NCERT Solutions Class 9 Maths Chapter 1: Number System

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 ≠ 0?

Answer: A number is considered to be rational, if it can be represented in the form p/q, where p, as well as q, are integers and q is less than zero.

In the instance of the number ‘0,’ Zero can be represented as 0/1, 0/2, 0/3… as well as 0/1, 0/2, 0/3… we may infer that 0 can be expressed in the p/q form, where q might be a positive or negative number, because it meets the required criteria. As a result, the number 0 is rational.

Q2. Find six rational numbers between 3 and 4.

Answer: Between 3 and 4, there exist infinite rational numbers. We’ll multiply both the integers, 3 and 4, by 6+1 = 7 since we need to find 6 rational numbers in between 3 and 4. that is,

therefore,

Between 21/7 and 28/7, the numbers will be logical and fall between 3 and 4.

As a result, the 6 rational numbers in between 3 and 4 are \dfrac{22}{7},\dfrac{23}{7},\dfrac{24}{7},\dfrac{25}{7},\dfrac{26}{7},\dfrac{27}{7} .

Q3. Find five rational numbers between 3/5 and 4/5.

Answer: Between 3/5 and 4/5, there exist infinite rational numbers.

Between 3/5 and 4/5, there exist infinite rational numbers. We can multiply both the numbers 3/5 and 4/5 to discover 5 rational numbers between them.

Using, 5+1=6(or any number that is greater than 5)

that is (\dfrac{3}{5})\times (\dfrac{6}{6})=\dfrac{18}{30}

hence, (\dfrac{4}{5})\times (\dfrac{6}{6})=\dfrac{24}{30}

Between 18/30 and 24/30, the numbers will be logical, falling in between 3/5 and 4/5.

As a result, the 5 rational numbers between 3/5 and 4/5 are \dfrac{19}{30},\dfrac{20}{30},\dfrac{21}{30},\dfrac{22}{30},\dfrac{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.

Answer:

1. True

Natural numbers are numbers that range from one to infinity (excluding fractions or decimals), for example, 1,2,3,4…

Whole numbers are numbers that range from 0 to infinity (without fractions or decimals): 0,1,2,3,…

Whole numbers, on the other hand, have all the constituents of natural numbers plus zero.

A whole number is a natural number, however not every whole number is a natural number.

2. False

Integers are numbers that contain positive, negative, and 0 but do not include fractional or decimal values.

Integers =…,-4,-3,-2,-1,0,1,2,3,4,…

Whole numbers are numbers that range from 0 to infinity (excluding fractions or decimals): 0,1,2,3,…

As a result, we may claim that integers include both positive and negative numbers. Every integer is a whole number, and not every whole number would be an integer.

3. False

All numbers with the format p/q, where p and q are the integers and

  q\neq 0 . Rational numbers are

0, \dfrac{19}{30}, 2, \dfrac{9}{-3}, \dfrac{-12}{7}, etc. Whole numbers are numbers that range from 0 to infinity (excluding fractions or decimals): 0,1,2,3,…

As a result, we may claim that integers include both positive and negative numbers.

Although all whole numbers is rational, not all rational numbers are whole numbers.

Exercise 1.2

Q1. State whether the following statements are true or false. Justify your answers.

1. Every irrational number is a real number.

2. Every point on the number line is of the form √m where m is a natural number.

3. Every real number is an irrational number.

Answer:

1. True

Irrational Numbers – A number is irrational if it cannot be represented in the p/q format, where p and q are integers and q is less than zero.

 (Irrational numbers= \pi, e, \sqrt{3}, 5+\sqrt{2}, 6.23146…, 0.101001001000… )

Real numbers are a category of numbers that includes both rational and irrational numbers.

(Real numbers =\sqrt{2}, ..,\sqrt{5}, 0.102)

Every irrational number is also a real number, although not all real numbers are irrational.

Therefore, the answer is True.

2. False

A negative number could not be expressed in form of square roots according to the rule. For instance, \sqrt{9}=3is indeed a natural number.

However, the number\sqrt{2}=1.414 is not a natural number. We also know that negative numbers exist on the number line, but when we extract the root of a negative number, it becomes a complex number rather than a natural number. For example, \sqrt{-7}=7i \;\text{where}\; i=\sqrt{-1}. It is incorrect to claim that every point situated on the number line has the form m, where m is a natural number.

3. False

Real numbers include both the irrational and rational numbers. As a result, no real number can be both rational and irrational. Real numbers are a category of numbers that includes both rational and irrational numbers. (Real numbers =\sqrt{2}, ..,\sqrt{5}, 0.102).

 Irrational Numbers – A number is irrational if it cannot be represented in the p/q format, where p and q are integers and q is less than zero.

 (Irrational numbers= \pi, e, \sqrt{3}, 5+\sqrt{2}, 6.23146…, 0.101001001000… ). Thus, every irrational number is also a real number, but not all real numbers are 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.

Answer: No, all positive integers’ square roots are not irrational.

For instance, \sqrt{4}=2\;\text{ rational number}\\ \sqrt{9}=3\;\text{rational} .

As a result, the square roots of the positive integers 4 and 9 that is 2 and 3 are rational.

Q3. Show how √5 can be represented on the number line.

Answer:

Step 1: Draw a number line with line AB of 2 units.

Step 2: Draw a 1 unit long perpendicular line BC at B.

Step 3: Join the points C and A.

Step 4: ABC has now been transformed into a right-angled triangle. Using Pythagoras’ theorem, ( AB^2 +BC^2=CA^2\\ 2^2+1^2=CA^2=5\\ =>CA=\sqrt{5} ). As a result, CA is a \sqrt{5} unit line.

Step 5: Draw an arc connecting the number line using CA as the radius and A as the centre. Because this is a radius of the circle whose centre was A, the location where the number line and arc connect is \sqrt{5} distance from 0.

Q4. Classroom activity (Constructing the ‘square root spiral’) : Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP_{1}  of unit length. Draw a line segment P_{1}P_{2} perpendicular to OP_{1}   of unit length (see Fig. 1.9). Now draw a line segment   P_{2}P_{3}  perpendicular to OP_{2} . Then draw a line segment P_{3}P_{4}  perpendicular to OP_{3} . Continuing in this manner, you can get the line segment P_{n–1}P_{n} by drawing a line segment of unit length perpendicular to OP_{n–1} . In this manner, you will have created the points P_{2}, P_{3},…., P_{n},… ., and joined them to create a beautiful spiral depicting \sqrt{2}, \sqrt{3}, \sqrt{4},…

Answer:

Step 1: On the paper, make a point O. The square root spiral will have O as its centre.

Step 2: Draw a 1cm horizontally straight-line OA from O.

Step 3: Draw a 1 cm perpendicular line from A to B.

Step 4: Join the points O and B. OB will be equal to \sqrt{2} .

Step 5: Draw a 1 cm perpendicular line from B to C and indicate the end point.

Step 6: Join the points O and C. OC will be equal to \sqrt{3} .

Step 7: Draw \sqrt{4}, \sqrt{5}, \text{ and } \sqrt{6} using the same procedures as before….

Exercise 1.3

Q1. Write the following in decimal form and say what kind of decimal expansion each has:

1. 36/100

Answer:

On dividing 36 by 100 we will get the quotient as 0.36 that is terminating.

2. 1/11

Answer:

On dividing 1 by 11 we will get the quotient as 0.90909 that is non terminating but is repeating.

3. 4 1/8

Answer:

4\dfrac{1}{8}= \dfrac{33}{8}

On dividing 33 by 8 we will get the quotient as 4.125 that is terminating.

4. 3/13

Answer:

On dividing 3 by 13 we will get 0.230769 that is not terminating but repeating.

5. 2/11

Answer:

On dividing 2 by 11 we will get 0.181818 that is not terminating but repeating.

6. 329/400

Answer:

On dividing 329 by 400 we will get 0.8225 that is terminating.

Q2. You know that 1/7 = 0.142857. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are, without actually doing the long division? If so, how? [Hint: Study the remainders while finding the value of 1/7 carefully.]

Answer: \dfrac{1}{7}=\overline{0.142857}\\ =>2\times\dfrac{1}{7}=2\times \overline{0.142857}=\overline{0.285714}\\ =>3\times\dfrac{1}{7}=2\times \overline{0.142857}=\overline{0.428571}\\ =>4\times\dfrac{1}{7}=2\times \overline{0.142857}=\overline{0.571428}\\ =>5\times\dfrac{1}{7}=2\times \overline{0.142857}=\overline{0.714285}\\ =>6\times\dfrac{1}{7}=2\times \overline{0.142857}=\overline{0.857142}\\

Q3. Express the following in the form p/q, where p and q are integers and q ≠ 0.

1. \overline{0.6}

Answer: Let us assume the following: x=0.6666….

Now, we can write:

10x=6.666…\\ 10x=6+x\\ 9x=6\\ X=\dfrac{2}{3}

2. \overline{0.47}

Answer: \overline{0.47} =0.4777…\\ =(\dfrac{4}{10})+(\dfrac{0.777}{10})\\

Assuming, x=0.7777…\\ \text{now,}\; 10x=7.777…\\ 10x=7+x\\ x=\dfrac{7}{9}

3. \<strong>overline</strong>{0.001}

Answer: Assuming, x=0.001001…\\ \text{now,}\;1000x=1.001001…\\ 1000x=1+x\\ 999x=1\\ X=\dfrac{1}{999}

Q4.  Express 0.99999…. in the form p/q. Are you surprised by your answer? With your teacher and classmates, discuss why the answer makes sense.

Answer: Assuming x=0.9999….(i)\\ \text{Multiplying both the sides with 10}\\ 10x=9.9999….(ii)\\ \text{subtracting equation (i) from (ii)}\\ 9x=9\\ x=1

The difference between 1 and 0.999999 is almost 0.000001, which is insignificant. As a result, we can conclude that 0.999 is too close to 1, hence 1 as the result is justified.

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.

Answer: On dividing 1 by 17 we will get 0.0588235294117647 as the quotient which gets repeated on further division.

Q6. Look at several examples of rational numbers in the form \dfrac{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: When q is 2, 4, 5, 8, 10, we notice that… The decimal expansion then comes to an end. For instance:

1/2 = 0.5, denominator q = 21

7/8 = 0. 875, denominator q =23

4/5 = 0.8, denominator q = 51

The terminating decimal can be found in situations where the prime factorization of the denominator of the provided fractions has the power of just 2 or 5 or both.

Q7. Write three numbers whose decimal expansions are non-terminating non-recurring.

Answer: All the irrational numbers are non-terminating non-recurring, as we know. The following are three non-terminating non-recurring numbers with decimal expansions:

\sqrt{3}=1.73205080\\ \sqrt{26}=5.0990195\\ \sqrt{101}=10.0498756\\

Q8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

Answer:

 5/7 =

9/11 =

The irrational numbers in between the given numbers are:

1. 0.7307300730007\\

2. 0.7507500750007\\

3. 0.7607600760007

Q9.  Classify the following numbers as rational or irrational according to their type

1. \sqrt{23}

Answer: \sqrt{23}=4.79583152331…

As the number is not terminating as well as not repeating, it is said to be an irrational number.

2. \sqrt{225}

Answer: \sqrt{225}=15=\dfrac{15}{1}

As the number can be written in p/q form, therefore it is a rational number.

3. 0.3796

Answer: As the number is terminating, it is said to be a rational number.

4. 7.478478

Answer: As the number is not- terminating but it is repeating, therefore it is said to be a rational number.

5. 1.101001000100001…

Answer: As the number is not terminating as well as not repeating, therefore it is an irrational number.

Exercise 1.4:

Q1. Visualise 3.765 on the number line, using successive magnification.

Answer:

Q2. Visualize 4.\overline{26} on the number line, up to 4 decimal places.

Answer: 4.\overline{26}=4.2626

Exercise 1.5:

Q1. Classify the following numbers as rational or irrational:

1. 2-\sqrt{5}

Answer: We already know that \sqrt{5}=2.236067…

is non-terminating and non-recurring in this case.

When we replace the value of 5 in 2-\sqrt{5}, we get 2-2.2360679… = -0.2360679

Therefore, 2-\sqrt{5} is an irrational number because the number – 0.2360679… is non-terminating non-recurring.

2. (3+\sqrt{23})-\sqrt{23}

Answer: (3+\sqrt{23}-\sqrt{23}=3+\sqrt{23}-\sqrt{23}\\ =3\\ =3/1

As the number that is 3/1 is in the form of p/q therefore, (3+\sqrt{23})-\sqrt{23} can be said as rational.

3. \dfrac{2\sqrt{7}}{2\sqrt{7}}

Answer: \dfrac{2\sqrt{7} }{2\sqrt{7}}=(\dfrac{2}{7})\times (\dfrac{\sqrt{7}}{\sqrt{7}})\\

Now, we already know: (\dfrac{\sqrt{7}}{\sqrt{7}})=1\\ 

Therefore, (\dfrac{2}{7})\times (\dfrac{7}{7})=(\dfrac{2}{7})\times 1=\dfrac{2}{7}\\

As the number that is 2/7 has the form of p/q therefore, we can say that it is a rational number.

4. \dfrac{1}{\sqrt{2}}

Answer: Firstly, we will multiply \sqrt{2} with the numerator as well as the denominator to get,

\dfrac{1}{\sqrt{2}}\times \dfrac{2}{2}=\dfrac{\sqrt{2}}{2}\\ \text{we already know that, }\; \sqrt{2}=1.4142\\ \text{so, we can write: }\dfrac{\sqrt{2}}{2}=\dfrac{1.4142}{2}=0.7071\\

As the number we got that is 0.7071… is neither terminating nor repeating therefore, we can say \dfrac{1}{\sqrt{2}} is an irrational number.

5. 2\pi

Answer: We know \pi =3.1415\\ \text{therefore, } 2\times \pi=6.2830\\

Because the number we got that is 6.2830.. is neither terminating nor repeating therefore, we can say 2\pi is an irrational number.

Q2. Simplify each of the following expressions:

1. (3+\sqrt{3})(2+\sqrt{2})

Answer: (3+\sqrt{3})(2+\sqrt{2}) = (3\times 2)+(3+\sqrt{2})+(\sqrt{3}\times 2)+(\sqrt{3}\times \sqrt{2})\\ =6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}

2. (3+\sqrt{3})(3-\sqrt{3})

Answer: (3+\sqrt{3})(3-\sqrt{3})= 3^2 \;-\;(\sqrt{3})^2\\=9-3\\ =6

3. (\sqrt{5}+\sqrt{2})^2

Answer: (\sqrt{5}+\sqrt{2})^2 =(\sqrt{5})^2+(2\times \sqrt{5}\times \sqrt{2})+(\sqrt{2})^2\\ =5+2\times \sqrt{10}+2=7+2\sqrt{10}

4. (\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})

Answer: (\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})=(\sqrt{5}^2 -\sqrt{2}^2)\\=5-2\\=3

Q3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter, (say d). That is, π =c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

Answer: There really is no contradiction. When we use a scale to measure something, we only get an estimate. We never get an exact number. As a result, we may be unaware of whether c or d is irrational. Almost equivalent to 22/7 or 3.142857 is the value of π.

Q4. Represent (√9.3) on the number line.

 Answer:

Step 1: Draw an AB line segment 9.3 units long. Extend AB to C so BC equals 1 unit.

Step 2: AC now equals 10.3 units. Let O be the center of AC.

Step 3: Create a semi-circle with a radius of OC and the center as O.

Step 4: At point B, draw a line BD perpendicular to AC that intersects the semicircle at D. Join OD.

Step 5: OBD is a right-angled triangle when achieved.

Now, in the diagram  OD will be equal to 10.3/2 that is the radius of semi circle,

OC= \dfrac{10.3}{2}, \; \;BC=1\\ OB \;= \;OC\;-\; BC =>(\dfrac{10.3}{2})-1\;=\dfrac{8.3}{2}\\

Now, we will be using the Pythagoras theorem:

OD^2\;=\;BD^2\;+OB^2\\ =>(\dfrac{10.3}{2})^2\;=\;BD^2\;+\;(\dfrac{8.3}{2})^2\\ =>BD^2\;=\;(\dfrac{10.3}{2})^2\;-\;(\dfrac{8.3}{2})^2\\ =>BD^2 \;=\;(\dfrac{10.3}{2})\;-\;(\dfrac{8.3}{2})(\dfrac{10.3}{2})+(\dfrac{8.3}{2})\\ =>BD\;=\;\sqrt{9.3}

Step 6: Draw an arc that meets the line segment using BD as the radius and B as the center. Its point of contact with the line segment is \sqrt{9.3} away from O.

Q5.  Rationalize the denominators of the following:

1. \dfrac{1}{\sqrt{7}}

Answer: We will multiply the given number with \sqrt{7} with both the numerator and denominator.

\dfrac{1}{\sqrt{7}}\times \dfrac{\sqrt{7}}{\sqrt{7}}\;=\;\dfrac{\sqrt{7}}{7}

2. \dfrac{1}{\sqrt{7}\;-\;\sqrt{6}}

Answer: We will multiply the given number with (\sqrt{7}\;+\;\sqrt{6}) with both the numerator and denominator.

[\dfrac{1}{\sqrt{7}\;-\;\sqrt{6}}]\times (\dfrac{(\sqrt{7}\;+\;\sqrt{6})}{(\sqrt{7}+\sqrt{6})})\;=\;\dfrac{(\sqrt{7}\;+\;\sqrt{6})}{ (\sqrt{7} \;-\;\sqrt{6})(\sqrt{7}\;+\;\sqrt{6})}\\ =\dfrac{(\sqrt{7}+\sqrt{6})}{\sqrt{7}^2-\sqrt{6}^2}\\ =\dfrac{(\sqrt{7}+\sqrt{6})}{(7-6)}\\ =\sqrt{7}+\sqrt{6}

3. \dfrac{1}{(\sqrt{5}+\sqrt{2})}

Answer: We will multiply the given number with (\sqrt{5}\;-\;\sqrt{2}) with both the numerator and denominator.

[\dfrac{1}{(\sqrt{5}+\sqrt{2})}]\times \dfrac{(\sqrt{5}-\sqrt{2})}{\sqrt{5}-\sqrt{2}}\;=\;\dfrac{(\sqrt{5}-\sqrt{2})}{ (\sqrt{5}+\sqrt{2}) (\sqrt{5}-\sqrt{2}) }\\ =\dfrac{(\sqrt{5}-\sqrt{2})}{\sqrt{5}^2-\sqrt{2}^2}\\ =\dfrac{\sqrt{5}-\sqrt{2}}{(5-2)}\\ =\dfrac{\sqrt{5}-\sqrt{2}}{3}

4. \dfrac{1}{(\sqrt{7}-2)}

Answer: We will multiply the given number with (\sqrt{7}\;+\;2) with both the numerator and denominator.

\dfrac{1}{\sqrt{7}-2}\times \dfrac{\sqrt{7}+2}{\sqrt{7}+2}\;=\; \dfrac{(\sqrt{7}+2)}{ (\sqrt{7}-2) (\sqrt{7}+2) }\\ =\dfrac{(\sqrt{7}+2)}{\sqrt{7}^2-\sqrt{2}^2}\\ =\dfrac{\sqrt{7}+2}{\sqrt{7}^2 -\sqrt{2}^2}\\ =\dfrac{\sqrt{7}+2}{3}

Exercise 1.6:

Q1. Find:

1. 64 ^{\dfrac{1}{2}}

Answer: 64 ^{\dfrac{1}{2}}=(8\times 8)^{\dfrac{1}{2}}\\ =(8^2)^{\dfrac{1}{2}}\\ =8^1\\ =8

2. 32^{\dfrac{1}{5}}

Answer: 32^{\dfrac{1}{5}} = (2^5)^{\dfrac{1}{5}}\\ =(2^5)^{\dfrac{1}{5}}\\ =2^1\\ =2

3. 125^{\dfrac{1}{3}}

Answer: 125^{\dfrac{1}{3}}=(5\times 5\times 5)^{\dfrac{1}{3}}\\ =(5^3)^{\dfrac{1}{3}}\\ =5^1\\ =5

Q2. Find:

1. 9^{\dfrac{3}{2}}

Answer: 9^{\dfrac{3}{2}}=(3\times 3)^{\dfrac{3}{2}}\\ =(3^2)^{\dfrac{3}{2}}\\ =3^3\\ =27

2. 32^{\dfrac{2}{5}}

Answer: 32^{\dfrac{2}{5}}=(2\times 2\times 2\times 2\times 2)^{\dfrac{2}{5}}\\ =(2^5)^{\dfrac{2}{5}}\\ =2^2\\ =4

3. 16^{\dfrac{3}{4}}

Answer: 16^{\dfrac{3}{4}}=(2\times 2\times 2\times 2)^{\dfrac{3}{4}}\\ =(2^4)^{\dfrac{3}{4}}\\ =2^3\\ =8

4. 125^{-\dfrac{1}{3}}

Answer: 125^{-\dfrac{1}{3}}= (5\times 5\times 5)^{-\dfrac{1}{3}}\\ =(5^3)^{-\dfrac{1}{3}}\\ =5^{-1}\\ =\dfrac{1}{5}

Q3. Simplify:

1. 2^{\dfrac{2}{3}}\times 2^{\dfrac{1}{5} }

Answer: \begin{align*} 2^{\dfrac{2}{3}}\times 2^{\dfrac{1}{5}}\;&=\;2^{(\dfrac{2}{3})+(\dfrac{1}{5})}\\ &=2^{\dfrac{13}{15}} \end{align*}  

2. (\dfrac{1}{3^3})^7

Answer: \begin{align*} (\dfrac{1}{3^3})^7 &=(3^{-3})^7\\ &=3^{-21} \end{align*}

3. \dfrac{11^{\dfrac{1}{2}}}{11^{\dfrac{1}{4}}}

Answer: \begin{align*} \dfrac{11^{\dfrac{1}{2}}}{11^{\dfrac{1}{4}}}\;&=\;11^{ (\dfrac{1}{2})-(\dfrac{1}{4})}\\ &=11^{\dfrac{1}{4}} \end{align*}

4. 7^{\dfrac{1}{2}} \times 8^{\dfrac{1}{2}}

Answer: \begin{align*} 7^{\dfrac{1}{2}} \times 8^{\dfrac{1}{2}}\;&=\; (7\times 8)^{\dfrac{1}{2}}\\ &=56^{\dfrac{1}{2}} \end{align*}