NCERT Solutions for Class 9 Maths chapter-1 Number Systems

Class 9 Maths chapter-1 Number Systems: Chapter 1 of the NCERT Class 9 Maths textbook, titled "Number Systems," introduces students to the fundamental concepts of real numbers and their properties. This chapter delves into the classification of numbers, including rational and irrational numbers, and explores their decimal expansions. Students learn about the representation of real numbers on the number line, operations involving real numbers, and the concept of the nth root. The chapter also covers important topics such as laws of exponents for real numbers and the intricacies of simplifying expressions involving surds. Overall, this chapter lays the groundwork for understanding advanced mathematical concepts and problem-solving techniques.

NCERT solutions for Class 9 Maths chapter 1 Number Systems is prepared by academic team of Physics Wallah. We have prepared NCERT Solutions for all exercise of chapter 1. Given below is step by step solutions of all questions given in NCERT textbook for chapter 1 . Read chapter-1 theory make sure you have gone through the theory part of chapter 1 from NCERT textbook and you have learned the formula of the given chapter. Physics Wallah prepared a detail notes and additional questions for class 9 maths with short notes of all maths formula of class 9 maths . Do read these contents before moving to solve the exercise of NCERT chapter 1.

NCERT Solutions for Class 9 Maths Exercise 1.1

Question 1. Is zero a rational number? Can you write it in the form, where p and q are integers and?

Solution : Consider the definition of a rational number.

A rational number is the one that can be written in the form of, where p and q are integers and.

Zero can be written as.

So, we arrive at the conclusion that 0 can be written in the form of, where q is any integer.

Therefore, zero is a rational number.

Question 2. Find six rational numbers between 3 and 4.
Solution : We know that there are infinite rational numbers between any two numbers.

A rational number is the one that can be written in the form of, where p and q are

Integers and .

We know that the numbers all lie between 3 and 4.

We need to rewrite the numbersin form to get the rational numbers between 3 and 4.

So, after converting, we get.

We can further convert the rational numbers into lowest fractions.

On converting the fractions into lowest fractions, we get

.

Therefore, six rational numbers between 3 and 4 are

.

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Question 3. Find five rational numbers between.
Solution : We know that there are infinite rational numbers between any two numbers.

A rational number is the one that can be written in the form of, where p and q are

Integers and.

We know that the numbers can also be written as.

We can conclude that the numbersall lie between

We need to rewrite the numbersin form to get the rational numbers between 3 and 4.

So, after converting, we get.

We can further convert the rational numbers into lowest fractions.

On converting the fractions, we get.

Therefore, six rational numbers between 3 and 4 are.

Question 4. 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.
Solution :
(i) Consider the whole numbers and natural numbers separately.

We know that whole number series is.

We know that natural number series is.

So, we can conclude that every number of the natural number series lie in the whole number series.

Therefore, we conclude that, yes every natural number is a whole number.

(ii) Consider the integers and whole numbers separately.

We know that integers are those numbers that can be written in the form of, where.

Now, considering the series of integers, we have.

We know that whole number series is.

We can conclude that all the numbers of whole number series lie in the series of integers. But every number of series of integers does not appear in the whole number series.

Therefore, we conclude that every integer is not a whole number.

(iii) Consider the rational numbers and whole numbers separately.

We know that rational numbers are the numbers that can be written in the form, where.

We know that whole number series is.

We know that every number of whole number series can be written in the form of as.

We conclude that every number of the whole number series is a rational number. But, every rational number does not appear in the whole number series.

Therefore, we conclude that every rational number is not a whole number.

NCERT Solutions for Class 9 Maths Exercise 1.2

Question 1. 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, where m is a natural number.
(iii) Every real number is an irrational number.
Solution :
(i) Consider the irrational numbers and the real numbers separately.

We know that irrational numbers are the numbers that cannot be converted in the form, where p and q are integers and.

We know that a real number is the collection of rational numbers and irrational numbers.

Therefore, we conclude that, yes every irrational number is a real number.

(ii) Consider a number line. We know that on a number line, we can represent negative as well as positive numbers.

We know that we cannot get a negative number after taking square root of any number.

Therefore, we conclude that not every number point on the number line is of the form, where m is a natural number.

(iii) Consider the irrational numbers and the real numbers separately.

We know that irrational numbers are the numbers that cannot be converted in the form, where p and q are integers and.

We know that a real number is the collection of rational numbers and irrational numbers.

So, we can conclude that every irrational number is a real number. But every real number is not an irrational number.

Therefore, we conclude that, every real number is not a rational number.

Question 2. 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.
Solution : We know that square root of every positive integer will not yield an integer.

We know that is 2, which is an integer. But,will give an irrational number.

Therefore, we conclude that square root of every positive integer is not an irrational number.

Question 3. Show how √5 can be represented on the number line
Solution :
Draw a number line and take point O and A on it such that OA = 1 unit. Draw BA ⊥ OA as BA = 1 unit. Join OB = √2 units.
Now draw BB1 ⊥ OB such that BB1 =1 unit. Join OB1 = √3 units.
Next, draw B1B2⊥ OB1such that B1B2 = 1 unit.
Join OB2 = units.
Again draw B2B3 ⊥OB2 such that B2B3 = 1 unit.
Join OB3 = √5 units.

Take O as centre and OB3 as radius, draw an arc which cuts the number line at D.
Point D
represents √5 on the number line.

NCERT Solutions for Class 9 Maths Exercise 1.3

Question 1. Write the following in decimal form and say what kind of decimal expansion each has:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Solution :
(i)

On dividing 36 by 100, we get

Therefore, we conclude that, which is a terminating decimal.

(ii)

On dividing 1 by 11, we get

We can observe that while dividing 1 by 11, we got the remainder as 1, which will continue to be 1.

Therefore, we conclude that, which is a non-terminating decimal and recurring decimal.

(iii)

On dividing 33 by 8, we get

We can observe that while dividing 33 by 8, we got the remainder as 0.

Therefore, we conclude that, which is a terminating decimal.

(iv)

On dividing 3 by 13, we get

We can observe that while dividing 3 by 13 we got the remainder as 3, which will continue to be 3 after carrying out 6 continuous divisions.

Therefore, we conclude that, which is a non-terminating decimal and recurring decimal.

(v)

On dividing 2 by 11, we get

We can observe that while dividing 2 by 11, first we got the remainder as 2 and then 9, which will continue to be 2 and 9 alternately.

Therefore, we conclude that, which is a non-terminating decimal and recurring decimal.

(vi)

On dividing 329 by 400, we get

We can observe that while dividing 329 by 400, we got the remainder as 0.

Therefore, we conclude that , which is a terminating decimal.

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Question 2.
You know that . Can you predict what the decimal expansions ofare, without actually doing the long division? If so, how?

[Hint: Study the remainders while finding the value of carefully.]

Solution :
We are given that.

We need to find the values of, without performing long division.

We know that, can be rewritten as

.

On substituting value of as , we get

Therefore, we conclude that, we can predict the values of, without performing long division, to get

Question 3. Express the following in the form , where p and q are integers and q0.

(i)

(ii)

(iii)

Solution :
(i)

We need to multiply both sides by 10 to get

We need to subtract (a)from (b), to get

We can also writeas or.

Therefore, on converting in theform, we get the answer a .

(ii)

We need to multiply both sides by 10 to get

We need to subtract (a)from (b), to get

We can also write as or.

Therefore, on converting in theform, we get the answer as .

(iii)

We need to multiply both sides by 1000 to get

We need to subtract (a)from (b), to get

We can also write as

Therefore, on converting in theform, we get the answer as4

Question 4. Express in the form. Are you surprised by your answer? Discuss why the answer makes sense with your teacher and classmates.
Solution :

We need to multiply both sides by 10 to get

We need to subtract (a)from (b), to get

We can also writeas.

Therefore, on converting in theform, we get the answer as.

Yes, at a glance we are surprised at our answer.

But the answer makes sense when we observe that 0.9999……… goes on forever. SO there is not gap between 1 and 0.9999……. and hence they are equal.

Question 5. What can the maximum number of digits be in the recurring block of digits in the decimal expansion of? Perform the division to check your answer.
Solution :
We need to find the number of digits in the recurring block of.

Let us perform the long division to get the recurring block of.

We need to divide 1 by 17, to get

We can observe that while dividing 1 by 17 we got the remainder as 1, which will continue to be 1 after carrying out 16 continuous divisions.

Therefore, we conclude that, which is a non-terminating decimal and recurring decimal.

Question 6. Look at several examples of rational numbers in the form (q ≠ 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?
Solution :
Let us consider the examples of the form that are terminating decimals.

We can observe that the denominators of the above rational numbers have powers of 2, 5 or both.

Therefore, we can conclude that the property, which q must satisfy in, so that the rational numberis a terminating decimal is that q must have powers of 2, 5 or both.

Question 7. Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution : The three numbers that have their expansions as non-terminating on recurring decimal are given below.

0.04004000400004  ….

0.07007000700007  ….

0.013001300013000013 ….

Question 8. Find three different irrational numbers between the rational numbersand.
Solution :
Let us convertinto decimal form, to get

and

Three irrational numbers that lie betweenare:

0.73073007300073 ….

0.74074007400074 ….

0.76076007600076 ….

Question 9. Classify the following numbers as rational or irrational :

(i) 23

(ii) 225

(iii) 0.3796

(iv)7.478478 …

(v)1.101001000100001 …

Solution :
(i)

We know that on finding the square root of 23, we will not get an integer.

Therefore, we conclude thatis an irrational number.

(ii)

We know that on finding the square root of 225, we get 15, which is an integer.

Therefore, we conclude thatis a rational number.

(iii) 0.3796

We know that 0.3796 can be converted into.

While, converting 0.3796 intoform, we get

.

The rational numbercan be converted into lowest fractions, to get.

We can observe that 0.3796 can be converted into a rational number.

Therefore, we conclude that 0.3796 is a rational number.

(iv) 7.478478….

We know that 7.478478…. is a non-terminating recurring decimal, which can be converted intoform.

While, converting 7.478478…. intoform, we get

While, subtracting (a) from (b), we get

We know thatcan also be written as.

Therefore, we conclude that 7.478478…. is a rational number.

(v)1.101001000100001  ….

We can observe that the number 1.101001000100001…. is a non-terminating on recurring decimal.

We know that non-terminating and non-recurring decimals cannot be converted intoform.

Therefore, we conclude that 1.101001000100001…. is an irrational number.

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NCERT Solutions for Class 9 Maths Exercise 1.4

Question 1. Visualise 3.765 on the number line, using successive magnification.
Solution :
3.765 lies between 3 and 4.

Question 2. Visualise 4. on the number line, upto 4 decimal places.
Solution :
4. or 4.2626 lies between 4 and 5.

 

NCERT Solutions for Class 9 Maths Exercise 1.5

Question 1. Classify the following numbers as rational or irrational:
(i)

(ii)

(iii)

(iv)

(v)

 

Solution :

(i)

We know that

which is also an irrational number.

Therefore, we conclude thatis an irrational number.

(ii)

= 3

Therefore, we conclude thatis a rational number.

(iii)

We can cancelin the numerator and denominator, asis the common number in numerator as well as denominator, to get

Therefore, we conclude thatis a rational number.

(iv)

We know that.

We can conclude that, when 1 is divided by, we will get an irrational number.

Therefore, we conclude thatis an irrational number.

(v)

We know that

We can conclude that will also be an irrational number.

Therefore, we conclude that is an irrational number.

Question 2. Simplify each of the following expressions

Solution :
(i) (3 + √3)(2 + √2)
= 2(3 + √3) + √2(3 + √3)
= 6 + 2√3 + 3√2 + √6
Thus, (3 + √3)(2 + √2) = 6 + 2√3 + 3√2 + √6
(ii) (3 + √3)(3 – √3) = (3)2 – (√3)2
= 9 – 3 = 6
Thus, (3 + √3)(3 – √3) = 6
(iii) (√5 + √2)2 = (√5)2 + (√2)2 + 2(√5)(√2)
= 5 + 2 + 2√10 = 7 + 2√10
Thus, (√5 + √2 )2 = 7 + 2√10
(iv) (√5 – √2)(√5 + √2) = (√5)2 – (√2)2 = 5 – 2 = 3
Thus, (√5 – √2) (√5 + √2) = 3

Question 3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is π = . This seems to contradict the fact that n is irrational. How will you resolve this contradiction?
Solution : When we measure the length of a line with a scale or with any other device, we only get an approximate ational value, i.e. c and d both are irrational.
∴ is irrational and hence π is irrational.
Thus, there is no contradiction in saying that it is irrational.

Question 4. Represent on the number line.
Solution :
Draw a line segment AB = 9.3 units and extend it to C such that BC = 1 unit.
Find mid point of AC and mark it as O.
Draw a semicircle taking O as centre and AO as radius. Draw BD ⊥ AC.
Draw an arc taking B as centre and BD as radius meeting AC produced at E such that BE = BD = units.

Question 5. Rationalise the denominator of the following

Solution :

NCERT Solutions for Class 9 Maths Exercise 1.6

Question 1. Find :
(i)

(ii)

(iii)

Solution :
(i)

We know that

We conclude thatcan also be written as

Therefore, the value ofwill be 8.

(ii)

We know that

We conclude thatcan also be written as

Therefore, the value ofwill be 2.

(iii)

We know that

We conclude thatcan also be written as

Therefore, the value ofwill be 5.

Question 2. Find:
(i)

(ii)

(iii)

(iv)

Solution :
(i)

We know that

We conclude thatcan also be written as

=27

Therefore, the value ofwill be 27.

(ii)

We know that

We conclude thatcan also be written as

= 4

Therefore, the value ofwill be 4.

(iii)

We know that

We conclude thatcan also be written as

= 8

Therefore, the value ofwill be 8.

(iv)

We know that

We conclude thatcan also be written as, or.

We know that

We know thatcan also be written as

Therefore, the value ofwill be.

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Question 3. Simplify :
(i)

(ii)

(iii)

(iv)

Solution :
(i)

We know that

.

We can conclude that

Therefore, the value ofwill be.

(ii)

We know that

We conclude that can also be written as.

(iii)

We know that

We conclude that

.

Therefore, the value of will be.

(iv)

We know that

.

We can conclude that

Therefore, the value of will be.

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