Chapter 1: Number System
Exercise 1.1
Q1
Is zero a rational number? Can you write it in the form \( p/q \), where \( p \) and \( q \) are integers and \( q \neq 0 \)?
Answer: Yes
Zero is a rational number as it can be represented as \(\frac{0}{1}\) or \(\frac{0}{2}\) or \(\frac{0}{3}\) etc.
Q2
Find six rational numbers between 3 and 4.
There are infinite rational numbers between 3 and 4.
To find 6 numbers, we can multiply and divide by \(6+1=7\) (or any number greater than 6).
3 and 4 can be represented as \(\frac{24}{8}\) and \(\frac{32}{8}\) respectively (multiplying by 8 here).
Therefore, six rational numbers between 3 and 4 are:
\[ \frac{25}{8}, \frac{26}{8}, \frac{27}{8}, \frac{28}{8}, \frac{29}{8}, \frac{30}{8} \]Q3
Find five rational numbers between \( \frac{3}{5} \) and \( \frac{4}{5} \).
There are infinite rational numbers between \( \frac{3}{5} \) and \( \frac{4}{5} \).
To find 5 numbers, multiply numerators and denominators by 6:
\[ \frac{3}{5} = \frac{3 \times 6}{5 \times 6} = \frac{18}{30} \] \[ \frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30} \]Therefore, five rational numbers between \( \frac{3}{5} \) and \( \frac{4}{5} \) are:
\[ \frac{19}{30}, \frac{20}{30}, \frac{21}{30}, \frac{22}{30}, \frac{23}{30} \]Q4
State whether the following statements are true or false. Give reasons for your answers.
(i) Every natural number is a whole number.
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.
(i) Every natural number is a whole number.
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.
(i) True; since the collection of whole numbers contains all natural numbers.
(ii) False; as integers may be negative but whole numbers are positive. For example: -3 is an integer but not a whole number.
(iii) False; as rational numbers may be fractional but whole numbers may not be. For example: \( \frac{1}{5} \) is a rational number but not a whole number.
Exercise 1.2
Q1
State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
(ii) Every point on the number line is of the form \( \sqrt{m} \), where \( m \) is a natural number.
(iii) Every real number is an irrational number.
(i) Every irrational number is a real number.
(ii) Every point on the number line is of the form \( \sqrt{m} \), where \( m \) is a natural number.
(iii) Every real number is an irrational number.
(i) True; since the collection of real numbers is made up of rational and irrational numbers.
(ii) False; as negative numbers cannot be expressed as the square root of any natural number (e.g., no natural number \(m\) gives \(\sqrt{m} = -2\)).
(iii) False; as real numbers include both rational and irrational numbers. Therefore, every real number cannot be an irrational number (e.g., 2 is real but not irrational).
Q2
Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
No. The square roots of all positive integers are not irrational.
For example:
\[ \sqrt{4} = 2 \quad \text{and} \quad \sqrt{9} = 3 \]Here, 2 and 3 are rational numbers.
Q3
Show how \( \sqrt{5} \) can be represented on the number line.
We know that:
\[ \sqrt{4} = 2 \] \[ \sqrt{5} = \sqrt{(2)^2 + (1)^2} \]Steps of Construction:
- Mark a point 'A' representing 2 on the number line (where O is 0 and A is 2).
- Construct AB of 1 unit length perpendicular to OA at A.
- Join O to B. By Pythagoras theorem, \( OB = \sqrt{2^2 + 1^2} = \sqrt{5} \).
- Taking O as centre and OB as radius, draw an arc intersecting the number line at C.
- Point C represents \( \sqrt{5} \) on the number line.
Exercise 1.3
Q1
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) \( \frac{329}{400} \)
(i) \( \frac{36}{100} \)
(ii) \( \frac{1}{11} \)
(iii) \( 4\frac{1}{8} \)
(iv) \( \frac{3}{13} \)
(v) \( \frac{2}{11} \)
(vi) \( \frac{329}{400} \)
(i) \( \frac{36}{100} = 0.36 \) (Terminating)
(ii) \( \frac{1}{11} = 0.090909... = 0.\overline{09} \) (Non-terminating repeating)
(iii) \( 4\frac{1}{8} = \frac{33}{8} = 4.125 \) (Terminating)
(iv) \( \frac{3}{13} = 0.230769230769... = 0.\overline{230769} \) (Non-terminating repeating)
(v) \( \frac{2}{11} = 0.181818... = 0.\overline{18} \) (Non-terminating repeating)
(vi) \( \frac{329}{400} = 0.8225 \) (Terminating)
Q2
You know that \( \frac{1}{7} = 0.\overline{142857} \). Can you predict what the decimal expansion 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?
Yes. It can be done by multiplying the decimal expansion of \( \frac{1}{7} \) by the respective numerator:
\[ \frac{2}{7} = 2 \times \frac{1}{7} = 2 \times 0.\overline{142857} = 0.\overline{285714} \] \[ \frac{3}{7} = 3 \times \frac{1}{7} = 3 \times 0.\overline{142857} = 0.\overline{428571} \] \[ \frac{4}{7} = 4 \times \frac{1}{7} = 4 \times 0.\overline{142857} = 0.\overline{571428} \] \[ \frac{5}{7} = 5 \times \frac{1}{7} = 5 \times 0.\overline{142857} = 0.\overline{714285} \] \[ \frac{6}{7} = 6 \times \frac{1}{7} = 6 \times 0.\overline{142857} = 0.\overline{857142} \]Q3
Express the following in the form \( p/q \), where \( p \) and \( q \) are integers and \( q \neq 0 \):
(i) \( 0.\overline{6} \)
(ii) \( 0.4\overline{7} \)
(iii) \( 0.\overline{001} \)
(i) \( 0.\overline{6} \)
(ii) \( 0.4\overline{7} \)
(iii) \( 0.\overline{001} \)
(i) \( 0.\overline{6} \)
Let \( x = 0.666... \)
\[ 10x = 6.666... \] \[ 10x = 6 + x \] \[ 9x = 6 \Rightarrow x = \frac{6}{9} = \frac{2}{3} \](ii) \( 0.4\overline{7} \)
Let \( x = 0.4777... \)
\[ 10x = 4.777... \] \[ 100x = 47.777... \]Subtracting \( 10x \) from \( 100x \):
\[ 90x = 43 \Rightarrow x = \frac{43}{90} \](iii) \( 0.\overline{001} \)
Let \( x = 0.001001... \)
\[ 1000x = 1.001001... \] \[ 1000x = 1 + x \] \[ 999x = 1 \Rightarrow x = \frac{1}{999} \]Q4
Express \( 0.99999... \) in the form \( p/q \). Are you surprised by your answer? Discuss why the answer makes sense.
Let \( x = 0.9999... \)
\[ 10x = 9.9999... \] \[ 10x = 9 + x \] \[ 9x = 9 \Rightarrow x = 1 \]The answer makes sense because \( 0.999... \) is infinitely close to 1, leaving no gap between them.
Q5
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of \( 1/17 \)? Perform the division to check your answer.
The maximum number of digits in the repeating block is \( n-1 \), where \( n \) is the divisor. Here, it is \( 17-1 = 16 \).
Performing the division:
\[ \frac{1}{17} = 0.\overline{0588235294117647} \]There are exactly 16 digits in the repeating block.
Q6
Look at several examples of rational numbers in the form \( p/q \) (\( q \neq 0 \)) where \( p \) and \( q \) are integers with no common factors other than 1 and having terminating decimal representations. Can you guess what property \( q \) must satisfy?
Examples:
\[ \frac{9}{4} = 2.25 \quad (4 = 2^2) \] \[ \frac{11}{8} = 1.375 \quad (8 = 2^3) \] \[ \frac{27}{5} = 5.4 \quad (5 = 5^1) \]Property: Terminating decimal expansion occurs when the prime factorisation of the denominator \( q \) contains only powers of 2, only powers of 5, or both.
Q7
Write three numbers whose decimal expansions are non-terminating non-recurring.
Three examples of irrational numbers:
- \( 0.505005000500005... \)
- \( 0.72072007200072000072... \)
- \( 0.08008000800008000008... \)
Q8
Find three different irrational numbers between the rational numbers \( \frac{5}{7} \) and \( \frac{9}{11} \).
First, convert to decimal:
\[ \frac{5}{7} = 0.\overline{714285} \] \[ \frac{9}{11} = 0.\overline{81} \]Three irrational numbers between \( 0.71... \) and \( 0.81... \) are:
- \( 0.73073007300073... \)
- \( 0.75075007500075... \)
- \( 0.79079007900079... \)
Q9
Classify the following numbers as rational or irrational:
(i) \( \sqrt{23} \)
(ii) \( \sqrt{225} \)
(iii) \( 0.3796 \)
(iv) \( 7.478478... \)
(v) \( 1.1010010001... \)
(i) \( \sqrt{23} \)
(ii) \( \sqrt{225} \)
(iii) \( 0.3796 \)
(iv) \( 7.478478... \)
(v) \( 1.1010010001... \)
(i) \( \sqrt{23} \) : Irrational. 23 is not a perfect square; its decimal is non-terminating non-recurring.
(ii) \( \sqrt{225} = 15 \) : Rational. Can be written as \( 15/1 \).
(iii) \( 0.3796 \) : Rational. Terminating decimal.
(iv) \( 7.478478... = 7.\overline{478} \) : Rational. Non-terminating recurring.
(v) \( 1.1010010001... \) : Irrational. Non-terminating non-recurring.
Exercise 1.4 (Operations and Rationalisation)
Q1
Classify the following numbers as rational or irrational:
(i) \( 2 - \sqrt{5} \)
(ii) \( (3 + \sqrt{23}) - \sqrt{23} \)
(iii) \( \frac{2\sqrt{7}}{7\sqrt{7}} \)
(iv) \( \frac{1}{\sqrt{2}} \)
(v) \( 2\pi \)
(i) \( 2 - \sqrt{5} \)
(ii) \( (3 + \sqrt{23}) - \sqrt{23} \)
(iii) \( \frac{2\sqrt{7}}{7\sqrt{7}} \)
(iv) \( \frac{1}{\sqrt{2}} \)
(v) \( 2\pi \)
(i) Irrational: Difference of a rational and an irrational is irrational.
(ii) Rational: \( 3 + \sqrt{23} - \sqrt{23} = 3 \), which is rational.
(iii) Rational: \( \frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7} \), which is rational.
(iv) Irrational: Quotient of a non-zero rational and an irrational is irrational.
(v) Irrational: Product of a non-zero rational and an irrational is irrational.
Q2
Simplify each of the following expressions:
(i) \( (3 + \sqrt{3})(2 + \sqrt{2}) \)
(ii) \( (3 + \sqrt{3})(3 - \sqrt{3}) \)
(iii) \( (\sqrt{5} + \sqrt{2})^2 \)
(iv) \( (\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2}) \)
(i) \( (3 + \sqrt{3})(2 + \sqrt{2}) \)
(ii) \( (3 + \sqrt{3})(3 - \sqrt{3}) \)
(iii) \( (\sqrt{5} + \sqrt{2})^2 \)
(iv) \( (\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2}) \)
(i)
\( = 3(2 + \sqrt{2}) + \sqrt{3}(2 + \sqrt{2}) \)
\( = 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6} \)
(ii)
Using \( (a+b)(a-b) = a^2 - b^2 \):
\( = 3^2 - (\sqrt{3})^2 = 9 - 3 = 6 \)
(iii)
Using \( (a+b)^2 = a^2 + b^2 + 2ab \):
\( = (\sqrt{5})^2 + (\sqrt{2})^2 + 2\sqrt{5}\sqrt{2} \)
\( = 5 + 2 + 2\sqrt{10} = 7 + 2\sqrt{10} \)
(iv)
Using \( (a-b)(a+b) = a^2 - b^2 \):
\( = (\sqrt{5})^2 - (\sqrt{2})^2 = 5 - 2 = 3 \)
Q3
Recall, \( \pi \) is defined as the ratio of the circumference (say \( c \)) of a circle to its diameter (say \( d \)). That is, \( \pi = c/d \). This seems to contradict the fact that \( \pi \) is irrational. How will you resolve this contradiction?
There is no contradiction. When we measure a length with a scale or any other instrument, we only obtain an approximate rational value. We never obtain an exact value. Therefore, we may not realize that either \( c \) or \( d \) is irrational. Since the quotient of an irrational and a rational (or vice versa) is irrational, \( \pi \) remains irrational.
Q4
Represent \( \sqrt{9.3} \) on the number line.
- Mark a line segment \( OB = 9.3 \) units on the number line.
- Extend the line to C such that \( BC = 1 \) unit. Total length \( OC = 10.3 \).
- Find the mid-point D of OC.
- Draw a semi-circle on OC taking D as its centre.
- Draw a perpendicular to OC at point B. Let it intersect the semi-circle at E.
- The length BE is \( \sqrt{9.3} \).
- Taking B as centre and BE as radius, draw an arc intersecting the number line at F. Point F represents \( \sqrt{9.3} \).
Q5
Rationalise the denominators of the following:
(i) \( \frac{1}{\sqrt{7}} \)
(ii) \( \frac{1}{\sqrt{7} - \sqrt{6}} \)
(iii) \( \frac{1}{\sqrt{5} + \sqrt{2}} \)
(iv) \( \frac{1}{\sqrt{7} - 2} \)
(i) \( \frac{1}{\sqrt{7}} \)
(ii) \( \frac{1}{\sqrt{7} - \sqrt{6}} \)
(iii) \( \frac{1}{\sqrt{5} + \sqrt{2}} \)
(iv) \( \frac{1}{\sqrt{7} - 2} \)
(i)
\( \frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7} \)
(ii)
\( \frac{1}{\sqrt{7} - \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} \)
\( = \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6} \)
(iii)
\( \frac{1}{\sqrt{5} + \sqrt{2}} \times \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} \)
\( = \frac{\sqrt{5} - \sqrt{2}}{5 - 2} = \frac{\sqrt{5} - \sqrt{2}}{3} \)
(iv)
\( \frac{1}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2} \)
\( = \frac{\sqrt{7} + 2}{(\sqrt{7})^2 - 2^2} = \frac{\sqrt{7} + 2}{7 - 4} = \frac{\sqrt{7} + 2}{3} \)
Exercise 1.5 (Laws of Exponents)
Q1
Find:
(i) \( 64^{1/2} \)
(ii) \( 32^{1/5} \)
(iii) \( 125^{1/3} \)
(i) \( 64^{1/2} \)
(ii) \( 32^{1/5} \)
(iii) \( 125^{1/3} \)
(i) \( 64^{1/2} = (2^6)^{1/2} = 2^{6 \times 1/2} = 2^3 = 8 \)
(ii) \( 32^{1/5} = (2^5)^{1/5} = 2^{5 \times 1/5} = 2^1 = 2 \)
(iii) \( 125^{1/3} = (5^3)^{1/3} = 5^{3 \times 1/3} = 5^1 = 5 \)
Q2
Find:
(i) \( 9^{3/2} \)
(ii) \( 32^{2/5} \)
(iii) \( 16^{3/4} \)
(iv) \( 125^{-1/3} \)
(i) \( 9^{3/2} \)
(ii) \( 32^{2/5} \)
(iii) \( 16^{3/4} \)
(iv) \( 125^{-1/3} \)
(i) \( 9^{3/2} = (3^2)^{3/2} = 3^3 = 27 \)
(ii) \( 32^{2/5} = (2^5)^{2/5} = 2^2 = 4 \)
(iii) \( 16^{3/4} = (2^4)^{3/4} = 2^3 = 8 \)
(iv) \( 125^{-1/3} = \frac{1}{125^{1/3}} = \frac{1}{(5^3)^{1/3}} = \frac{1}{5} \)
Q3
Simplify:
(i) \( 2^{2/3} \cdot 2^{1/5} \)
(ii) \( \left(\frac{1}{3^3}\right)^7 \)
(iii) \( \frac{11^{1/2}}{11^{1/4}} \)
(iv) \( 7^{1/2} \cdot 8^{1/2} \)
(i) \( 2^{2/3} \cdot 2^{1/5} \)
(ii) \( \left(\frac{1}{3^3}\right)^7 \)
(iii) \( \frac{11^{1/2}}{11^{1/4}} \)
(iv) \( 7^{1/2} \cdot 8^{1/2} \)
(i)
\( 2^{2/3} \cdot 2^{1/5} = 2^{2/3 + 1/5} = 2^{(10+3)/15} = 2^{13/15} \)
(ii)
\( \left(\frac{1}{3^3}\right)^7 = (3^{-3})^7 = 3^{-21} = \frac{1}{3^{21}} \)
(iii)
\( \frac{11^{1/2}}{11^{1/4}} = 11^{1/2 - 1/4} = 11^{(2-1)/4} = 11^{1/4} \)
(iv)
\( 7^{1/2} \cdot 8^{1/2} = (7 \cdot 8)^{1/2} = 56^{1/2} \)