NCERT Solutions of Class 9 Maths Ch-1 Number Systems
NCERT Solutions of Class 9 Maths Ch-1 Number Systems
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Important Questions for Class 9 Maths Ch-1 Number System
Table of Contents
Exercise 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 ≠ 0?
प्रश्न 1. क्या शून्य एक परिमेय संख्या है ? क्या इसे आप \(\frac{p}{q}\) के रूप में लिख सकते हैं, जहाँ p और q पूर्णांक हैं और q ≠ 0 है ?
Answer
Yes, zero is a rational number as we can write it in the form \(\frac{p}{q}\).
हाँ, शून्य एक परिमेय संख्या है, क्योंकि इसे हम \(\frac{p}{q}\) के रूप में लिख सकते हैं |
Question 2:
Find six rational numbers between 3 and 4.
प्रश्न 2. 3 और 4 के बीच में छ: परिमेय संख्याएँ ज्ञात कीजिए |
Answer
3 and 4
3 = \(\frac{3 \times 7}{1 \times 7} = \frac{21}{7}\)
4 = \(\frac{4 \times 7}{1 \times 7} = \frac{28}{7}\)
Six rational numbers between 3 and 4 are:
\(\frac{22}{28}, \frac{23}{28}, \frac{24}{28}, \frac{25}{28}, \frac{26}{28}, \frac{27}{28}\)
Question 3:
Find five rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\) .
प्रश्न 3. \(\frac{3}{5}\) और \(\frac{4}{5}\) के बीच पाँच परिमेय संख्या ज्ञात कीजिए ।
Answer
\(\frac{3}{5}\) and \(\frac{4}{5}\)
\(\frac{3}{5}\) = \(\frac{3 \times 6}{5 \times 6} = \frac{18}{30}\)
\(\frac{4}{5}\) = \(\frac{4 \times 6}{5 \times 6} = \frac{24}{30}\)
Six 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
प्रश्न 4. नीचे दिए गए कथन सत्य हैं या असत्य? कारण के साथ अपने उत्तर दीजिए।
(i) प्रत्येक प्राकृत संख्या एक पूर्ण संख्या होती है।
(ii) प्रत्येक पूर्णांक एक पूर्ण संख्या होती है।
(iii) प्रत्येक परिमेय संख्या एक पूर्ण संख्या होती है।
Answer
(i) True, since the collection of whole number contains all natural numbers.
(ii) False, since integers may be negative but whole number are always positive.
(iii) False, since rational number may be fractional but whole numbers may not be.
(i) सत्य, क्योंकि पूर्ण संख्या में सभी प्राकृत संख्याएँ शामिल हैं |
(ii) असत्य, क्योंकि पूर्णांक में ऋणात्मक पूर्णांक भी होते हैं जबकि पूर्ण संख्याओं में कोई भी संख्या ऋणात्मक नहीं होता हैं |
(iii) असत्य, क्योंकि परिमेय संख्या भिन्न संख्याएँ हो सकती है जबकि पूर्ण संख्या नहीं हो सकती है |
NCERT Solutions of Class 9 Maths Ch-1 Number Systems
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 m , where m is a natural number.
(iii) Every real number is an irrational number.
प्रश्न 1. नीचे दिए गए कथन सत्य हैं या असत्य हैं? कारण के साथ अपने उत्तर दीजिए।
(i) प्रत्येक अपरिमेय संख्या एक वास्तविक संख्या होती है।
(ii) संख्या रेखा का प्रत्येक बिन्दु √m के रूप का होता है, जहाँ m एक प्राकृत संख्या है।
(iii) प्रत्येक वास्तविक संख्या एक अपरिमेय संख्या होती है।
Answer
(i) True, since the collection of real numbers is made up of rational and irrational numbers.
(ii) False, since positive number cannot be expressed as square roots.
(iii) False, as real numbers include both rational and irrational numbers. Therefore, every real number cannot be an irrational number.
(i) सत्य, क्योंकि वास्तविक संख्याओं में अपरिमेय संख्याएँ भी होती है |
(ii) असत्य, संख्या रेखा पर दोनों ऋणात्मक एवं धनात्मक संख्याएँ होती है, परन्तु प्रत्येक बिंदु पर एक वर्गमूल संख्या हो यह संभव नहीं है |
(iii) असत्य, क्योंकि वास्तविक संख्याओं के समूह में परिमेय सा संख्याएँ एवं अपरिमेय संख्याएँ दोनों होती हैं | केवल अपरिमेय संख्या नहीं होती हैं |
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.
प्रश्न 2. क्या सभी धनात्मक पूर्णांकों के वर्गमूल अपरिमेय होते हैं? यदि नहीं, तो एक ऐसी संख्या के वर्गमूल का उदाहरण दीजिए जो एक परिमेय संख्या है।
Answer
No, the square roots of all positive integers are not irrational. For example √4 = 2.
सभी धनात्मक पूर्णांकों के वर्गमूल अपरिमेय नहीं होते हैं, उदाहरण के लिए √4 = 2
Question 3:
Show how \(\sqrt{5}\) can be represented on the number line.
Answer
Step 1: Let AB be a line of length 2 unit on number line.
Step 2: At B, draw a perpendicular line BC of length 1 unit. Join CA.
Step 3: Now, ABC is a right angled triangle. Applying Pythagoras theorem,
AB2 + BC2 = CA2
⇒ 22 + 12 = CA2
⇒ CA2 = 5
⇒ CA = √5
Thus, CA is a line of length √5 unit.
Step 4: Taking CA as a radius and A as a centre draw an arc touching
the number line. The point at which number line get intersected by
arc is at √5 distance from 0 because it is a radius of the circle
whose centre was A.
Thus, √5 is represented on the number line as shown in the figure.
NCERT Solutions of Class 9 Maths Ch-1 Number Systems
Exercise 1.3
Question 1:
Write the following in decimal form and say what kind of decimal expansion each has :
प्रश्न 1. निम्नलिखित भिन्नों को दशमलव रूप में लिखिए और बताइए कि प्रत्येक का दशमलव प्रसार किस प्रकार का है:
\({(i)} \frac{36}{100}\), \({(ii)} \frac{1}{11}\), \({(iii)} 4\frac{1}{8}\), \({(iv)} \frac{3}{13}\), \({(v)} \frac{2}{11}\), \({(vi)} \frac{329}{400}\)
Answer
\({(i)} \frac{36}{100}\) = 0.36 (Terminating)
\({(ii)} \frac{1}{11}\) = 0.0909… = \(0.\overline{09}\) (Terminating)
\({(iii)} 4\frac{1}{8}\) = \(\frac{33}{8}\) = 4.125 (Terminating)
\({(iv)} \frac{3}{13}\) = 0.230769230769… = \(0.\overline{230769}\) (Non-terminating repeating)
\({(v)} \frac{2}{11}\) = 0.18181818.. = \(0.\overline{18}\) (Non-terminating repeating)
\({(vi)} \frac{329}{400}\) = 0.8225 (Terminating)
Question 2:
You know that \(\frac {1}{7} \)=\(0.\overline{142857}\). Can you predict what the decimal expansions of \(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, and \frac{6}{7}\) are, without actually doing the long division? If so, how?
प्रश्न 2. आप जानते हैं कि \(\frac {1}{7} \)= \(0.\overline{142857}\) हैं | वास्तव में, लंबा भाग दिए बिना क्या आप यह बता सकते हैं कि \(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, तथा \frac{6}{7}\) के दशमलव प्रसार क्या है और यदि हाँ, तो कैसे ?
Answer
\(\frac {1}{7} \) = \(0.\overline{142857}\)
\(\frac {2}{7} = 2 \times \frac{1}{7}\) = \(2 \times 0.\overline{142857} = 0.\overline{285714}\)
\(\frac {3}{7} = 3 \times \frac{1}{7}\) = \(3 \times 0.\overline{142857} = 0.\overline{428571}\)
\(\frac {4}{7} = 4 \times \frac{1}{7}\) = \(4 \times 0.\overline{142857} = 0.\overline{571428}\)
\(\frac {5}{7} = 5 \times \frac{1}{7}\) = \(5 \times 0.\overline{142857} = 0.\overline{714285}\)
\(\frac {6}{7} = 6 \times \frac{1}{7}\) = \(6 \times 0.\overline{142857} = 0.\overline{857142}\)
Question 3:
Express the following in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0.
प्रश्न 3. निम्नलिखित को \(\frac{p}{q}\) के रूप में व्यक्त कीजिए, जहाँ p और q पूर्णांक हैं तथा q ≠ 0 है:
(i) \(0.\overline{6}\), (ii) \(0.4\overline{7}\), (iii) \(0.\overline{001}\),
Answer
(i) Let x = \(0.\overline{6}\) = 0.666… –(1)
Multiply by 10 on both sides,
10x = 6.666… –(2)
Subtracting (1) from (2)
10x = 6.666…
x = 0.666…
– –
————-
9x = 6
\(\Rightarrow \) x = \(\frac{6}{9} = \frac{2}{3}\)
(ii) Let x = \(0.4\overline{7}\) = 0.4777…
Multiply by 10 on both sides,
10x = 4.777… –(1)
Again, multiply by 10 on both sides,
100x = 47.777… –(2)
Subtracting (1) from (2)
100x = 47.777…
10x = 4.777…
– –
————-
90x = 43
\(\Rightarrow \) x = \(\frac{43}{90}\)
(iii) Let x = \(0.\overline{001}\) = 0.001001… –(1)
Multiply by 1000 on both sides,
1000x = 1.001001… –(2)
Subtracting (1) from (2)
1000x = 1.001001…
x = 0.001001…
– –
————-
999x = 1
\(\Rightarrow \) x = \(\frac{1}{999}\)
Question 4:
Express 0.99999 …. in the form \(\frac{p}{q}\). Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
0.99999….. को \(\frac{p}{q}\) के रूप में व्यक्त कीजिए। क्या आप अपने उत्तर से आश्चर्यचकित है? अपने अध्यापक और कक्षा के सहोगियों के साथ उत्तर की सार्थकता पर चर्चा कीजिए |
Answer
Let x = 0.99999….. …. (i)
Multiplying (i) by 10 on both sides, we get
10x = 9.9999 … (ii)
Subtracting (i) from (ii), we get
10x = 9.9999…
\(\quad\)x = 0.9999…
\(\quad\)- \(\quad\)-
___________________
9x = 9
⇒ x = 9/9 = 1
Thus, 0.9999 =1
As 0.9999… goes on forever, there is no such a big difference between 1 and 0.9999
Hence, both are equal.
Question 5:
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.
प्रश्न 5. \(\frac{1}{17}\) के दशमलव प्रसार में अंकों के पुनरावृति खंड में अंकों की अधिकतम संख्या क्या हो सकती है ? अपने उत्तर की जाँच करने के लिए विभाजन क्रिया कीजिए |
Answer
\(\frac{1}{17} = 0.\overline{0588235294117647}\)
Thus, there are 16 digits in the repeating block in the decimal expansion of 117.
Hence, our answer is verified.
Question 6:
Look at several examples of rational numbers in the form \(\frac{p}{q}\) (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?
प्रश्न 6. \(\frac{p}{q}\) (q ≠ 0) के रूप की परिमेय संख्याओं के अनेक उदहारण लीजिए, जहाँ p और q पूर्णांक हैं, जिनका 1 के अतिरिक्त अन्य कोई उभयनिष्ठ गुणनखंड नहीं है और जिसका सांत दशमलव निरूपण (प्रसार) है | क्या आप यह अनुमान लगा सकते हैं कि १ को कौन-सा गुण अवश्य संतुष्ट करना चाहिए ?
Answer
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 = 21
\(\frac{7}{8}\) = 0. 875, denominator q =23
\(\frac{4}{5}\) = 0. 8, denominator q = 51
We can observe that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.
Question 7:
Write three numbers whose decimal expansions are non-terminating non-recurring.
प्रश्न 7. ऐसी तीन संख्याएँ लिखिए जिनके दशमलव प्रसार अनवसानी अनावर्ती हों |
Answer
We know that all irrational numbers are non-terminating non-recurring.
Three numbers with decimal expansions that are non-terminating non-recurring are:
\(\sqrt{3}\) = 1.732050807568
\(\sqrt{26}\) =5.099019513592
\(\sqrt{101}\) = 10.04987562112
Question 8:
Find three different irrational numbers between the rational numbers \(\frac{5}{7} and \frac{9}{11}\).
प्रश्न 8. परिमेय संख्याओं \(\frac{5}{7} and \frac{9}{11}\) के बीच की तीन अलग-अलग अपरिमेय संख्याएँ ज्ञात कीजिए |
Answer
\(\frac{5}{7} = 0.\overline{714285} \;and \;\frac{9}{11} = 0.\overline{81}\)
Three irrational numbers between \(\frac{5}{7} = 0.\overline{714285} \;and \; \frac{9}{11} = 0.\overline{81}\):
(i) 0.750750075000 …..
(ii) 0.767076700767000 ……
(iii) 0.78080078008000 ……
Question 9:
Classify the following numbers as rational or irrational :
प्रश्न 9. बताइए कि निम्नलिखित संख्याओं में कौन-कौन संख्याएँ परिमेय और कौन-कौन संख्याएँ अपरिमेय हैं |
(i) \(\sqrt{23}\), (ii) \(\sqrt{225}\), (iii) 0.3796, (iv) 7.478478…, (v) 1.101001000100001…
Answer
(1) ∵ 23 is not a perfect square.
∴ \(\sqrt{23}\) is an irrational number.
(ii) ∵ 225 = 15 x 15 = 152
∴ 225 is a perfect square.
Thus, \(\sqrt{225}\) is a rational number.
(iii) ∵ 0.3796 is a terminating decimal.
∴ It is a rational number.
(iv) 7.478478… = \(7.\overline{478}\)
Since, \(7.\overline{478}\) is a non-terminating recurring (repeating) decimal.
∴ It is a rational number.
(v) Since, 1.101001000100001… is a non terminating, non-repeating decimal number.
∴ It is an irrational number.
NCERT Solutions of Class 9 Maths Ch-1 Number Systems
Exercise 1.4
Question 1:
Classify the following numbers as rational or irrational:
(i) 2 – \(\sqrt{5}, (ii) (3 + \sqrt{23})-\sqrt{23}, (iii) \frac{2\sqrt{7}}{7\sqrt{7}}, (iv) \frac{1}{\sqrt{2}}, (v) 2\pi\)
Answer
(i) 2 – √5 is an irrational number.
(ii) (3 + \(\sqrt{23})-\sqrt{23} = 3 + \sqrt{23} – \sqrt{23} = 3\)
which is a rational number.
(iii) Since, \(\frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}\) , which is a rational number.
(iv) ∵ The quotient of rational and irrational number is an irrational number.
∴ \(\frac{1}{\sqrt{2}}\) is an irrational number.
(v) ∵ 2π = 2 x π = Product of a rational and an irrational number is an irrational number.
∴ 2π is an irrational number.
Question 2:
Simplify each of the following expressions:
\((i) (3+\sqrt{3})(2+\sqrt{2}),\; (ii) (3+\sqrt{3})(3-\sqrt{3}),\; (iii) (\sqrt{5} +\sqrt{2})^2,\; (iv) (\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})\)
Answer
\((i) (3+\sqrt{3})(2+\sqrt{2}) \\= 3(2+\sqrt{2})+\sqrt{3}(2+\sqrt{2}) \\ = 6 + 3\sqrt{2} + 2\sqrt{3} +\sqrt{6}
\\.
\\ (ii) (3+\sqrt{3})(3-\sqrt{3})\\= 3(3 – \sqrt{3}) + \sqrt{3}(3 – \sqrt{3}) \\ = 9 – 3\sqrt{3} + 3\sqrt{3} -3 \\ = 6
\\.
\\ (iii) (\sqrt{5} +\sqrt{2})^2 \\ = (\sqrt{5}+\sqrt{2})(\sqrt{5} + \sqrt{2}) \\= \sqrt{5}(\sqrt{5} + \sqrt{2})+\sqrt{2}(\sqrt{5}+\sqrt{2}) \\ = 5 + \sqrt{10} + \sqrt{10} + 2 \\= 7 + 2\sqrt{10}
\\.
\\ (iv) (\sqrt{5} – \sqrt{2})(\sqrt{5} + \sqrt{2}) \\= \sqrt{5}(\sqrt{5} + \sqrt{2}) – \sqrt{2}(\sqrt{5} + \sqrt{2}) \\ = 5 + \sqrt{10} – \sqrt{10} – 2 \\= 3\)
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 = \(\frac{c}{d}\)⋅ This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
आपको याद होगा कि π को एक वृत्त की परिधि (मान लीजिए c) और उसके व्यास (मान लीजिए d) के अनुपात से परिभाषित किया जाता है, अर्थात π = c/d है | यह इस तथ्य का अन्तर्विरोध करता हुआ प्रतीत होता है कि π एक अपरिमेय संख्या है | इस अन्तर्विरोध का निराकरण आप किस प्रकार करेंगे ?
Answer
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 realise that either c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…
π = c/d, दरअसल यह वृत्त के परिधि और व्यास का अनुपात है।
जहाँ, c/d = 22/7, सिर्फ π का अनुमानित मान होता है और जिसका दशमलव मान अनवसानी अनावर्ती प्रसार होता है |
Question 4:
Represent \(\sqrt{9.3}\) on the number line.
संख्या रेखा पर \(\sqrt{9.3}\) को निरुपित कीजिए |
Answer
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 = \(\sqrt{9.3}\) units.
Question 5:
Rationalise the denominators of the following:
निम्नलिखित के हरों का परिमेयकरण कीजिए :
\((i) \frac{1}{\sqrt{7}}\; (ii) \frac{1}{\sqrt{7} – \sqrt{6}} \;(iii) \frac{1}{\sqrt{5} + \sqrt{2}} \;(iv) \frac{1}{\sqrt{7} – {2}}\)
Answer
\((i) \frac{1}{\sqrt{7}}= \frac{1}{\sqrt{7}} \times\frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7}\)
\((ii) \frac{1}{\sqrt{7} – \sqrt{6}} \\= \frac{1}{\sqrt{7} – \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}}
\\= \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 – (\sqrt{6})^2} \quad [(A+B)(A-B) = A^2 – B^2]
\\= \frac{\sqrt{7} + \sqrt{6}}{7 – 6} = \frac{\sqrt{7} + \sqrt{6}}{1}\)
\( (iii) \frac{1}{\sqrt{5} + \sqrt{2}} \\= \frac{1}{\sqrt{5} + \sqrt{2}} \times \frac{\sqrt{5} – \sqrt{2}}{\sqrt{5} – \sqrt{2}}
\\= \frac{\sqrt{5} – \sqrt{2}}{(\sqrt{5})^2 – (\sqrt{2})^2} \quad [(A+B)(A-B) = A^2 – B^2]
\\= \frac{\sqrt{5} – \sqrt{2}}{5 – 2} = \frac{\sqrt{5} – \sqrt{2}}{3}\)
\( (iv) \frac{1}{\sqrt{7} – {2}} \\= \frac{1}{\sqrt{7} – 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2}
\\= \frac{\sqrt{7} + 2}{(\sqrt{7})^2 – (2)^2} \quad [(A+B)(A-B) = A^2 – B^2]
\\= \frac{\sqrt{7} + 2}{7 – 4} = \frac{\sqrt{7} + 2}{3}\)
NCERT Solutions of Class 9 Maths Ch-1 Number Systems
Exercise 1.5
Question 1:
Find:
ज्ञात कीजिए :
\((i) 64^{\frac{1}{2}}, \;(ii) 32^{\frac{1}{5}}, \;(iii) 125^{\frac{1}{3}}\)
Answer
\((i) 64^{\frac{1}{2}} \\= (2\times2\times2\times2\times2\times2)^{\frac{1}{2}} = (2^6)^{\frac{1}{2}} = 2^3 = 8\)
\((ii) 32^{\frac{1}{5}} \\= (2\times2\times2\times2\times2)^{\frac{1}{5}} = (2^5)^{\frac{1}{5}} = 2^1 = 2\)
\((iii) 125^{\frac{1}{3}} \\ = (5\times5\times5)^{\frac{1}{3}} = (5^3)^{\frac{1}{3}} = 5^1 = 5\)
Question 2:
Find :
ज्ञात कीजिए :
\((i) 9^{\frac{3}{2}},\; (ii) 32^{\frac{2}{5}}, \; (iii) 16^{\frac{3}{4}}, \; (iv) 125^{\frac{-1}{3}}\)
Answer
\((i) 9^{\frac{3}{2}} = (3\times3)^{\frac{3}{2}} = 3^{2\times\frac{3}{2}} = 3^3 = 27\)
\((ii) 32^{\frac{2}{5}} = (2\times2\times2\times2\times2)^{\frac{2}{5}} = 2^{5\times\frac{2}{5}} = 2^2 = 4\)
\((iii) 16^{\frac{3}{4}} = (2\times2\times2\times2)^{\frac{3}{4}} = 2^{4\times\frac{3}{4}} = 2^3 = 8\)
\((iv) 125^{\frac{-1}{3}} = (5\times5\times5)^{\frac{-1}{3}} = 5^{3\times\frac{-1}{3}} = 5^{-1} = \frac{1}{5}\)
Question 3:
Simplify:
सरल कीजिए :
\((i) 2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}\quad (ii) \left( \frac{1}{3^3} \right)^7\quad (iii) \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}\quad (iv) 7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} \)
Answer
\((i) 2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}} = 2^{\frac{2}{3}+\frac{1}{5}} = 2^{\frac{10+3}{15}} = 2^{\frac{13}{15}}\)
\((ii) \left( \frac{1}{3^3} \right)^7 = \left( 3^{-3} \right)^7 = 3^{-21}\)
\((iii) \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}= 11^{\frac{1}{2}-\frac{1}{4}} = 11^{\frac{2-1}{4}} = 11^{\frac{1}{4}}\)
\((iv) 7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} = (7\times8)^\frac{1}{2} = 56^\frac{1}{2}\)
NCERT Solutions of Class 9 Maths Ch-1 Number Systems
