CBSE Class 9 Mathematics Number System Assignment Set 05

Chapter - 1 ( Term-I)

(Number System)

Key Concepts

- Natural numbers are 1, 2, 3, ... denoted by \(N\).
- Whole numbers are 0, 1, 2, 3, ... denoted by \(W\).
- Integers are ..., -3, -2, -1, 0, 1, 2, 3, ... denoted by \(Z\).
- Rational numbers - All the numbers which can be written in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\), are called rational numbers.
- Irrational numbers - A number \(s\) is called irrational if it cannot be written in the form \(\frac{p}{q}\) where \(p\) and \(q\) are integers and \(q \neq 0\).
- The decimal expansion of a rational number is either terminating or non-terminating recurring. Thus we say that a number whose decimal expansion is either terminating or non-terminating recurring is a rational number.
- The decimal expansion of an irrational number is non-terminating non-recurring.
- All the rational numbers and irrational numbers taken together make a collection of real numbers.
- A real number is either rational or irrational.
- If \(r\) is rational and \(s\) is irrational then \(r+s\), \(r-s\), \(r \cdot s\) are always irrational numbers, but \(\frac{r}{s}\) may be rational or irrational.
- Every irrational number can be represented on a number line using Pythagoras theorem.
- Rationalization means to remove square root from the denominator.
To remove \(\sqrt{2}\) from \(\frac{3+\sqrt{5}}{\sqrt{2}}\), we will multiply both numerator and denominator by \(\sqrt{2}\).
For \(\frac{1}{a \pm \sqrt{b}}\), its rationalization factor is \(a \mp \sqrt{b}\).

Section - A

Q.1 Is zero a rational number? Can you write in the form p/q, where p and q are integer and \(q \neq 0\)?
Answer: Yes, zero is a rational number.
It can be written as \(\frac{0}{1}\), \(\frac{0}{2}\), \(\frac{0}{-3}\), etc., where \(p = 0\) and \(q\) is any non-zero integer.
In simple words: Zero can be expressed as a fraction with a denominator of 1, satisfying the definition of a rational number.

Exam Tip: Remember that the only condition for a rational number \(\frac{p}{q}\) is that \(q\) must not be zero; the numerator \(p\) can be zero.

Q.2 Find five rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\)?
Answer: To find five rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\), multiply the numerator and denominator of both numbers by \(5 + 1 = 6\).
\(\frac{3}{5} = \frac{3 \times 6}{5 \times 6} = \frac{18}{30}\)
\(\frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30}\)
Five rational numbers between \(\frac{18}{30}\) and \(\frac{24}{30}\) are:
\(\frac{19}{30}, \frac{20}{30}, \frac{21}{30}, \frac{22}{30}, \frac{23}{30}\)
In simple words: Expand the denominators by multiplying by a number greater than 5 to create enough space between them.

Exam Tip: If you need \(n\) rational numbers, always multiply both numbers by \(n+1\) or higher.

Q.3 State whether the following statements are true or false give reasons for your answers.
(i) Every natural no. is whole number.
Answer: True. Because the collection of whole numbers contains all natural numbers plus zero.
In simple words: All counting numbers are inside the whole numbers set.
(ii) Every integer is a whole number.
Answer: False. Negative integers like -2, -3 are not whole numbers.
In simple words: Whole numbers start from zero and do not include negative numbers.
(iii) Every rational number is a whole number.
Answer: False. Fractions like \(\frac{1}{2}\) are rational numbers but not whole numbers.
In simple words: Rational numbers include fractions and decimals, which are not whole numbers.
(iv) Every irrational number is a real number.
Answer: True. The collection of real numbers is made up of rational and irrational numbers.
In simple words: Real numbers include every type of number except complex numbers.
(v) Every real number is an irrational number.
Answer: False. Real numbers also include rational numbers like 2, 5, etc.
In simple words: Real numbers contain both rational and irrational numbers, not just irrational ones.
(vi) Every point on the number line is of the form \(\sqrt{m}\) where m is a natural no's.
Answer: False. No negative number can be the square root of any natural number \(m\), but negative numbers exist on the number line.
In simple words: Square roots of natural numbers are always positive, but the number line has negative numbers too.

Exam Tip: Learn number system subsets thoroughly: Natural \(\subset\) Whole \(\subset\) Integers \(\subset\) Rational \(\subset\) Real.

Q.4 Show how \(\sqrt{5}\) can be represented on the number line?
Answer: 1. Draw a line and mark point 0 as origin O.
2. Mark 2 units from 0 to A along the positive X-axis.
3. At point A, draw a perpendicular line AB of length 1 unit.
4. Join OB using Pythagoras theorem: \(OB = \sqrt{2^2 + 1^2} = \sqrt{4+1} = \sqrt{5}\).
5. Taking O as center and OB as radius, draw an arc intersecting the number line at point C. Point C represents \(\sqrt{5}\).
In simple words: Use a right-angled triangle with base 2 and perpendicular 1 to construct a hypotenuse of length \(\sqrt{5}\), then transfer it to the number line.

Exam Tip: Write down the Pythagoras theorem calculation clearly while showing construction steps.

Section - B

Q.5 Find the decimal expansion of \(\frac{10}{3}\), \(\frac{7}{8}\) and \(\frac{1}{7}\)? What kind of decimal expansion each has.
Answer:
(i) \(\frac{10}{3} = 3.333... = 3.\bar{3}\) (Non-terminating recurring decimal)
(ii) \(\frac{7}{8} = 0.875\) (Terminating decimal)
(iii) \(\frac{1}{7} = 0.142857142857... = 0.\overline{142857}\) (Non-terminating recurring decimal)
In simple words: Divide the numerator by the denominator to see if the remainder becomes zero (terminating) or repeats in a cycle (recurring).

Exam Tip: Fractions with denominators whose prime factors are only 2 and/or 5 always have terminating decimal expansions.

Q.6 Show that \(1.272727 = 1.\overline{27}\) can be expressed in the form p/q, where p and q are integers and \(q \neq 0\).
Answer: Let \(x = 1.272727...\) --- (1)
Since two digits are repeating, multiply both sides by 100:
\(100x = 127.272727...\) --- (2)
Subtract equation (1) from equation (2):
\(100x - x = 127.272727... - 1.272727...\)
\(99x = 126\)
\(x = \frac{126}{99}\)
Simplifying by dividing numerator and denominator by 9:
\(x = \frac{14}{11}\)
In simple words: Set the recurring decimal as \(x\), multiply by 10 powers depending on repeating digits, and subtract to eliminate the decimal part.

Exam Tip: Multiply by 10 for 1 repeating digit, 100 for 2 repeating digits, and 1000 for 3 repeating digits.

Q.7 Write three numbers whose decimal expressions are non-terminating & non-recurring?
Answer: Three examples of non-terminating and non-recurring numbers (irrational numbers) are:
1. \(0.101001000100001...\)
2. \(0.202002000200002...\)
3. \(0.503003000300003...\)
In simple words: Create patterns that never terminate and never repeat the exact same block of digits.

Exam Tip: Introduce an increasing number of zeros between non-zero digits to easily generate non-terminating non-recurring decimals.

Q.8 Find three different rational between 3/5 and 4/7.
Answer: First, make the denominators equal by finding the LCM of 5 and 7, which is 35.
\(\frac{3}{5} = \frac{3 \times 7}{5 \times 7} = \frac{21}{35}\)
\(\frac{4}{7} = \frac{4 \times 5}{7 \times 5} = \frac{20}{35}\)
To find three rational numbers, multiply numerator and denominator of both fractions by \(3 + 1 = 4\) (or a larger multiplier like 10):
\(\frac{21}{35} = \frac{21 \times 10}{35 \times 10} = \frac{210}{350}\)
\(\frac{20}{35} = \frac{20 \times 10}{35 \times 10} = \frac{200}{350}\)
Three different rational numbers between \(\frac{200}{350}\) and \(\frac{210}{350}\) are:
\(\frac{201}{350}, \frac{202}{350}, \frac{203}{350}\)
In simple words: Convert to equivalent fractions with common denominators and expand them to insert numbers in between.

Exam Tip: Always make denominators identical before comparing or finding numbers between two fractions.

Q.9 Classify the following numbers as rational or irrational.
(a) \(\sqrt{23}\)
Answer: Irrational. (Since 23 is a prime number, its square root cannot be simplified to a fraction.)
(b) \(\sqrt{225}\)
Answer: Rational. (Since \(\sqrt{225} = 15\), which can be written as \(\frac{15}{1}\).)
(c) \(0.6796\)
Answer: Rational. (It is a terminating decimal which can be expressed as \(\frac{6796}{10000}\).)
(d) \(1.101001000100001...\)
Answer: Irrational. (It has a non-terminating and non-recurring decimal expansion.)
In simple words: Perfect squares and terminating decimals are rational, while non-perfect square roots and non-terminating non-repeating decimals are irrational.

Exam Tip: Check if numbers under roots are perfect squares. If yes, they are rational.

Section - C

Q.10 Visualize 3.765 on the number line using successive magnification.
Answer: 1. Locate 3.765 between 3 and 4 on the number line. Divide the interval into 10 equal parts (3.1, 3.2, ..., 3.9).
2. Visualize the interval between 3.7 and 3.8 and divide it into 10 equal parts (3.71, 3.72, ..., 3.8).
3. Locate 3.76 between 3.76 and 3.77 and divide this interval into 10 equal parts (3.761, 3.762, ..., 3.77).
4. The point 3.765 is precisely located as the fifth mark between 3.76 and 3.77.
In simple words: Zoom in progressively by one decimal place at a time between the two closest bounds until the exact number is reached.

Exam Tip: Draw clear magnified number line segments for each decimal place step.

Q.11 Visualize \(4.\bar{26}\) on the number line upto 4 decimal places.
Answer: \(4.\bar{26} = 4.2626...\) upto 4 decimal places is 4.2626.
1. Locate 4.2626 between 4 and 5.
2. Magnify the interval between 4.2 and 4.3 to locate 4.26 between 4.26 and 4.27.
3. Magnify the interval between 4.26 and 4.27 to locate 4.262 between 4.262 and 4.263.
4. Magnify the interval between 4.262 and 4.263 to finally locate 4.2626.
In simple words: Expand the repeating bar into digits (4.2626) and perform 4 successive magnification steps.

Exam Tip: Always expand recurring decimals to the specified number of decimal places before beginning magnification.

Q.12 simplify the following expressions.
(i) \((5 + \sqrt{7})(2 + \sqrt{5})\)
Answer:
\(= 5(2) + 5\sqrt{5} + \sqrt{7}(2) + \sqrt{7}\sqrt{5}\)
\(= 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35}\)
In simple words: Multiply each term of the first bracket with every term of the second bracket.
(ii) \((5 + \sqrt{5})(5 - \sqrt{5})\)
Answer:
Using identity \((a+b)(a-b) = a^2 - b^2\):
\(= 5^2 - (\sqrt{5})^2\)
\(= 25 - 5 = 20\)
In simple words: Apply the difference of squares formula directly.
(iii) \((\sqrt{3} + \sqrt{7})^2\)
Answer:
Using identity \((a+b)^2 = a^2 + 2ab + b^2\):
\(= (\sqrt{3})^2 + 2(\sqrt{3})(\sqrt{7}) + (\sqrt{7})^2\)
\(= 3 + 2\sqrt{21} + 7\)
\(= 10 + 2\sqrt{21}\)
In simple words: Expand using the square of a binomial algebraic identity.
(iv) \((\sqrt{11} - \sqrt{7})(\sqrt{11} + \sqrt{7})\)
Answer:
Using identity \((a-b)(a+b) = a^2 - b^2\):
\(= (\sqrt{11})^2 - (\sqrt{7})^2\)
\(= 11 - 7 = 4\)
In simple words: Simplify using the algebraic formula for the product of sum and difference.

Exam Tip: Memorize standard algebraic identities like \((a+b)(a-b) = a^2 - b^2\) and \((a+b)^2 = a^2 + 2ab + b^2\) to solve surds quickly.

Q.13 Rationalize the denominator of \(\frac{5}{\sqrt{3} - \sqrt{5}}\).
Answer:
\(= \frac{5}{\sqrt{3} - \sqrt{5}} \times \frac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}}\)
\(= \frac{5(\sqrt{3} + \sqrt{5})}{(\sqrt{3})^2 - (\sqrt{5})^2}\)
\(= \frac{5(\sqrt{3} + \sqrt{5})}{3 - 5}\)
\(= \frac{5(\sqrt{3} + \sqrt{5})}{-2} = -\frac{5}{2}(\sqrt{3} + \sqrt{5})\)
In simple words: Multiply both numerator and denominator by the conjugate of the denominator, which is \(\sqrt{3} + \sqrt{5}\).

Exam Tip: The conjugate of \(a - \sqrt{b}\) is \(a + \sqrt{b}\).

Section - D

Q.1 Represent \(\sqrt{9.3}\) on number line.
Answer: 1. Draw a line segment \(AB = 9.3\) units.
2. Extend AB by 1 unit to C, so that \(AC = 10.3\) units.
3. Find the midpoint O of AC.
4. Taking O as center and OC as radius, draw a semi-circle.
5. Draw a perpendicular line from point B intersecting the semi-circle at point D. The length \(BD = \sqrt{9.3}\).
6. Taking B as center and BD as radius, draw an arc on the number line to locate \(\sqrt{9.3}\).
In simple words: Add 1 to the given number, find the semicircle radius, construct a perpendicular at the original point, and swing it onto the number line.

Exam Tip: Remember to extend the line segment by exactly 1 unit before finding the midpoint.

Q.2 Recall, \(\pi\) is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is \(\pi = \frac{c}{d}\). This seems to contradict the fact that \(\pi\) is irrational. How will you resolve this contradiction?
Answer: There is no contradiction. When we measure a length with a scale or any other device, we only obtain an approximate rational value. Therefore, either \(c\) or \(d\) or both are irrational. Thus, \(\frac{c}{d}\) is irrational, making \(\pi\) irrational.
In simple words: Measurement values are always approximations, so a fraction of measured values can still result in an irrational number.

Exam Tip: State clearly that fractional representations from physical measurements are only approximations.

Q.3 Simplify
(i) \(2^{2/3} \cdot 2^{1/5}\)
Answer:
Using law of exponents \(a^m \cdot a^n = a^{m+n}\):
\(= 2^{(2/3 + 1/5)}\)
\(= 2^{(10/15 + 3/15)}\)
\(= 2^{13/15}\)
In simple words: Add the fractional exponents since the bases are identical.
(ii) \(\left(\frac{1}{3^7}\right)^7\)
Answer:
Using law of exponents \((a^m)^n = a^{m \cdot n}\) and \(\left(\frac{1}{a^m}\right) = a^{-m}\):
\(= (3^{-7})^7\)
\(= 3^{-49}\)
In simple words: Multiply the outer exponent with the inner exponent.
(iii) \((16)^{3/4}\)
Answer:
\(= (2^4)^{3/4}\)
\(= 2^{4 \times 3/4}\)
\(= 2^3 = 8\)
In simple words: Express 16 as \(2^4\) and multiply the powers.
(iv) \(7^{1/2} \cdot 8^{1/2}\)
Answer:
Using law of exponents \(a^m \cdot b^m = (ab)^m\):
\(= (7 \times 8)^{1/2}\)
\(= 56^{1/2} = \sqrt{56}\)
In simple words: Multiply the bases together when the exponents are the same.

Exam Tip: Revise all laws of exponents for rational powers before attempting simplification problems.

Self Evaluation

Q.1 Write the value of \(\left(\frac{x^a}{x^b}\right)^{a+b} \times \left(\frac{x^b}{x^c}\right)^{b+c} \times \left(\frac{x^c}{x^a}\right)^{c+a}\)
Answer:
\(= (x^{a-b})^{a+b} \times (x^{b-c})^{b+c} \times (x^{c-a})^{c+a}\)
\(= x^{a^2 - b^2} \times x^{b^2 - c^2} \times x^{c^2 - a^2}\)
Using \(a^m \cdot a^n = a^{m+n}\):
\(= x^{(a^2 - b^2 + b^2 - c^2 + c^2 - a^2)}\)
\(= x^0 = 1\)
In simple words: Use exponent rules to subtract powers in division, multiply brackets using identity \((a-b)(a+b) = a^2-b^2\), and sum them up to get zero.

Exam Tip: All terms in the final exponent cancel out, yielding 1 as the answer.

Q.2 \(\left\{5 \left(8^{1/3} + 27^{1/3}\right)^3\right\}^{1/4}\)
Answer:
\(= \left\{5 \left((2^3)^{1/3} + (3^3)^{1/3}\right)^3\right\}^{1/4}\)
\(= \left\{5 (2 + 3)^3\right\}^{1/4}\)
\(= \left\{5 (5)^3\right\}^{1/4}\)
\(= \left\{5^1 \times 5^3\right\}^{1/4}\)
\(= \left(5^4\right)^{1/4}\)
\(= 5^{4 \times 1/4} = 5^1 = 5\)
In simple words: Simplify cube roots first, add the terms, combine the powers of 5, and apply the outer fourth root.

Exam Tip: Convert numbers into exponential form with matching powers to simplify cube roots easily.

Q.3 If a & b are rational number, find the value of a & b in each of the following equalities.
(a) \(\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = a + b\sqrt{3}\)
Answer:
Rationalize the LHS: \(\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}\)
\(= \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2}\)
\(= \frac{3 - 2\sqrt{3} + 1}{3 - 1}\)
\(= \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}\)
Comparing with \(a + b\sqrt{3}\):
\(a = 2, b = -1\)
In simple words: Rationalize the denominator of the left side, expand, and compare rational and irrational coefficients with the right side.
(ii) \(\frac{3 + \sqrt{7}}{3 - \sqrt{7}} = a + b\sqrt{7}\)
Answer:
Rationalize the LHS: \(\frac{3 + \sqrt{7}}{3 - \sqrt{7}} \times \frac{3 + \sqrt{7}}{3 + \sqrt{7}}\)
\(= \frac{(3 + \sqrt{7})^2}{3^2 - (\sqrt{7})^2}\)
\(= \frac{9 + 6\sqrt{7} + 7}{9 - 7}\)
\(= \frac{16 + 6\sqrt{7}}{2} = 8 + 3\sqrt{7}\)
Comparing with \(a + b\sqrt{7}\):
\(a = 8, b = 3\)
In simple words: Multiply numerator and denominator by the conjugate of the denominator, simplify, and equate terms.

Exam Tip: Equate the rational part to \(a\) and the coefficient of the surd part to \(b\).

Q.4 Prove that \(\sqrt{2}\) is an irrational number using long division method?
Answer: When we find the square root of 2 by the long division method, the decimal expansion goes on infinitely without terminating or repeating:
\(\sqrt{2} = 1.41421356...\)
Since the decimal representation is non-terminating and non-recurring, \(\sqrt{2}\) is an irrational number.
In simple words: Long division shows that the decimal digits of \(\sqrt{2}\) never end and never form a repeating pattern, which proves it is irrational.

Exam Tip: Mention the non-terminating and non-recurring nature of the decimal expansion obtained from long division.