Chapter 1: Number Systems – Quick Notes & Practice (Class 9, NCERT/CBSE)
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I can’t reproduce the exact NCERT textbook exercise questions verbatim due to copyright. Below, I’ve included original, textbook-style practice questions (with full solutions) that cover every skill from the chapter. If you provide the exercise text, I’ll solve them exactly as given.
I can’t reproduce the exact NCERT textbook exercise questions verbatim due to copyright. Below, I’ve included original, textbook-style practice questions (with full solutions) that cover every skill from the chapter. If you provide the exercise text, I’ll solve them exactly as given.
1-Mark Questions (20)
1) State whether \( \sqrt{49} \) is rational or irrational.
Rational, because \( \sqrt{49}=7 \in \mathbb{Z} \subset \mathbb{Q}\).
2) Classify \( \sqrt{2} \): rational or irrational.
Irrational; \( \sqrt{2} \notin \mathbb{Q}\).
3) Is \(0\) a natural number?
By CBSE convention for Class 9, \( \mathbb{N}=\{1,2,3,\dots\}\). So, \(0\) is not natural (it is a whole number).
4) Give the additive inverse of \(-\dfrac{5}{7}\).
It is \(+\dfrac{5}{7}\).
5) Write the multiplicative identity in \( \mathbb{R} \).
\(1\).
6) Decide if \( \dfrac{14}{35} \) is in lowest form.
No. \( \dfrac{14}{35}=\dfrac{2}{5}\) (divide by 7).
7) Decimal expansion of \( \dfrac{1}{8} \) is terminating or non-terminating?
Terminating: \( \dfrac{1}{8}=0.125\).
8) Is \(0.\overline{3}\) rational?
Yes; \(0.\overline{3}=\dfrac{1}{3}\in \mathbb{Q}\).
9) Name the set to which \( -\sqrt{9} \) belongs.
\( -\sqrt{9}=-3 \in \mathbb{Z} \subset \mathbb{Q}\subset \mathbb{R}\).
10) Write two integers between \(-2\) and \(3\).
\(-1,0,1,2\) (any two).
11) Which is greater: \( \sqrt{5} \) or \(2.2\)?
\( \sqrt{5}\approx 2.236 > 2.2\).
12) Is every integer a rational number?
Yes; \(n=\dfrac{n}{1}\in\mathbb{Q}\).
13) Identify: terminating or non-terminating repeating: \( \dfrac{7}{20}\).
Terminating; \(20=2^{2}\cdot 5\) only primes 2 and/or 5.
14) Simplify \( \left(\dfrac{1}{2}\right)^{-3} \).
\( \left(\dfrac{1}{2}\right)^{-3}=2^{3}=8\).
15) True/False: \( \sqrt{49}=\pm 7\).
False. Principal square root \( \sqrt{49}=+7\).
16) Value of \(5^{0}\).
\(1\) (for \(5\neq 0\)).
17) Find the reciprocal of \(-\dfrac{3}{5}\).
\(-\dfrac{5}{3}\).
18) Write a number that is irrational and lies between \(1\) and \(2\).
\( \sqrt{2}\approx 1.414\) is one example.
19) State the set relation: \( \mathbb{N}\subset \mathbb{W}\subset \mathbb{Z}\subset \mathbb{Q}\subset \mathbb{R}\).
True.
20) Write \(0.7\overline{2}\) as a fraction (in lowest form).
Let \(x=0.7\overline{2}\). Then \(10x=7.\overline{2}\), \(100x=72.\overline{2}\). Subtract: \(90x=65\Rightarrow x=\dfrac{65}{90}=\dfrac{13}{18}\).
2-Mark Questions (20)
1) Without division, decide if \( \dfrac{13}{40} \) has terminating decimal expansion.
Yes. \(40=2^{3}\cdot 5\). Denominator in lowest form has only 2’s and/or 5’s \(\Rightarrow\) terminating.
2) Express \(0.125\) as a rational number in lowest form.
\(0.125=\dfrac{125}{1000}=\dfrac{1}{8}\).
3) Convert \(1.0\overline{6}\) to a fraction.
Let \(x=1.0\overline{6}\). Then \(10x=10.\overline{6},\ 100x=106.\overline{6}\). Subtract: \(90x=96\Rightarrow x=\dfrac{96}{90}=\dfrac{16}{15}\).
4) Find three rational numbers between \(\dfrac{1}{5}\) and \(\dfrac{1}{3}\).
Equalize denominators: \(\dfrac{1}{5}=\dfrac{3}{15}\), \(\dfrac{1}{3}=\dfrac{5}{15}\). Choose \(\dfrac{7}{35},\dfrac{2}{9},\dfrac{4}{15}\) etc. One systematic way: write as decimals \(0.2\) and \(0.333...\); pick \(0.21, 0.25, 0.3\).
5) Place \( \sqrt{10} \) on the number line (explain construction briefly).
Construct a right triangle with legs 1 and 3 (\(1^{2}+3^{2}=10\)). Using the Pythagorean theorem, hypotenuse \(=\sqrt{10}\). Place it via semicircle method from 0 to 3 then add 1-unit perpendicular; the hypotenuse from 0 gives the point at \(\sqrt{10}\).
6) Compare \( \sqrt{3} \) and \( \dfrac{7}{4}\).
\( \sqrt{3}\approx 1.732 < 1.75=\dfrac{7}{4}\). Hence \( \sqrt{3}<\dfrac{7}{4}\).
7) Rationalize the denominator: \( \dfrac{5}{\sqrt{7}} \).
\( \dfrac{5}{\sqrt{7}}\cdot\dfrac{\sqrt{7}}{\sqrt{7}}=\dfrac{5\sqrt{7}}{7}\).
8) Simplify: \( \sqrt{72} \).
\( \sqrt{72}=\sqrt{36\cdot 2}=6\sqrt{2}\).
9) Determine whether \(0.1010010001\ldots\) is rational.
No. The block lengths of 1’s increase irregularly; the decimal is non-terminating and non-repeating \(\Rightarrow\) irrational.
10) Write \( \dfrac{3^{4}\cdot 3^{-2}}{3} \) as a single power of 3.
\(3^{4-2-1}=3^{1}=3\).
11) If \(a>0\), compare \(a\) and \(\dfrac{1}{a}\).
If \(a>1\Rightarrow a>\dfrac{1}{a}\); if \(0
12) Find the product: \( \sqrt{5}\cdot \sqrt{20} \).
\( \sqrt{100}=10\).
13) Show that \( \dfrac{2}{3}\) has a non-terminating repeating decimal.
Denominator \(3\) has prime factor 3 (not 2 or 5) \(\Rightarrow\) non-terminating repeating: \(0.\overline{6}\).
14) Find two irrational numbers between \(2\) and \(3\).
\( \sqrt{5}\approx 2.236\), \( \sqrt{7}\approx 2.646\) (both between 2 and 3).
15) Simplify \( \left(\dfrac{5}{\sqrt{2}}\right)\left(\dfrac{\sqrt{8}}{5}\right) \).
\( \dfrac{5}{\sqrt{2}}\cdot \dfrac{2\sqrt{2}}{5}=2\).
16) Write the set of integers between \(-4\) and \(2\).
\(-3,-2,-1,0,1\).
17) Evaluate \( \left(2^{3}\right)^{4} \).
\(2^{12}=4096\).
18) Find the value of \( \dfrac{7^{5}}{7^{2}}\cdot 7^{-1}\).
\(7^{5-2-1}=7^{2}=49\).
19) Express \( \dfrac{9}{2^{3}\cdot 5^{2}} \) as a decimal and say its type.
Denominator has only \(2,5\Rightarrow\) terminating. Value \(=\dfrac{9}{200}=0.045\).
20) True/False with reason: The sum of a rational and an irrational is irrational.
True (in general). If \(r\in \mathbb{Q},\ i\notin \mathbb{Q}\), then \(r+i\notin \mathbb{Q}\). (Proof by contradiction: if \(r+i\in\mathbb{Q}\Rightarrow i\) would be rational.)
3-Mark Questions (20)
1) Prove: A rational number \( \dfrac{p}{q}\) (in lowest form) has a terminating decimal iff prime factorization of \(q\) is of the form \(2^{m}5^{n}\) (with \(m,n\ge 0\)).
(\(\Rightarrow\)) If decimal terminates at \(k\) places, \( \dfrac{p}{q}=\dfrac{N}{10^{k}}=\dfrac{N}{2^{k}5^{k}}\Rightarrow q\mid 2^{k}5^{k}\Rightarrow q=2^{m}5^{n}\). (\(\Leftarrow\)) If \(q=2^{m}5^{n}\), multiply numerator and denominator by \(2^{n-m}\) or \(5^{m-n}\) to get denominator \(10^{\max(m,n)}\Rightarrow\) terminating.
2) Show that \( \sqrt{3} \) is irrational (outline).
Assume \( \sqrt{3}=\dfrac{p}{q}\) in lowest terms. Then \(3q^{2}=p^{2}\Rightarrow 3\mid p\Rightarrow p=3k\Rightarrow p^{2}=9k^{2}\Rightarrow q^{2}=3k^{2}\Rightarrow 3\mid q\). Contradiction to lowest terms. Hence irrational.
3) Find five rational numbers between \( \dfrac{2}{7}\) and \( \dfrac{3}{7}\).
Write as \( \dfrac{20}{70}\) and \( \dfrac{30}{70}\). Choose equal-denominator fractions: \( \dfrac{21}{70}, \dfrac{22}{70}, \dfrac{23}{70}, \dfrac{24}{70}, \dfrac{25}{70}\) \((=\dfrac{3}{14},\dfrac{11}{35},\dfrac{23}{70},\dfrac{12}{35},\dfrac{5}{14})\).
4) Simplify and rationalize: \( \dfrac{3}{\sqrt{5}-\sqrt{3}} \).
\(\dfrac{3}{\sqrt{5}-\sqrt{3}}\cdot \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{3(\sqrt{5}+\sqrt{3})}{5-3}=\dfrac{3(\sqrt{5}+\sqrt{3})}{2}\).
5) If \(x=\sqrt{2}+\sqrt{3}\), find \(x^{2}\) and deduce whether \(x\) is rational.
\(x^{2}=2+3+2\sqrt{6}=5+2\sqrt{6}\) (irrational), so \(x\) is irrational.
6) Convert \( 3.2\overline{45} \) to fraction in lowest form.
Let \(x=3.2\overline{45}\). Then \(10x=32.\overline{45}\), \(1000x=3245.\overline{45}\). Subtract: \(990x=3213 \Rightarrow x=\dfrac{3213}{990}=\dfrac{1071}{330}\).
7) Show that the sum of two irrational numbers can be rational, with an example.
Example: \(\sqrt{2} + (2-\sqrt{2}) = 2\in \mathbb{Q}\). Both addends irrational, sum rational.
8) Evaluate: \( (\sqrt{5}-\sqrt{2})^{2} + 2\sqrt{10} \).
\((5+2-2\sqrt{10})+2\sqrt{10}=7\).
9) If \(a=\sqrt{7}-\sqrt{5}\), compute \( \dfrac{1}{a} \) in the form \(p\sqrt{7}+q\sqrt{5}\).
\(\dfrac{1}{\sqrt{7}-\sqrt{5}}\cdot\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\dfrac{\sqrt{7}+\sqrt{5}}{2}\Rightarrow p=q=\dfrac{1}{2}\).
10) Prove that between any two distinct reals there is a rational number.
Let \(a
11) Find \(x\) if \( \sqrt{2x+1} - \sqrt{x-2} = 1 \) (principal roots).
Let \(u=\sqrt{2x+1}, v=\sqrt{x-2}\). Then \(u=v+1\Rightarrow u^{2}=v^{2}+2v+1\Rightarrow 2x+1=(x-2)+2\sqrt{x-2}+1\Rightarrow x=2\sqrt{x-2}\). So \(x\ge 2\). Square: \(x^{2}=4(x-2)\Rightarrow x^{2}-4x+8=0\) has discriminant \(16-32<0\). Check initial steps: try plugging \(x=5\): \( \sqrt{11}-\sqrt{3}\approx 3.317-1.732\approx 1.585\neq 1\). Conclude: no real solution (domains satisfied but equality not met).
12) Simplify: \( \dfrac{\sqrt{18}+\sqrt{8}}{\sqrt{2}} \).
\( \dfrac{3\sqrt{2}+2\sqrt{2}}{\sqrt{2}}=5\).
13) If \(x= \dfrac{1}{\sqrt{3}-1} - \dfrac{1}{\sqrt{3}+1}\), find \(x\).
Rationalize each: \( \dfrac{\sqrt{3}+1}{2} - \dfrac{\sqrt{3}-1}{2} = 1\).
14) Prove: \( \sqrt{2}+\sqrt{3} \) is irrational.
If \( \sqrt{2}+\sqrt{3}\in\mathbb{Q}\Rightarrow (\sqrt{2}+\sqrt{3})^{2}=5+2\sqrt{6}\in\mathbb{Q}\Rightarrow \sqrt{6}\in\mathbb{Q}\) (contradiction). Hence irrational.
15) Find the decimal expansion type of \( \dfrac{77}{2^{2}\cdot 5^{3}} \) and write the value.
Terminating (only 2’s and 5’s). \( \dfrac{77}{500}=0.154\).
16) Evaluate: \( \left(\dfrac{3}{4}\right)^{-2} + \left(\dfrac{4}{3}\right)^{-2} \).
\( \left(\dfrac{4}{3}\right)^{2}+ \left(\dfrac{3}{4}\right)^{2}=\dfrac{16}{9}+\dfrac{9}{16}=\dfrac{256+81}{144}=\dfrac{337}{144}\).
17) Show \( \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{8}}=\dfrac{3}{2\sqrt{2}} \) and rationalize.
\( \dfrac{1}{\sqrt{2}}+\dfrac{1}{2\sqrt{2}}=\dfrac{3}{2\sqrt{2}}=\dfrac{3\sqrt{2}}{4}\) after rationalizing.
18) If \(x=\sqrt{50}-\sqrt{8}+\sqrt{2}\), simplify.
\(x=5\sqrt{2}-2\sqrt{2}+\sqrt{2}=4\sqrt{2}\).
19) Prove: \( \sqrt{5} \) is not a terminating or repeating decimal.
If it terminated or repeated, it would be rational. But \(\sqrt{5}\) is irrational (standard proof like \(\sqrt{2}\)). Hence neither terminating nor repeating.
20) Let \(a=\sqrt{2}+\sqrt{5}\). Compute \(a^{2}-14\).
\(a^{2}=2+5+2\sqrt{10}=7+2\sqrt{10}\Rightarrow a^{2}-14=2\sqrt{10}-7\).
Exercise-Style Practice (Paraphrased) + Full Solutions
These are original problems that mirror the skills in NCERT Class 9 Chapter 1 (Number Systems): decimal expansions, rational/irrational classification, rationalizing surds, locating surds on the number line, and exponent laws. Every problem below includes a complete solution in green.
E1) Classify as terminating / non-terminating repeating / non-terminating non-repeating: \( \dfrac{17}{64},\ \dfrac{5}{12},\ \dfrac{2}{11}\).
\(\dfrac{17}{64}\): \(64=2^{6}\Rightarrow\) terminating.
\(\dfrac{5}{12}\): \(12=2^{2}\cdot 3\Rightarrow\) non-terminating repeating.
\(\dfrac{2}{11}\): prime 11 (not 2/5)\(\Rightarrow\) non-terminating repeating.
\(\dfrac{5}{12}\): \(12=2^{2}\cdot 3\Rightarrow\) non-terminating repeating.
\(\dfrac{2}{11}\): prime 11 (not 2/5)\(\Rightarrow\) non-terminating repeating.
E2) Write \( \dfrac{7}{15} \) as a decimal up to 6 places and classify.
Long division gives \(0.466666\ldots=0.4\overline{6}\Rightarrow\) non-terminating repeating.
E3) Insert four rational numbers between \( \dfrac{1}{6} \) and \( \dfrac{1}{5}\).
As decimals: \(0.1666\ldots\) and \(0.2\). Choose \(0.17, 0.175, 0.18, 0.19\) \(\Rightarrow\) \( \dfrac{17}{100},\dfrac{7}{40},\dfrac{9}{50},\dfrac{19}{100}\).
E4) Rationalize and simplify: \( \dfrac{4}{\sqrt{3}+\sqrt{2}} \).
Multiply by \( \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \Rightarrow \dfrac{4(\sqrt{3}-\sqrt{2})}{3-2}=4(\sqrt{3}-\sqrt{2})\).
E5) Show that \( 0.10\overline{9}=0.11\).
Since \(0.\overline{9}=1\), \(0.10\overline{9}=0.10+0.0\overline{9}=0.10+0.01=0.11\).
E6) Locate \( \sqrt{13} \) on the number line via geometric construction (brief steps).
Take segment \(OA=3\) and \(AB=2\Rightarrow OB=\sqrt{3^{2}+2^{2}}=\sqrt{13}\). With \(O\) as origin, mark the hypotenuse on the line using compass to locate \(\sqrt{13}\).
E7) Convert \( 2.4\overline{07}\) to a rational number.
Let \(x=2.4\overline{07}\). Then \(10x=24.\overline{07}\), \(1000x=2407.\overline{07}\). Subtract: \(990x=2383\Rightarrow x=\dfrac{2383}{990}\).
E8) Simplify: \( \sqrt{32} - \sqrt{18} + \sqrt{8} \).
\(4\sqrt{2}-3\sqrt{2}+2\sqrt{2}=3\sqrt{2}\).
E9) Decide if \( 0.202002000200002\ldots \) is rational.
The pattern gap increases; not repeating \(\Rightarrow\) irrational.
E10) Find three irrational numbers between \( \sqrt{2} \) and \( \sqrt{3} \).
Take convex combos: \( (1-t)\sqrt{2}+t\sqrt{3}\) for \(t=0.2,0.5,0.8\) (still irrational) or simpler: \( \sqrt{2.2}, \sqrt{2.4}, \sqrt{2.6}\).
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