, you can often find the others by multiplying the known decimal and observing the cycle shift.

Question 3. Express the following in the form \( \frac { p }{ q } \), where p and q are integers and q \( \neq \) 0.
(i) \( 0.\overline {6} \)
(ii) \( 0.4\overline {7} \)
(iii) \( 0.\overline { 001 } \)
Answer:
(i) \( 0.\overline {6} \)
Let \( x = 0.\overline{6} \)
\( x = 0.66666... \) ...(1)
Multiply equation (1) by 10 (because one digit is repeating):
\( 10x = 6.66666... \) ...(2)
Subtract equation (1) from equation (2):
\( 10x - x = 6.66666... - 0.66666... \)
\( 9x = 6 \)
\( \implies x = \frac{6}{9} \)
\( \implies x = \frac{2}{3} \)
So, \( 0.\overline{6} \) can be expressed as \( \frac{2}{3} \).
(ii) \( 0.4\overline {7} \)
Let \( x = 0.4\overline{7} \)
\( x = 0.477777... \) ...(1)
Multiply equation (1) by 10 (to move the non-repeating part to the left of the decimal):
\( 10x = 4.77777... \) ...(2)
Now, multiply equation (1) by 100 (to move the repeating and non-repeating parts past the decimal):
\( 100x = 47.77777... \) ...(3)
Subtract equation (2) from equation (3):
\( 100x - 10x = 47.77777... - 4.77777... \)
\( 90x = 43 \)
\( \implies x = \frac{43}{90} \)
So, \( 0.4\overline{7} \) can be expressed as \( \frac{43}{90} \).
(iii) \( 0.\overline { 001 } \)
Let \( x = 0.\overline{001} \)
\( x = 0.001001001... \) ...(1)
Multiply equation (1) by 1000 (because three digits are repeating):
\( 1000x = 1.001001001... \) ...(2)
Subtract equation (1) from equation (2):
\( 1000x - x = 1.001001001... - 0.001001001... \)
\( 999x = 1 \)
\( \implies x = \frac{1}{999} \)
So, \( 0.\overline{001} \) can be expressed as \( \frac{1}{999} \).
In simple words: To convert a repeating decimal into a fraction, we set the decimal equal to 'x'. Then, we multiply 'x' by powers of 10 to shift the decimal point so that the repeating part aligns. Subtracting these equations helps us get rid of the repeating decimals and solve for 'x' as a simple fraction.

Exam Tip: The key to converting repeating decimals to fractions is to choose the correct powers of 10 for multiplication, ensuring that when you subtract, the repeating part cancels out completely.

Question 4. Express 0.99999... in the form of \( \frac { p }{ q } \). Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Answer: Let \( x = 0.99999... \) ...(1)
Multiply both sides by 10:
\( 10x = 9.99999... \) ...(2)
Subtract equation (1) from equation (2):
\( 10x - x = 9.99999... - 0.99999... \)
\( 9x = 9 \)
\( \implies x = \frac{9}{9} \)
\( \implies x = 1 \)
Thus, \( 0.99999... = 1 = \frac { 1 }{ 1 } \)
Here, we get \( p = 1 \) and \( q = 1 \).
Yes, this answer might seem surprising at first. The reason it makes sense is that \( 0.99999... \) represents a number that is infinitely close to 1, with no measurable gap between them. As the number of nines increases, the value gets closer and closer to 1, effectively becoming 1 at infinity. Therefore, \( 0.99999... \) and 1 represent the exact same point on the number line.
In simple words: We can show that \( 0.99999... \) is actually the same as the number 1. Even though it looks like it's just under 1, the endless nines make it equal to 1. This is a bit surprising, but mathematically, it's correct because there's no space between them.

Exam Tip: This question often confuses students. Remember that \( 0.999... \) is a special case where an infinitely repeating decimal precisely equals the next whole number due to the concept of limits.

Question 5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of \( \frac {1}{ 17 } \)? Perform the division to check your answer.
Answer: Long division method for \( \frac {1}{17} \):
17) 1.0000000000000000 (0.0588235294117647... -0.85 ----- 150 -136 ----- 140 -136 ----- 40 -34 ---- 60 -51 ---- 90 -85 ---- 50 -34 ---- 160 -153 ---- 70 -68 ---- 20 -17 ---- 30 -17 ---- 130 -119 ---- 110 -102 ---- 80 -68 ---- 120 -119 ---- 1
Thus, \( \frac { 1 }{ 17 } = 0.\overline { 0588235294117647 } \)
We observe that by the long division method, the maximum number of digits in the repeating block in the decimal expansion of \( \frac {1}{17} \) is 16. This verifies our answer.
In simple words: When we divide 1 by 17, the decimal part repeats after many digits. We found that 16 digits make up the repeating section. This happens because the remainder repeats during the division process.

Exam Tip: For a fraction \( \frac{1}{n} \), the maximum number of digits in the repeating block is at most \( n-1 \). This property is useful for predicting the length of the repetend without performing the full division.

Question 6. Look at several examples of rational numbers in the form \( \frac { p }{q} \) (q \( \neq \) 0) where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Answer: Let's look at a few examples:
(i) \( \frac {1}{2 } = \frac { 1 \times 5 }{ 2 \times 5 } = \frac { 5 }{ 10 } = 0.5 \)
(ii) \( \frac { 3 }{ 4 } = \frac {3 \times 5 \times 5 }{ 2 \times 2 \times 5 \times 5 } = \frac { 75 }{ 100 } = 0.75 \)
(iii) \( \frac {7}{8} = \frac { 7 \times 5 \times 5 \times 5 }{ 2 \times 2 \times 2 \times 5 \times 5 \times 5 } = \frac { 875 }{ 1000 } = 0.875 \)
(iv) \( \frac { 13 }{ 25 } = \frac { 13 \times 2 \times 2 }{ 5 \times 5 \times 2 \times 2 } = \frac{52}{5^{2} \times 2^{2}} = \frac{52}{(10)^{2}} = \frac { 52 }{ 100 } = 0.52 \)
(v) \( \frac { 3 }{ 125 } = \frac { 3 }{ 5 \times 5 \times 5 } = \frac{3}{5^{3}} = \frac{3 \times 2^{3}}{5^{3} \times 2^{3}} = \frac{3 \times 8}{(5 \times 2)^{3}} = \frac { 24 }{ 1000 } = 0.024 \)
(vi) \( \frac { 27 }{ 16 } = \frac{27 \times 5^{4}}{2^{4} \times 5^{4}} = \frac{27 \times 5^{4}}{(2 \times 5)^{4}} = \frac{27 \times 625}{(10)^{4}} = \frac{16875}{(10)^{4}} = 1.6875 \)
We observe that the denominator of all the above rational numbers are of the form \( 2^m \times 5^n \). This means the prime factorization of the denominators only has powers of 2, or powers of 5, or both. For a rational number \( \frac{p}{q} \) (where p and q are coprime) to have a terminating decimal expansion, the prime factorization of the denominator \( q \) must only contain powers of 2 and/or powers of 5.
In simple words: For a fraction to have a decimal that stops (terminating decimal), the bottom number (denominator) must only be made up of 2s and/or 5s when you break it down into its prime factors. If it has any other prime factors, the decimal will keep going forever.

Exam Tip: A rational number \( \frac{p}{q} \) (in simplest form) has a terminating decimal expansion if and only if the prime factorization of \( q \) is of the form \( 2^m \cdot 5^n \), where \( m \) and \( n \) are non-negative integers.

Question 7. Write three numbers whose decimal expansions are non-terminating non-recurring.
Answer: Three numbers whose decimal expansions are non-terminating and non-recurring (irrational numbers) are:
(i) \( 0.012012001200012... \)
(ii) \( 0.21021002100021000021... \)
(iii) \( 0.32032003200032000032... \)
These numbers show no repeating pattern and continue infinitely without stopping.
In simple words: These are numbers where the digits after the decimal point never end and never show a repeating pattern. Each new digit seems random, preventing any repeating block.

Exam Tip: Irrational numbers like \( \sqrt{2} \), \( \pi \), or carefully constructed non-repeating decimal sequences are good examples of non-terminating, non-recurring decimals. Make sure the pattern is truly non-repeating for full marks.

Question 8. Find three different irrational numbers between the rational numbers \( \frac {5}{7} \) and \( \frac { 9 }{ 11 } \).
Answer: First, let's find the decimal expansions of the given rational numbers:
For \( \frac {5}{7} \):
7) 5.000000 (0.714285... -4.9 ---- 10 -7 -- 30 -28 ---- 20 -14 ---- 60 -56 ---- 40 -35 ---- 5
So, \( \frac {5}{7} = 0.714285... \implies \frac {5}{7} = 0.\overline { 714285 } \).

For \( \frac {9}{11} \):
11) 9.00 (0.8181... -8.8 ---- 20 -11 ---- 90 -88 ---- 2
So, \( \frac { 9 }{ 11 } = 0.8181... \implies \frac { 9 }{ 11 } = 0.\overline { 81 } \).

We need to find three different irrational numbers between \( 0.714285... \) and \( 0.8181... \). Irrational numbers are non-terminating and non-recurring.
We can select numbers that start with digits between 7 and 8, and ensure they have no repeating pattern.
Three possible irrational numbers are:
1. \( 0.73073007300073000073... \)
2. \( 0.757075700757000757... \)
3. \( 0.808008000800008... \)
These numbers are all between \( 0.714285... \) and \( 0.8181... \) and do not have a repeating decimal pattern.
In simple words: First, we change the fractions into decimals to see their values. Then, we think of numbers that fall between those two decimals. To make them irrational, we create a decimal that never ends and never repeats, like a unique pattern of increasing zeros.

Exam Tip: To create an irrational number between two given numbers, ensure its decimal expansion lies within the range and has no repeating block or termination. A common way is to make an increasing sequence of zeros (e.g., \( 0.1010010001... \)).

Question 9. Classify the following numbers as rational or irrational.
(i) \( \sqrt{23} \)
(ii) \( \sqrt{225} \)
(iii) \( 0.3796 \)
(iv) \( 7.478478... \)
(v) \( 1.101001000100001... \)
Answer:
(i) \( \sqrt{23} \)
Since 23 is not a perfect square, \( \sqrt{23} \) will not give an integer value. Its decimal expansion would be non-terminating and non-recurring. Hence, \( \sqrt{23} \) is an irrational number.
(ii) \( \sqrt{225} \)
To find \( \sqrt{225} \), we can perform long division or recognize it as \( 15 \times 15 = 225 \).
15 --- 1|0225 -1 ---- 25| 125 -125 ---- 0
\( \sqrt{225} = 15 \)
We can write 15 as \( \frac{15}{1} \), which is in the form of \( \frac{p}{q} \) where \( p=15 \) and \( q=1 \) (with \( q \neq 0 \)). Therefore, \( \sqrt{225} \) is a rational number.
(iii) \( 0.3796 \)
The decimal expansion \( 0.3796 \) is terminating, meaning it ends. Any terminating decimal can be written as a fraction \( \frac{p}{q} \). Hence, \( 0.3796 \) is a rational number.
(iv) \( 7.478478... \)
The decimal expansion \( 7.478478... \) can be written as \( 7.\overline{478} \). This is a non-terminating but recurring decimal, as the block '478' repeats. Any non-terminating recurring decimal can be expressed as a fraction \( \frac{p}{q} \). Therefore, \( 7.478478... \) is a rational number.
(v) \( 1.101001000100001... \)
The decimal expansion \( 1.101001000100001... \) is non-terminating and non-recurring. The pattern of zeros between the ones increases (one zero, then two, then three, etc.), so there is no fixed block of digits that repeats. Hence, \( 1.101001000100001... \) is an irrational number.
In simple words: We check if the number is a perfect square, if its decimal ends, or if its decimal repeats in a pattern. If it's a perfect square, ends, or repeats, it's rational. If it's not a perfect square and its decimal never ends or repeats, it's irrational.

Exam Tip: Remember the definitions: rational numbers can be expressed as \( \frac{p}{q} \) (terminating or repeating decimals), while irrational numbers cannot (non-terminating, non-recurring decimals, like square roots of non-perfect squares).