NCERT Solutions for Class 9 Maths Chapter 1 Number Systems
Number Systems Class 9: NCERT Solutions for Chapter 1
Number System Class 9 is one of the foundations of mathematics. Having a strong knowledge of the concepts will help students to ace their examinations. SimplyAcad has brought you the solutions of all the exercises of the NCERT textbook for Maths Chapter 1: Number Systems Class 9. The solutions are explained here in a step by step way for clear understanding.
Scroll below to find the answers!
Overview of the exercises of Chapter 1 – Number Systems Class 9
Exercise 1.1: The first exercise deals with helping students distinguish between natural, integer, and whole numbers, and represent the correct format of real numbers.
Exercise 1.2: The second exercise focuses on forming the basics of irrational numbers.
Exercise 1.3: The third exercise deals with the concept of rational and irrational numbers.
Exercise 1.4: The fourth exercise focuses on explaining the process of successive magnification.
Exercise 1.5: The fifth exercise deals with addition and subtraction of a rational and irrational number resulting in an irrational number.
Exercise 1.6: The final exercise of the number systems focuses on the application of the law of exponents for real numbers.
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.1
Number Systems Class 9 Ex 1.1
Question 1: Is zero a rational number? Can you write it in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\)?
Solution:
Step: Express \(0\) as a rational number.
A number is said to be rational if it can be expressed in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).
\(0\) can be expressed in the form \(\frac{0}{1}, \frac{0}{2},\frac{0}{3},\) and so on.
Since it satisfies the definition of rational numbers and can be expressed in the form \(\frac{p}{q}\), where \(q \neq 0\).
Hence, \(0\) is a rational number.
Question 2. Find six rational numbers between \(3\) and \(4\).
Solution:
Step: Write rational numbers
Since, the numbers \(3\) and \(4\) can be expressed as the rational numbers with denominator \(‘6 + 1’ = 7\)
\(\therefore\) \(3 = \frac{21}{7}\) and \(4 = \frac{28}{7}\)
So, the numbers between \(\frac{21}{7}\) and \(\frac{28}{7}\) will fall between \(3\) and \(4\).
The rational numbers \(\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7},\frac{26}{7}\) and \(\frac{27}{7}\) lie between \(3\) and \(4\).
Hence, the required numbers are \(\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7},\frac{26}{7}\) and \(\frac{27}{7}.\)
Question 3 : Find five rational numbers between \(\frac{3}{5}~ and~\frac{4}{5}\).
Solution:
Solution:
There are infinite rational numbers between \(\frac{3}{5}~ and~\frac{4}{5}\).
\(~\left ( \frac{3}{5} \right ) \times \left ( \frac{6}{6} \right )=\frac{18}{30}~and~\left ( \frac{4}{5} \right ) \times \left ( \frac{6}{6} \right )=\frac{24}{30}\)
Now, 18 < 19 < 20 < 21 < 22 < 23 < 24
\(So, ~\frac{18}{30}\)<\(\frac{19}{30}\)<\(\frac{20}{30}\)<\(\frac{21}{30}\)<\(\frac{22}{30}\)<\(\frac{23}{30}\)<\(\frac{24}{30}\)
Therefore, 5 rational numbers between \(\frac{3}{5}~ and~\frac{4}{5}\) are :
\(\frac{19}{30}\),\(\frac{20}{30}\),\(\frac{21}{30}\),\(\frac{22}{30}\),\(\frac{23}{30}\)
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) True.
Since the natural numbers are a subset of whole numbers.
Whole numbers are 0, 1, 2, …
Natural numbers are 1, 2, 3, …
(ii) False.
An integer can either be negative or positive but whole numbers are always non negative.
-1 is an integer but not a whole number.
(iii) False.
Rational numbers may be fractional but whole numbers are not.
For example, \(\frac{1}{2}\) is a rational number but not a whole number.
Question 5 : 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) True.
Since the natural numbers are a subset of whole numbers.
Whole numbers are 0, 1, 2, …
Natural numbers are 1, 2, 3, …
(ii) False.
An integer can either be negative or positive but whole numbers are always non negative.
-1 is an integer but not a whole number.
(iii) False.
Rational numbers may be fractional but whole numbers are not.
For example, \(\frac{1}{2}\) is a rational number but not a whole number.
NCERT Solutions for Class 9 Number System Chapter 1
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.2
Number Systems Class 9 Ex 1.2
Question 1 : (i) State whether the following statement is true or false. Justify your answer.
Every irrational number is a real number.
(ii) State whether the following statement is true or false. Justify your answer.
Every point on the number line is of the form \(\sqrt{m}\), where m is a natural number.
(iii) State whether the following statement is true or false. Justify your answer.
Every real number is an irrational number.
Solution:
(i) True.
Since the collection of real numbers is made up of rational and irrational numbers.
For example,
\(\sqrt2\) is a real number.
(ii) The given statement is false.
All negative numbers cannot be expressed as the square root of any other number.
(iii) False.
As real numbers include both rational and irrational numbers.
Therefore, every real number cannot be an irrational number.
For example, \(\frac{2}{5}\) is a real number but not an irrational 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:
If numbers such as \(\sqrt{25}=5, \sqrt{36}=6\) are considered, then 5 and 6 are rational numbers.
Therefore, the square roots of all positive integers are not irrational.
Question 3 : 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:
If numbers such as \(\sqrt{25}=5, \sqrt{36}=6\) are considered, then 5 and 6 are rational numbers.
Therefore, the square roots of all positive integers are not irrational.
Question 4 : Show how \(\sqrt{5}\) can be represented on the number line.
Solution:
Draw a line segment \(AB\) of \(2\) units on the number line.
Draw a perpendicular BC of \(1\) unit from \(B\).
Join \(CA\).
By using Pythagoras theorem,
\(AC = \sqrt{2^{2} + 1^{2}} = \sqrt{5}~\text{units}\)
Taking \(AC\) as the radius and \(A\) as the centre, draw an arc touching the number line.
Thus, we get \(\sqrt{5}\) on the number line.
Question 5 :
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 OP1 of unit length. Draw a line segment P1P2 perpendicular to OP1 of unit length. Now draw a line segment P2P3 perpendicular to OP2. Then draw a line segment P3P4 perpendicular to OP3. Continuing in this manner, you can get the line segment Pn−1Pn by drawing a line segment of unit length perpendicular to OPn−1. In this manner, you will have created the points P2,P3,.......,Pn,....., and joined them to create a beautiful spiral depicting √2,√3,√4,.......
Solution:
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.3
Number Systems Class 9 Ex 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) \(\frac{329}{400}\)
Solution:
(i) Perform long division and identify the type of decimal
\(\therefore\) \(\frac{36}{100} = 0.36\)
Therefore, it is a terminating decimal.
(ii) Perform long division and identify the type of decimal
\(\therefore\) \(\frac{1}{11} = 0.09090 = 0. \overline{09}\)
Therefore, it is a non-terminating repeating decimal.
(iii) \(4 \frac{1}{8} = \frac{33}{8}\)
In decimal form,
\(\frac{33}{8}= 4.125\)
4.125 is a terminating decimal.
(iv) In decimal form,
\(\frac{3}{13} = 0.230769230769…\)
\(=0.\overline{230769}\)
It is a non-terminating repeating number.
(v) In decimal form,
\(\frac{2}{11} = 0.181818181818…\)
\(= 0. \overline{18}\)
It is a non terminating repeating number.
(vi) In decimal form,
\(\frac{329}{400}=0.8225\)
It is a terminating number.
Question 2 : You know that \(\frac{1}{7}=0.\overline{142587}\) .Can you predict the decimal expansion of \(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\) without actually doing the long division? If so, how?
Solution: Yes, we can know the decimal expansions of the given numbers by doing the following :
\(\frac{2}{7}=2 \times \frac{1}{7}= 2 \times 0.\overline{142587}=0.\overline{285714}\\ \frac{3}{7}= 3 \times \frac{1}{7} = 3 \times 0. \overline{142587}= 0.\overline{428571}\\ \frac{4}{7}= 4 \times \frac{1}{7} = 4 \times 0.\overline{142587}=0.\overline{571428}\\ \frac{5}{7}= 5 \times \frac{1}{7} = 5 \times 0.\overline{142587}= 0.\overline{714285}\\ \frac{6}{7}= 6 \times \frac{1}{7} = 6 \times 0.\overline{142587}= 0.\overline{857142}\)
Question 3 : Express the following in the form \(\frac{p}{q}\), where p and q are integers and \(q \neq 0\).
(i) \(0.\bar{6}\)
(ii) \(0.4 \bar{7}\)
(iii) \(0.\overline{001}\)
Solution:
(i) \(0.\bar{6}\) = 0.666…
Let x = 0.666…
\(\Rightarrow\) 10x = 6.666… [Since only one digit is repeating, so multiply x with 10.]
\(\Rightarrow\) 10x = 6 + .666
\(\Rightarrow\) 10x = 6 + x
\(\Rightarrow\) 9x = 6
\(\Rightarrow\) \(x = \frac{2}{3}\)
(ii) \(\textbf{Solution:}\)
Step: Express in the form \(\frac{p}{q}\)
\(0.4 \bar7= 0.4777 ~\ldots\)
Let us assume that \(x = 0.4777 ~\ldots\)
\(10 x = 4.7777 ~\ldots\)
\(10 x = 4.3 + 0.4777 ~\ldots\)
\(10 x = 4.3 + x\)
\(9 x = 4.3\)
\( x = \frac{4.3}{9} = \frac{43}{90}\)
Hence, \(0.4 \bar7\) can be expressed as \(\frac{43}{90}\).
(iii) \(0.\overline{001}\)= 0.001001…
Let x = 0.001001…
\(\Rightarrow\) 1000x = 1.001001… [Since, there are three repeating decimal digits, so multiply x with 1000.]
\(\Rightarrow\) 1000x = 1+ 0.001001
\(\Rightarrow\) 1000x = 1 + x
\(\Rightarrow\) 999x = 1
\(\Rightarrow x = \frac{1}{999}\)
Question 4 : Express the following in the form \(\frac{p}{q}\), where p and q are integers and \(q \neq 0\).
(i) \(0.\bar{6}\)
(ii) \(0.4 \bar{7}\)
(iii) \(0.\overline{001}\)
Solution:
(i) \(0.\bar{6}\) = 0.666…
Let x = 0.666…
\(\Rightarrow\) 10x = 6.666… [Since only one digit is repeating, so multiply x with 10.]
\(\Rightarrow\) 10x = 6 + .666
\(\Rightarrow\) 10x = 6 + x
\(\Rightarrow\) 9x = 6
\(\Rightarrow\) \(x = \frac{2}{3}\)
(ii) \(\textbf{Solution:}\)
Step: Express in the form \(\frac{p}{q}\)
\(0.4 \bar7= 0.4777 ~\ldots\)
Let us assume that \(x = 0.4777 ~\ldots\)
\(10 x = 4.7777 ~\ldots\)
\(10 x = 4.3 + 0.4777 ~\ldots\)
\(10 x = 4.3 + x\)
\(9 x = 4.3\)
\( x = \frac{4.3}{9} = \frac{43}{90}\)
Hence, \(0.4 \bar7\) can be expressed as \(\frac{43}{90}\).
(iii) \(0.\overline{001}\)= 0.001001…
Let x = 0.001001…
\(\Rightarrow\) 1000x = 1.001001… [Since, there are three repeating decimal digits, so multiply x with 1000.]
\(\Rightarrow\) 1000x = 1+ 0.001001
\(\Rightarrow\) 1000x = 1 + x
\(\Rightarrow\) 999x = 1
\(\Rightarrow x = \frac{1}{999}\)
Question 5 : Express 0.99999…in the form \(\frac{p}{q}\). Are you surprised by your answer? Discuss why the answer makes sense.
Solution: Let x = 0.9999…
10x = 9.9999… (Since there is only one repeating digit after the decimal, thus multiply x by 10)
10x = 9 + x
9x = 9
x = 1
The difference between 1 and 0.999999 is 0.000001 which is negligible.
Thus, 0.999… is too near to 1.
Therefore, 1 can be justified as the answer.
Question 6 : 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.
Solution:
\(\frac{1}{17}=0.\overline{0588235294117647}\)
Division Check:
\(\overset{~~~~~~0.0588235294117647}{17\sqrt{~~~~~~100~~~~~~~~~}}\)
\(~~~~~~~-85\\
\overline{~~~~~~~~~~~~~~}\\
~~~~150\\
-136\\
\overline{~~~~~~~~~~~~~~}\\
~~~~140\\
-136\\
\overline{~~~~~~~~~~~~~~}\\
~~~~40\\
-34\\
\overline{~~~~~~~~~~~~~~}\\
~~~~60\\
-51\\
\overline{~~~~~~~~~~~~~~}\\
~~~~90\\
-85\\
\overline{~~~~~~~~~~~~~~}\\
~~~~50\\
-34\\
\overline{~~~~~~~~~~~~~~}\\
~~~~160\\
-153\\
\overline{~~~~~~~~~~~~~~}\\
~~~~70\\
-68\\
\overline{~~~~~~~~~~~~~~}\\
~~~~20\\
-17\\
\overline{~~~~~~~~~~~~~~}\\
~~~~30\\
-17\\
\overline{~~~~~~~~~~~~~~}\\
~~~~130\\
-119\\
\overline{~~~~~~~~~~~~~~}\\
~~~~110\\
-102\\
\overline{~~~~~~~~~~~~~~}\\
~~~~80\\
-68\\
\overline{~~~~~~~~~~~~~~}\\
~~~~120\\
\overline{~~~~~~~~~~~~~~}\\
=0.0588235294117647\)
There are 16 digits in the repeating block of the decimal expansion of \(\frac{1}{17}\).
Question 7 : 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?
Solution: We observe that when q is 2, 4, 5, 8, 10… then the decimal expansion is terminating. For example:
\(\frac{1}{2}= 0.5\), denominator \(q = 2^1\)
\(\frac{7}{8} = 0.875\), denominator \(q = 2^3\)
\(\frac{4}{5} = 0.8\), denominator \(q = 5^1\)
We can observe that terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of 2 only or 5 only or both.
Question 8 : Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution: Three numbers whose decimal expansions are non-terminating non-recurring are:
0.303003000300003…
0.505005000500005…
0.7207200720007200007200000…
Question 9 : Find three different irrational numbers between the rational numbers \(\frac{5}{7}~ and~\frac{9}{11}\).
Solution: \(\frac{5}{7}=0.\overline{714285}\\ \frac{9}{11}=0.\overline{81}\)
Three different irrational numbers are:
0.73073007300073000073…
0.75075007300075000075…
0.76076007600076000076…
Question 10 : Classify the following numbers as rational or irrational :
(i) \(\sqrt{23}\)
(ii) \(\sqrt{225}\)
(iii) 0.3796
(iv) 7.478478
(v) 1.101001000100001…
Solution:
(i) \(\sqrt{23}\) = 4.79583152331…
The number is non-terminating non-recurring.
Therefore, it is an irrational number.
(ii) \(\sqrt{225} = 15 = \frac{15}{1}\)
The number can be represented in \(\frac{p}{q}\) form.
Therefore, it is a rational number.
(iii) 0.3796 is a terminating number.
Therefore, it is a rational number.
(iv) 7.478478 = \(7.\overline{478}\)
The number is non-terminating recurring.
Therefore, it is a rational number.
(v) 1.101001000100001…
Since the number is non-terminating non-repeating, it is an irrational number.
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.4
Number Systems Class 9 Ex 1.4
Question 1 : Visualise \(3.765\) on the number line using successive magnification.
(i) \(2-\sqrt{5}\)
(ii) \((3+\sqrt{23})-\sqrt{23}\)
(iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)
(iv) \(\frac{1}{\sqrt{2}}\)
(v) \(2 \pi\)
Solution:
(i) \(2-\sqrt{5}\) = 2 – 2.2360679… = – 0.2360679…
Since the number is non-terminating non-recurring, it is an irrational number.
Alternate method :
Since this equation contains an irrational number \(\sqrt{5}\). Thus, given number is an irrational number.
(ii) Step: Classify the number
As \((3+\sqrt{23}) – \sqrt{23} = 3+\sqrt{23} – \sqrt{23} = 3\)
Now, \(3\) can be written as \(\frac{3}{1}\) which is in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).
Hence, \((3+\sqrt{23})-\sqrt{23}\) is a rational number.
(iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}=\frac{2}{7}\)
It is a rational number as it can be represented in \(\frac{p}{q}\) form.
(iv) \(\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} = 0.7071067811…\)
Since the number is non-terminating non-recurring, it is an irrational number.
(v) \(2 \pi = 2 \times 3.1415 \cdots = 6.280 \cdots\)
Since the number is non-terminating non-recurring, it is an irrational number.
Question 2 : Simplify each of the following expressions:
(i) \((3+\sqrt{3})(2+\sqrt{2})\)
(ii) \((3+\sqrt{3})(3-\sqrt{3})\)
(iii) \(\left ( \sqrt{5}+\sqrt{2} \right )^2\)
(iv) \((\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})\)
Solution:
(i) (3+√3)(2+√2)
(3+√3)(2+√2 )
Opening the brackets, we get, (3×2)+(3×√2)+(√3×2)+(√3×√2)
= 6+3√2+2√3+√6
(ii) (3+√3)(3-√3 )
(3+√3)(3-√3 ) = 32-(√3)2 = 9-3
= 6
(iii) (√5+√2)2
(√5+√2)2 = √52+(2×√5×√2)+ √22
= 5+2×√10+2 = 7+2√10
(iv) (√5-√2)(√5+√2)
(√5-√2)(√5+√2) = (√52-√22) = 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 =c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Solution: There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…
Question 4 : Represent \((\sqrt{9.3})\) on the number line.
Solution: \((i)\) Draw a line segment of \(9.3~ \text{units}\) long. Mark it as \(AB\). Now extend \(AB\) to \(C\) such that \(BC\) is \(1\) unit.
\((ii)\) Now, \(AC = 10.3~ \text{units}\). Let the centre of \(AC\) be \(O\).
\((iii)\) Draw a semi\(-\)circle of radius \(OC \) taking centre as \(O\).
\(iv)\) Draw a line segment \(BD\) perpendicular to \(AC\) from \(B\) such that it intersects the semicircle at \(D\). Join \(OD\).
\((v)\) Now we get a right-angled triangle \(OBD\).
where, \(OB = OC – BC = \frac{10.3}{2} – 1 = \frac{8.3}{2}\)
Now, using Pythagoras theorem,
\(OD^{2}=BD^{2}+OB^{2}\)
\(BD^{2} = \sqrt{ \left ( \frac{10.3}{2} \right )^{2} – \left ( \frac{8.3}{2} \right )^{2}}\)
\( = \sqrt{ \left ( \frac{10.3}{2} + \frac{8.3}{2} \right ) \times \left ( \frac{10.3}{2} – \frac{8.3}{2} \right )} = \sqrt{9.3}\)
Therefore, length of \(BD = \sqrt{9.3}\).
\((vi)\) Taking \(BD\) as the radius and \(B\) as the centre draw an arc which cuts the number line at exactly, \(\sqrt{9.3}\) from \(B\).
Question 5 : Rationalize the denominator of the following
(i) \(\frac{1}{\sqrt{7}}\)
(ii) \(\frac{1}{(\sqrt{7}-\sqrt{6})}\)
(iii) \(\frac{1}{(\sqrt{5}+\sqrt{2})}\)
(iv) \(\frac{1}{(\sqrt{7}-2)}\)
Solution:
(i) \(\frac{1}{\sqrt{7}}\)
Multiplying \(\sqrt7\) to numerator and denominator, we get,
\(=\frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}}\)
\(=\frac{\sqrt{7}}{7}\)
(ii) Step: Rationalise
\(\frac{1}{(\sqrt{7}-\sqrt{6})}\)
Multiply the numerator and denominator with \((\sqrt{7}-\sqrt{6})\)
\(\therefore\) \(\frac{1}{(\sqrt{7}-\sqrt{6})} = \frac{1}{(\sqrt{7}-\sqrt{6})} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}}\)
\( = \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^{2} – (\sqrt{6})^{2}}\)
\( = \sqrt{7} + \sqrt{6}\)
Hence, \(\frac{1}{\sqrt{7}-\sqrt{6}} = \sqrt{7} + \sqrt{6}\)
(iii) Multiply the numerator and denominator of \(\frac{1}{(\sqrt{5}+\sqrt{2})}\) with \((\sqrt{5}-\sqrt{2}).\)
We get,
\(\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{1(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}~~~~~~~~[(a+b)(a-b)=(a^2-b^2)]\\
= \frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^2-(\sqrt{2})^2} ~~~~~~~~[{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2}) =\sqrt{5}^2-\sqrt{2}^2=5-2]}
\\=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}\)
(iv) Multiply and divide \(\frac{1}{(\sqrt{7}-2)}\) with \((\sqrt{7}+2).\)
\(\frac{1}{\sqrt{7}-2}=\frac{1(\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}\\ =\frac{\sqrt{7}+2}{(\sqrt{7})^2-(2)^2} ~~~~~~~[\text{using identity,}(a+b)(a-b)=a^2-b^2]\\=\frac{\sqrt{7}+2}{7-4}\\=\frac{\sqrt{7}+2}{3}\)
NCERT Solutions for Class 9 Number System Chapter 1
Exercise 1.5
Solve The Following Questions NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.5 Number system:
Question 1 : Find:
(i) \(64^{\frac{1}{2}}\)
(ii) \(32^{\frac{1}{5}}\)
(iii) \(125^{\frac{1}{3}}\)
Solution:
(i) \(64^{\frac{1}{2}}\)
\( =\left ( 2^6 \right )^{\frac{1}{2}}~~~~[\because\ 64=2^6] \)
\(=2^{6 \times \frac{1}{2}}=2^3=8~~~~[\because\ (a^m)^n= a^{mn}]\)
(ii) \((32)^{\frac{1}{5}}\)
\( =( 2^{5})^{\frac{1}{5}}\)
\(= 2^{5 \times \frac{1}{5}}\) \([\text{Using} \ (a^m)^{n}= a^{m \times n}]\)
\(=2\)
Hence, \(~32^{\frac{1}{5}} = 2\)
(iii) \(125^{\frac{1}{3}}\)
\(= \left ( 5^3 \right )^{\frac{1}{3}} ~~~~[\because\ 125=5^3]\\
=5^{3 \times \frac{1}{3}}~~~~~[\because\ (a^m)^n= a^{mn}]\\
=5\)
Question 2 : Find
(i) \(9^{\frac{3}{2}}\)
(ii) \(32^{\frac{2}{5}}\)
(iii) \(16^{\frac{3}{4}}\)
(iv) \(125^{\frac{-1}{3}}\)
Solution:
(i) \(9^{\frac{3}{2}}\)
\(= \left (^{2} \sqrt{9} \right )^{3}\)
\( \left[\text{Using} ~, a^{\frac{m}{n}} = (^{n} \sqrt{a})^{m} = ^{n} \sqrt{a^{m}} \right]\)
\(= \left (^{2} \sqrt{3^{2}} \right )^{3}\)
\( = 3^{3}\)
\( = 27\)
Hence, \(9^{\frac{3}{2}} = 27\)
(ii) \(32^{\frac{2}{5}}=\left ( 2^5 \right )^{\frac{2}{5}} ~~~~~~[\because\ 32=2^5] \)
\(=(2)^{5 \times \frac{2}{5}} ~~~~~~~~~ [\because\ (a^m)^n= a^{mn}]\)
\(=2^2=4\)
(iii) \(16^{\frac{3}{4}}\)
\(=\left ( 2^4 \right )^{\frac{3}{4}} ~~~~~~~~~~ [\because\ 16=2^4] \\
=2^{4 \times \frac{3}{4}} ~~~~~~~ [\because\ (a^m)^n= a^{mn}]\\
=2^3 =8\)
(iv) \((125)^{\frac{-1}{3}}\)
\(=\frac{1}{(125)^{\frac{1}{3}}} ~~~~~~ [ \because a^{-1}=\frac{1}{a}]\)
\(=\frac{1}{\left ( 5^3 \right )^{\frac{1}{3}}} ~~~~~~[\because\ 125=5^3] \\
=\frac{1}{5^{3 \times \frac{1}{3}}}~~~~~~~~~ [\because\ (a^m)^n= a^{mn}]\\\)
\(=\frac{1}{5}\)
Question 3 : Simplify
(i) \(2^{\frac{2}{3}}.2^{\frac{1}{5}}\)
(ii) \(\left ( \frac{1}{3^3} \right )^7\)
(iii) \(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}\)
(iv) Simplify \(7^{\frac{1}{2}}.8^\frac{1}{2}\)
Solution:
(i) \(2^{\frac{2}{3}}.2^{\frac{1}{5}}\)
\(2^{\frac{2}{3}}.2^{\frac{1}{5}} = 2^\left({\frac{2}{3}+\frac{1}{5}} \right)\)
\(\left[\text{Using}, ~a^{p}.a^{q} = a^{(p + q)}\right]\)
\(=2^{\left(\frac{10+3}{15}\right)}\)
\(=2^{\frac{13}{15}}\)
Hence, \(2^{\frac{2}{3}}.2^{\frac{1}{5}} = 2^{\frac{13}{15}}\)
(ii) \(\left ( \frac{1}{3^3} \right )^7\)
\(\left ( \frac{1}{3^3} \right )^7 = \frac{1}{3^{3 \times 7}}\)
\([\text{Using}~(a^{p})^{q} = a^{(pq)}]\)
\(=\frac{1}{3^{21}}\)
Also, \(a^{-n} = \frac{1}{a^{n}}\)
\(\frac{1}{3^{21}} = 3^{-21} \)
Hence, \(\left ( \frac{1}{3^3} \right )^7 = 3^{-21} \)
(iii) \(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}\)
\(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} =11^{\left(\frac{1}{2}-\frac{1}{4}\right)}\)
\([\text{Using}~ \frac{a^{p}}{a^{q}}=a^{p-q}]\)
\(=11^{\frac{2-1}{4}}\)
\(=11^{\frac{1}{4}}\)
Hence, \(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} = 11^{\frac{1}{4}}\)
(iv) \(7^{\frac{1}{2}}.8^\frac{1}{2}\)
\(= (7 \times 8)^{\frac{1}{2}}\)
\(=56^{\frac{1}{2}}\)
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