Rational and Irrational Numbers Class 9 Selina Solutions

Class 9 ICSE Mathematics · Selina

Rational and Irrational Numbers

Complete, clean and step-by-step solutions for Exercise 1(A), Exercise 1(B), Exercise 1(C), Test Yourself and the case-study questions.

Fractions & decimalsSurdsRationalisationProofsConstructions

Exercise 1(A)Exercise 1(B)Exercise 1(C)Test YourselfCase Study

Rational Number

\(p/q\), where \(p,q\in\mathbb Z\) and \(q\ne0\).

Terminating Test

In lowest form, the denominator must be \(2^m5^n\).

Irrational Number

A non-terminating, non-recurring decimal.

Conjugate

For \(a+b\sqrt c\), use \(a-b\sqrt c\).

No matching question was found. Try a different keyword.

Complete Solutions

Exercise 1(A)

Rational numbers, decimal expansions, ordering and the terminating-decimal test.

Question 1(a)

Solved

Let \(0=\dfrac{p}{q}\), where \(p\) and \(q\) are integers. What additional condition makes it a rational number?

1. \(q=0\)
2. \(p\ne0\)
3. \(q\ne0\)
4. \(p\ne0\) and \(q\ne0\)

View step-by-step solution＋

A rational number has the form \(\dfrac{p}{q}\), where \(p,q\in\mathbb Z\) and the denominator is non-zero.

\[0=\frac{0}{q},\qquad q\ne0\]

Correct option: C — \(q\ne0\).

Question 1(b)

Solved

Every non-terminating decimal number is a:

1. recurring decimal
2. real number
3. non-recurring decimal
4. circulating decimal

View step-by-step solution＋

A non-terminating decimal may be recurring (rational) or non-recurring (irrational), but in both cases it is a real number.

Correct option: B — real number.

Question 1(c)

Solved

\(7.478478478\ldots\) is a:

1. terminating rational decimal
2. recurring decimal
3. neither rational nor non-terminating
4. not a real number

View step-by-step solution＋

The block \(478\) repeats continuously:

\[7.\overline{478}\]

Therefore, it is a recurring decimal and hence rational.

Correct option: B.

Question 1(d)

Solved

Classify \(\dfrac{71}{75}\).

1. terminating
2. non-terminating
3. periodic decimal
4. not a rational number

View step-by-step solution＋

\[\frac{71}{75}=0.94666\ldots=0.94\overline{6}\]

The decimal is recurring (periodic).

Correct option: C.

Question 1(e)

Solved

Which of the following fractions has a terminating decimal expansion?

1. \(\dfrac{13}{85}\)
2. \(\dfrac{9}{524}\)
3. \(\dfrac{51}{405}\)
4. None of these

View step-by-step solution＋

After reducing a fraction, its decimal terminates only when the denominator contains no prime factor other than \(2\) and \(5\).

\[ 85=5\times17,\qquad \frac{51}{405}=\frac{17}{135},\;135=3^3\times5,\qquad 524=2^2\times131 \]

Each denominator has a prime factor other than \(2\) or \(5\).

Correct option: D — none of these.

Question 2

Solved

State whether each statement is true or false, with a reason.

1. Every whole number is a natural number.
2. Every whole number is a rational number.
3. Every integer is a rational number.
4. Every rational number is a whole number.

View step-by-step solution＋

1. False. \(0\) is a whole number but, under the convention used here, it is not a natural number.
2. True. Every whole number \(n\) can be written as \(\dfrac{n}{1}\).
3. True. Every integer \(z\) can be written as \(\dfrac{z}{1}\).
4. False. For example, \(\dfrac25\) is rational but is not a whole number.

Question 3

Solved

Arrange \(-\dfrac59,\;\dfrac7{12},\;-\dfrac23,\;\dfrac{11}{18}\) in ascending order. Also find the difference between the largest and smallest numbers, correct to one decimal place.

View step-by-step solution＋

The LCM of \(9,12,3,18\) is \(36\).

\[ -\frac59=-\frac{20}{36},\quad \frac7{12}=\frac{21}{36},\quad -\frac23=-\frac{24}{36},\quad \frac{11}{18}=\frac{22}{36} \]

\[-24<-20<21<22\]

Therefore,

\[-\frac23<-\frac59<\frac7{12}<\frac{11}{18}\]

Difference between the largest and smallest:

\[ \frac{11}{18}-\left(-\frac23\right)=\frac{11}{18}+\frac{12}{18}=\frac{23}{18}=1.277\ldots\approx1.3 \]

Answer: \(-\dfrac23<-\dfrac59<\dfrac7{12}<\dfrac{11}{18}\); difference \(=1.3\).

Question 4

Solved

Arrange \(\dfrac58,\;-\dfrac3{16},\;-\dfrac14,\;\dfrac{17}{32}\) in descending order. Find the sum of the largest and smallest numbers, correct to two decimal places.

View step-by-step solution＋

The LCM of \(8,16,4,32\) is \(32\).

\[ \frac58=\frac{20}{32},\quad -\frac3{16}=-\frac6{32},\quad -\frac14=-\frac8{32},\quad \frac{17}{32}=\frac{17}{32} \]

\[\frac58>\frac{17}{32}>-\frac3{16}>-\frac14\]

Sum of the largest and smallest:

\[ \frac58+\left(-\frac14\right)=\frac58-\frac28=\frac38=0.375\approx0.38 \]

Answer: \(\dfrac58>\dfrac{17}{32}>-\dfrac3{16}>-\dfrac14\); sum \(=0.38\).

Question 5

Solved

Without actual division, identify which fractions have terminating decimal representations:

1. \(\dfrac7{16}\)
2. \(\dfrac{23}{125}\)
3. \(\dfrac9{14}\)
4. \(\dfrac{32}{45}\)
5. \(\dfrac{43}{50}\)

View step-by-step solution＋

In lowest form, a rational number terminates exactly when its denominator is \(2^m5^n\), where \(m,n\) are non-negative integers.

(i) \(16=2^4\)Terminating

(ii) \(125=5^3\)Terminating

(iii) \(14=2\times7\)Non-terminating recurring

(iv) \(45=3^2\times5\)Non-terminating recurring

(v) \(50=2\times5^2\)Terminating

Terminating fractions: \(\dfrac7{16},\dfrac{23}{125},\dfrac{43}{50}\).

Complete Solutions

Exercise 1(B)

Irrational numbers, surd expressions, number sets and proof-based questions.

Question 1(a)

Solved

The negative of an irrational number is:

1. a rational number
2. an irrational number
3. both rational and irrational
4. a whole number

View step-by-step solution＋

If \(x\) is irrational and \(-x\) were rational, then \(x=-(-x)\) would also be rational, a contradiction.

Correct option: B — an irrational number.

Question 1(b)

Solved

\(\sqrt8(\sqrt8-1)\) is always:

1. rational
2. irrational
3. a whole number
4. a natural number

View step-by-step solution＋

\[\sqrt8(\sqrt8-1)=8-\sqrt8\]

Since \(\sqrt8\) is irrational, \(8-\sqrt8\) is irrational.

Correct option: B.

Question 1(c)

Solved

In the given right-triangle figure, \(OB=1\) unit and \(AB=2\) units. Find \(OA\).

1. \(\sqrt5\)
2. \(\sqrt3\)
3. \(\sqrt5\) or \(\sqrt3\)
4. neither

View step-by-step solution＋

By Pythagoras’ theorem:

\[OA^2=OB^2+AB^2=1^2+2^2=5\]

\[OA=\sqrt5\]

Correct option: A.

Question 1(d)

Solved

Classify \(2\sqrt3\times3\sqrt8\).

1. rational
2. irrational
3. neither rational nor irrational
4. \(12\sqrt5\)

View step-by-step solution＋

\[2\sqrt3\times3\sqrt8=6\sqrt{24}=6\cdot2\sqrt6=12\sqrt6\]

Because \(\sqrt6\) is irrational and the coefficient is non-zero rational, the product is irrational.

Correct option: B.

Question 1(e)

Solved

Choose two irrational numbers lying between \(8\) and \(11\).

1. \(\sqrt{65}\) and \(\sqrt{120}\)
2. \(\sqrt{69}\) and \(10.5\)
3. \(8.2\) and \(\sqrt{125}\)
4. \(3\) and \(\sqrt{110}\)

View step-by-step solution＋

\[\sqrt{65}\approx8.062,\qquad \sqrt{120}\approx10.954\]

Both radicands are non-perfect squares, so both numbers are irrational and lie between \(8\) and \(11\).

Correct option: A.

Question 2

Solved

State whether each expression is rational or irrational.

1. \((2+\sqrt2)^2\)
2. \((3-\sqrt3)^2\)
3. \((5+\sqrt5)(5-\sqrt5)\)
4. \((\sqrt3-\sqrt2)^2\)

View step-by-step solution＋

1. \((2+\sqrt2)^2=4+2+4\sqrt2=6+4\sqrt2\): irrational.
2. \((3-\sqrt3)^2=9+3-6\sqrt3=12-6\sqrt3\): irrational.
3. \((5+\sqrt5)(5-\sqrt5)=25-5=20\): rational.
4. \((\sqrt3-\sqrt2)^2=3+2-2\sqrt6=5-2\sqrt6\): irrational.

Question 3

Solved

Find the square of each expression.

1. \(\dfrac{3\sqrt5}{5}\)
2. \(\sqrt3+\sqrt2\)
3. \(\sqrt5-2\)
4. \(3+2\sqrt5\)

View step-by-step solution＋

1. \(\left(\dfrac{3\sqrt5}{5}\right)^2=\dfrac{45}{25}=\dfrac95=1\dfrac45\).
2. \((\sqrt3+\sqrt2)^2=3+2+2\sqrt6=5+2\sqrt6\).
3. \((\sqrt5-2)^2=5+4-4\sqrt5=9-4\sqrt5\).
4. \((3+2\sqrt5)^2=9+20+12\sqrt5=29+12\sqrt5\).

Question 4

Solved

State whether each statement is true or false.

1. \(\sqrt2+\sqrt3=\sqrt5\)
2. \(2\sqrt4+2=6\)
3. \(3\sqrt7-2\sqrt7=\sqrt7\)
4. \(\dfrac27\) is irrational.
5. \(\dfrac5{11}\) is rational.
6. All rational numbers are real numbers.
7. All real numbers are rational numbers.
8. Some real numbers are rational numbers.

View step-by-step solution＋

1. False. \(\sqrt2+\sqrt3\approx3.146\), whereas \(\sqrt5\approx2.236\).
2. True. \(2\sqrt4+2=2(2)+2=6\).
3. True. Like surds combine: \((3-2)\sqrt7=\sqrt7\).
4. False. It is of the form \(p/q\), with \(q\ne0\).
5. True. It is a rational number.
6. True. Rational numbers are a subset of real numbers.
7. False. Real numbers also include irrational numbers.
8. True. In fact, every rational number is real.

Question 5

Solved

Given the universal set

\[U=\left\{-6,-5\frac34,-\sqrt4,-\frac35,-\frac38,0,\frac45,1,1\frac23,\sqrt8,3.01,\pi,8.47\right\}\]

Find the sets of (i) rational numbers, (ii) irrational numbers, (iii) integers and (iv) non-negative integers.

View step-by-step solution＋

1. Rational numbers: \(\left\{-6,-5\frac34,-\sqrt4,-\frac35,-\frac38,0,\frac45,1,1\frac23,3.01,8.47\right\}\). Note that \(-\sqrt4=-2\).
2. Irrational numbers: \(\{\sqrt8,\pi\}\).
3. Integers: \(\{-6,-\sqrt4,0,1\}=\{-6,-2,0,1\}\).
4. Non-negative integers: \(\{0,1\}\).

Question 6

Solved

Prove that each number is irrational.

1. \(\sqrt3+\sqrt2\)
2. \(3-\sqrt2\)

View step-by-step solution＋

(i) Assume \(x=\sqrt3+\sqrt2\) is rational. Squaring,

\[x^2=5+2\sqrt6\quad\Longrightarrow\quad\sqrt6=\frac{x^2-5}{2}\]

The right side would be rational, but \(\sqrt6\) is irrational. This contradiction proves \(\sqrt3+\sqrt2\) is irrational.

(ii) Assume \(x=3-\sqrt2\) is rational. Then

\[\sqrt2=3-x\]

The right side would be rational, contradicting the irrationality of \(\sqrt2\). Hence \(3-\sqrt2\) is irrational.

Question 7

Solved

Write a pair of irrational numbers whose sum is irrational.

View step-by-step solution＋

One suitable pair is \(\sqrt3+2\) and \(\sqrt2-3\).

\[(\sqrt3+2)+(\sqrt2-3)=\sqrt3+\sqrt2-1\]

The result is irrational.

Question 8

Solved

Write a pair of irrational numbers whose sum is rational.

View step-by-step solution＋

Take \(\sqrt3+2\) and \(5-\sqrt3\).

\[(\sqrt3+2)+(5-\sqrt3)=7\]

Thus, the sum is rational.

Question 9

Solved

Write a pair of irrational numbers whose difference is irrational.

View step-by-step solution＋

Take \(\sqrt5+5\) and \(\sqrt2+5\).

\[(\sqrt5+5)-(\sqrt2+5)=\sqrt5-\sqrt2\]

The difference is irrational.

Question 10

Solved

Write a pair of irrational numbers whose difference is rational.

View step-by-step solution＋

Take \(\sqrt3+5\) and \(\sqrt3+2\).

\[(\sqrt3+5)-(\sqrt3+2)=3\]

The difference is rational.

Question 11

Solved

Write a pair of irrational numbers whose product is irrational.

View step-by-step solution＋

Take \(\sqrt2\) and \(\sqrt3\).

\[\sqrt2\cdot\sqrt3=\sqrt6\]

Since \(6\) is not a perfect square, \(\sqrt6\) is irrational.

Question 12

Solved

Write a pair of irrational numbers whose product is rational.

View step-by-step solution＋

Take the conjugate pair \(5+\sqrt2\) and \(5-\sqrt2\).

\[(5+\sqrt2)(5-\sqrt2)=25-2=23\]

The product is rational.

Complete Solutions

Exercise 1(C)

Rationalisation, conjugates, surds and algebraic simplification.

Question 1(a)

Solved

If \(x=\sqrt5-2\), find \(x+\dfrac1x\).

1. \(2\sqrt5\)
2. \(4\)
3. \(4\sqrt5\)
4. \(-4\)

View step-by-step solution＋

\[\frac1x=\frac1{\sqrt5-2}\cdot\frac{\sqrt5+2}{\sqrt5+2}=\sqrt5+2\]

\[x+\frac1x=(\sqrt5-2)+(\sqrt5+2)=2\sqrt5\]

Correct option: A.

Question 1(b)

Solved

If \(x=1+\sqrt2\), find \(\left(x+\dfrac1x\right)^2\).

1. \(2\sqrt2\)
2. \(8\)
3. \(4\)
4. \(4\sqrt2\)

View step-by-step solution＋

\[\frac1x=\frac1{1+\sqrt2}=\sqrt2-1\]

\[x+\frac1x=(1+\sqrt2)+(\sqrt2-1)=2\sqrt2\]

\[\left(2\sqrt2\right)^2=8\]

Correct option: B.

Question 1(c)

Solved

Evaluate \(\dfrac{2\sqrt{27}+3\sqrt{12}}{4\sqrt3}\).

1. \(2\sqrt3\)
2. \(3\sqrt2\)
3. \(3\)
4. \(\sqrt3+\sqrt2\)

View step-by-step solution＋

\[2\sqrt{27}+3\sqrt{12}=2(3\sqrt3)+3(2\sqrt3)=12\sqrt3\]

\[\frac{12\sqrt3}{4\sqrt3}=3\]

Correct option: C.

Question 1(d)

Solved

Expand \((\sqrt5-\sqrt3)^2\).

1. \(8+2\sqrt{15}\)
2. \(8+\sqrt{15}\)
3. \(8-\sqrt{15}\)
4. \(8-2\sqrt{15}\)

View step-by-step solution＋

\[(\sqrt5-\sqrt3)^2=5+3-2\sqrt{15}=8-2\sqrt{15}\]

Correct option: D.

Question 1(e)

Solved

Rationalize \(\dfrac3{4+\sqrt7}\).

1. \(\dfrac13(4-\sqrt7)\)
2. \(3(4-\sqrt7)\)
3. \(\dfrac13(4+\sqrt7)\)
4. \(3(4+\sqrt7)\)

View step-by-step solution＋

\[\frac3{4+\sqrt7}\cdot\frac{4-\sqrt7}{4-\sqrt7}=\frac{3(4-\sqrt7)}{16-7}=\frac{4-\sqrt7}{3}\]

Correct option: A.

Question 1(f)

Solved

Rationalize \(\dfrac1{7-\sqrt5}\).

1. \(4(7+\sqrt5)\)
2. \(\dfrac1{44}(7+\sqrt5)\)
3. \(\dfrac1{44}(7-\sqrt5)\)
4. \(4(7-\sqrt5)\)

View step-by-step solution＋

\[\frac1{7-\sqrt5}\cdot\frac{7+\sqrt5}{7+\sqrt5}=\frac{7+\sqrt5}{49-5}=\frac{7+\sqrt5}{44}\]

Correct option: B.

Question 1(g)

Solved

If \(x=\sqrt2-1\), find \(\left(x-\dfrac1x\right)^2\).

1. \(2\sqrt2\)
2. \(8\)
3. \(4\)
4. \(2-\sqrt2\)

View step-by-step solution＋

\[\frac1x=\frac1{\sqrt2-1}=\sqrt2+1\]

\[x-\frac1x=(\sqrt2-1)-(\sqrt2+1)=-2\]

\[(-2)^2=4\]

Correct option: C.

Question 1(h)

Solved

Evaluate \(\dfrac{5-\sqrt7}{5+\sqrt7}-\dfrac{5+\sqrt7}{5-\sqrt7}\).

1. \(10\sqrt7\)
2. \(\dfrac{\sqrt7}{9}\)
3. \(\dfrac{10\sqrt7}{9}\)
4. \(-\dfrac{10\sqrt7}{9}\)

View step-by-step solution＋

\[ \frac{(5-\sqrt7)^2-(5+\sqrt7)^2}{25-7} =\frac{-20\sqrt7}{18} =-\frac{10\sqrt7}{9} \]

Correct option: D.

Question 2

Solved

State, with reasons, which expressions are surds.

1. \(\sqrt{180}\)
2. \(\sqrt[4]{27}\)
3. \(\sqrt[5]{128}\)
4. \(\sqrt[3]{64}\)
5. \(\sqrt[3]{25}\cdot\sqrt[3]{40}\)
6. \(\sqrt[3]{-125}\)
7. \(\sqrt\pi\)
8. \(\sqrt{3+\sqrt2}\)

View step-by-step solution＋

A surd is an irrational root of a rational number.

(i) \(\sqrt{180}=6\sqrt5\)Surd

(ii) \(\sqrt[4]{27}=3^{3/4}\)Surd

(iii) \(\sqrt[5]{128}=2\sqrt[5]{4}\)Surd

(iv) \(\sqrt[3]{64}=4\)Not a surd

(v) \(\sqrt[3]{25\cdot40}=\sqrt[3]{1000}=10\)Not a surd

(vi) \(\sqrt[3]{-125}=-5\)Not a surd

(vii) The radicand \(\pi\) is irrational.Not a surd

(viii) The radicand \(3+\sqrt2\) is irrational.Not a surd

Question 3

Solved

Write the lowest rationalizing factor of each expression.

1. \(5\sqrt2\)
2. \(\sqrt{24}\)
3. \(\sqrt5-3\)
4. \(7-\sqrt7\)
5. \(\sqrt{18}-\sqrt{50}\)
6. \(\sqrt5-\sqrt2\)
7. \(\sqrt{13}+3\)

View step-by-step solution＋

1. \(\sqrt2\), since \(5\sqrt2\cdot\sqrt2=10\).
2. \(\sqrt6\), since \(\sqrt{24}=2\sqrt6\) and \(2\sqrt6\cdot\sqrt6=12\).
3. \(\sqrt5+3\), the conjugate.
4. \(7+\sqrt7\), the conjugate.
5. \(\sqrt2\), because \(\sqrt{18}-\sqrt{50}=3\sqrt2-5\sqrt2=-2\sqrt2\).
6. \(\sqrt5+\sqrt2\), the conjugate.
7. \(\sqrt{13}-3\), the conjugate.

Question 4

Solved

Rationalize the denominators.

1. \(\dfrac{2\sqrt3}{\sqrt5}\)
2. \(\dfrac{\sqrt6-\sqrt5}{\sqrt6+\sqrt5}\)

View step-by-step solution＋

(i)

\[\frac{2\sqrt3}{\sqrt5}\cdot\frac{\sqrt5}{\sqrt5}=\frac{2\sqrt{15}}5\]

(ii)

\[\frac{\sqrt6-\sqrt5}{\sqrt6+\sqrt5}\cdot\frac{\sqrt6-\sqrt5}{\sqrt6-\sqrt5}=\frac{(\sqrt6-\sqrt5)^2}{6-5}=11-2\sqrt{30}\]

Question 5(i)

Solved

If \(\dfrac{2+\sqrt3}{2-\sqrt3}=a+b\sqrt3\), find \(a\) and \(b\).

View step-by-step solution＋

\[\frac{2+\sqrt3}{2-\sqrt3}\cdot\frac{2+\sqrt3}{2+\sqrt3}=\frac{(2+\sqrt3)^2}{4-3}=7+4\sqrt3\]

Comparing with \(a+b\sqrt3\):

\(a=7,\quad b=4\).

Question 5(ii)

Solved

If \(\dfrac{\sqrt7-2}{\sqrt7+2}=a\sqrt7+b\), find \(a\) and \(b\).

View step-by-step solution＋

\[\frac{\sqrt7-2}{\sqrt7+2}\cdot\frac{\sqrt7-2}{\sqrt7-2}=\frac{(\sqrt7-2)^2}{7-4}=\frac{11-4\sqrt7}{3}\]

\[= -\frac43\sqrt7+\frac{11}{3}\]

\(a=-\dfrac43,\quad b=\dfrac{11}{3}\).

Question 5(iii)

Solved

If \(\dfrac3{\sqrt3-\sqrt2}=a\sqrt3-b\sqrt2\), find \(a\) and \(b\).

View step-by-step solution＋

\[\frac3{\sqrt3-\sqrt2}\cdot\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}=3\sqrt3+3\sqrt2\]

Comparing with \(a\sqrt3-b\sqrt2\), we get

\(a=3,\quad b=-3\).

Question 6(i)

Solved

Simplify \(\dfrac{22}{2\sqrt3+1}+\dfrac{17}{2\sqrt3-1}\).

View step-by-step solution＋

\[ \frac{22(2\sqrt3-1)+17(2\sqrt3+1)}{(2\sqrt3+1)(2\sqrt3-1)} =\frac{78\sqrt3-5}{12-1} =\frac{78\sqrt3-5}{11} \]

Question 6(ii)

Solved

Simplify \(\dfrac{\sqrt2}{\sqrt6-\sqrt2}-\dfrac{\sqrt3}{\sqrt6+\sqrt2}\).

View step-by-step solution＋

\[ \frac{\sqrt2(\sqrt6+\sqrt2)-\sqrt3(\sqrt6-\sqrt2)}{(\sqrt6-\sqrt2)(\sqrt6+\sqrt2)} \]

\[=\frac{2\sqrt3+2-3\sqrt2+\sqrt6}{4}\]

Question 7

Solved

If

\[x=\frac{\sqrt5-2}{\sqrt5+2},\qquad y=\frac{\sqrt5+2}{\sqrt5-2}\]

find (i) \(x^2\), (ii) \(y^2\), (iii) \(xy\), and (iv) \(x^2+y^2+xy\).

View step-by-step solution＋

Since \((\sqrt5+2)(\sqrt5-2)=1\),

\[x=(\sqrt5-2)^2=9-4\sqrt5,\qquad y=(\sqrt5+2)^2=9+4\sqrt5\]

1. \(x^2=(9-4\sqrt5)^2=161-72\sqrt5\).
2. \(y^2=(9+4\sqrt5)^2=161+72\sqrt5\).
3. \(xy=1\).
4. \(x^2+y^2+xy=(161-72\sqrt5)+(161+72\sqrt5)+1=323\).

Question 8

Solved

If

\[m=\frac1{3-2\sqrt2},\qquad n=\frac1{3+2\sqrt2}\]

find (i) \(m^2\), (ii) \(n^2\), and (iii) \(mn\).

View step-by-step solution＋

Because \((3-2\sqrt2)(3+2\sqrt2)=1\),

\[m=3+2\sqrt2,\qquad n=3-2\sqrt2\]

1. \(m^2=17+12\sqrt2\).
2. \(n^2=17-12\sqrt2\).
3. \(mn=1\).

Question 9

Solved

If \(x=2\sqrt3+2\sqrt2\), find (i) \(1/x\), (ii) \(x+1/x\), and (iii) \((x+1/x)^2\).

View step-by-step solution＋

(i)

\[\frac1x=\frac1{2(\sqrt3+\sqrt2)}=\frac{\sqrt3-\sqrt2}{2}\]

(ii)

\[x+\frac1x=2\sqrt3+2\sqrt2+\frac{\sqrt3-\sqrt2}{2}=\frac{5\sqrt3+3\sqrt2}{2}\]

(iii)

\[\left(x+\frac1x\right)^2=\frac{(5\sqrt3+3\sqrt2)^2}{4}=\frac{93+30\sqrt6}{4}\]

Question 10

Solved

If \(x=1-\sqrt2\), find \(\left(x-\dfrac1x\right)^3\).

View step-by-step solution＋

\[\frac1x=\frac1{1-\sqrt2}=-(1+\sqrt2)\]

\[x-\frac1x=(1-\sqrt2)-(-1-\sqrt2)=2\]

\[\left(x-\frac1x\right)^3=2^3=8\]

Question 11

Solved

If \(x=5-2\sqrt6\), find \(x^2+\dfrac1{x^2}\).

View step-by-step solution＋

\[\frac1x=5+2\sqrt6\]

\[x+\frac1x=10\]

\[x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2=100-2=98\]

Question 12

Solved

Using \(\sqrt2=1.4\) and \(\sqrt3=1.7\), evaluate each expression correct to one decimal place.

1. \(\dfrac1{\sqrt3-\sqrt2}\)
2. \(\dfrac1{3+2\sqrt2}\)
3. \(\dfrac{2-\sqrt3}{\sqrt3}\)

View step-by-step solution＋

1. \(\dfrac1{\sqrt3-\sqrt2}=\sqrt3+\sqrt2\approx1.7+1.4=3.1\).
2. \(\dfrac1{3+2\sqrt2}=3-2\sqrt2\approx3-2.8=0.2\).
3. \(\dfrac{2-\sqrt3}{\sqrt3}=\dfrac{2\sqrt3-3}{3}\approx\dfrac{3.4-3}{3}=0.133\ldots\approx0.1\).

Question 13

Solved

Evaluate \(\dfrac{4-\sqrt5}{4+\sqrt5}+\dfrac{4+\sqrt5}{4-\sqrt5}\).

View step-by-step solution＋

\[ \frac{(4-\sqrt5)^2+(4+\sqrt5)^2}{16-5} =\frac{42}{11}=3\frac9{11} \]

Question 14

Solved

If

\[x=\frac{2+\sqrt5}{2-\sqrt5},\qquad y=\frac{2-\sqrt5}{2+\sqrt5}\]

find \(x^2-y^2\).

View step-by-step solution＋

\[x=-(9+4\sqrt5),\qquad y=-(9-4\sqrt5)\]

\[x^2=161+72\sqrt5,\qquad y^2=161-72\sqrt5\]

\[x^2-y^2=144\sqrt5\]

Complete Solutions

Test Yourself

MCQs, assertions, constructions, identities and mixed revision.

Question 1(a)

Solved

Since \(90=2\times3\times3\times5\), the fraction \(\dfrac{23}{90}\) is not a terminating decimal.

1. True
2. False
3. None of these

View step-by-step solution＋

The denominator contains the prime factor \(3\), so it is not of the form \(2^m5^n\).

Correct option: A — True.

Question 1(b)

Solved

\(\sqrt{27}\) and \(\sqrt3\) are irrational. Which expression is rational?

1. \(\sqrt{27}-\sqrt3\)
2. \(\sqrt{27}+\sqrt3\)
3. \(\sqrt{27}\times\sqrt3\)
4. None of these

View step-by-step solution＋

\[\sqrt{27}\times\sqrt3=\sqrt{81}=9\]

Correct option: C.

Question 1(c)

Solved

If \(x=\sqrt6-\sqrt5\), find \(x-\dfrac1x\).

1. \(1\)
2. \(11\)
3. \(2\sqrt6\)
4. \(-2\sqrt5\)

View step-by-step solution＋

\[\frac1x=\sqrt6+\sqrt5\]

\[x-\frac1x=(\sqrt6-\sqrt5)-(\sqrt6+\sqrt5)=-2\sqrt5\]

Correct option: D.

Question 1(d)

Solved

Evaluate \(\dfrac{2+\sqrt3}{2-\sqrt3}-\dfrac{2-\sqrt3}{2+\sqrt3}\).

1. \(4\)
2. \(2\sqrt3\)
3. \(1\)
4. \(8\sqrt3\)

View step-by-step solution＋

\[ \frac{(2+\sqrt3)^2-(2-\sqrt3)^2}{4-3}=8\sqrt3 \]

Correct option: D.

Question 1(e)

Solved

Statement 1: If \(a=3\sqrt3\) and \(b=\dfrac5{\sqrt{12}}\), then \(ab\) is irrational.

Statement 2: \(ab=\dfrac{15}{2}\).

1. Both statements are true.
2. Both statements are false.
3. Statement 1 is true and Statement 2 is false.
4. Statement 1 is false and Statement 2 is true.

View step-by-step solution＋

\[ab=3\sqrt3\cdot\frac5{\sqrt{12}}=3\sqrt3\cdot\frac5{2\sqrt3}=\frac{15}{2}\]

The product is rational. Thus Statement 1 is false and Statement 2 is true.

Correct option: D.

Question 1(f)

Solved

Statement 1: If \(x=\sqrt5+2\), then \(x-\dfrac1x=4\).

Statement 2: \(\dfrac1x=\sqrt5-2\).

1. Both statements are true.
2. Both statements are false.
3. Statement 1 is true and Statement 2 is false.
4. Statement 1 is false and Statement 2 is true.

View step-by-step solution＋

\[\frac1{\sqrt5+2}=\sqrt5-2\]

\[x-\frac1x=(\sqrt5+2)-(\sqrt5-2)=4\]

Both statements are true.

Correct option: A.

Question 1(g)

Solved

Assertion (A): If \(x+\dfrac1x=4\) and \(\dfrac1x=2+\sqrt3\), then \(x=2-\sqrt3\).

Reason (R): Substituting \(1/x=2+\sqrt3\) into \(x+1/x=4\) gives \(x=2-\sqrt3\).

1. A is true but R is false.
2. A is false but R is true.
3. Both A and R are true, and R correctly explains A.
4. Both A and R are true, but R does not explain A.

View step-by-step solution＋

\[x+(2+\sqrt3)=4\quad\Longrightarrow\quad x=2-\sqrt3\]

Also, \(1/(2+\sqrt3)=2-\sqrt3\). Both A and R are true, and R is the correct explanation.

Correct option: C.

Question 1(h)

Solved

Assertion (A): \(\sqrt{22},\sqrt{23},\sqrt{24},\sqrt{25},\sqrt{26},\sqrt{27}\) are irrational numbers between \(\sqrt{21}\) and \(\sqrt{28}\).

Reason (R): \(\sqrt{25}=5\), which is rational.

1. A is true but R is false.
2. A is false but R is true.
3. Both A and R are true, and R correctly explains A.
4. Both A and R are true, but R does not explain A.

View step-by-step solution＋

Although all six values lie between \(\sqrt{21}\) and \(\sqrt{28}\), \(\sqrt{25}=5\) is rational. Therefore A is false and R is true.

Correct option: B.

Question 2

Solved

Simplify

\[ \frac{\sqrt{x^2+y^2}-y}{x-\sqrt{x^2-y^2}} \div \frac{\sqrt{x^2-y^2}+x}{\sqrt{x^2+y^2}+y} \]

View step-by-step solution＋

\[ \frac{\sqrt{x^2+y^2}-y}{x-\sqrt{x^2-y^2}} \times \frac{\sqrt{x^2+y^2}+y}{\sqrt{x^2-y^2}+x} \]

\[ =\frac{(x^2+y^2)-y^2}{x^2-(x^2-y^2)}=\frac{x^2}{y^2} \]

Question 3

Solved

Evaluate \(\dfrac5{\sqrt{20}-\sqrt{10}}\), correct to one decimal place, using \(\sqrt5=2.2\) and \(\sqrt{10}=3.2\).

View step-by-step solution＋

Use the stated approximations directly:

\[\sqrt{20}=2\sqrt5\approx2(2.2)=4.4\]

\[\frac5{\sqrt{20}-\sqrt{10}}\approx\frac5{4.4-3.2}=\frac5{1.2}=4.166\ldots\]

Answer: \(4.2\) correct to one decimal place.

Because the supplied square-root values are rounded, rationalising first and then substituting can produce a different rounded result. Direct substitution follows the wording of the question.

Question 4

Solved

If \(x=\sqrt3-\sqrt2\), find (i) \(x+1/x\), (ii) \(x^2+1/x^2\), (iii) \(x^3+1/x^3\), and (iv) \(x^3+1/x^3-3(x^2+1/x^2)+x+1/x\).

View step-by-step solution＋

Since \(1/x=\sqrt3+\sqrt2\):

1. \(x+1/x=2\sqrt3\).
2. \(x^2+1/x^2=(2\sqrt3)^2-2=10\).
3. \(x^3+1/x^3=(x+1/x)^3-3(x+1/x)=24\sqrt3-6\sqrt3=18\sqrt3\).
4. \(18\sqrt3-3(10)+2\sqrt3=20\sqrt3-30=10(2\sqrt3-3)\).

Question 5

Solved

State true or false.

1. The negative of an irrational number is irrational.
2. The product of a non-zero rational number and an irrational number is rational.

View step-by-step solution＋

1. True. Example: \(\sqrt3\) and \(-\sqrt3\) are both irrational.
2. False. For example, \(2\) is non-zero rational and \(\sqrt2\) is irrational, while \(2\sqrt2\) is irrational.

Question 6

Solved

Construct a line segment of length \(\sqrt3\) cm.

View step-by-step solution＋

1. Draw a straight line \(XY\) and mark a point \(O\).
2. At \(O\), draw \(OB\perp XY\) with \(OB=1\) cm.
3. With centre \(B\) and radius \(2\) cm, cut the line at \(A\), so \(AB=2\) cm.
4. Join \(OA\).

\[OA^2=AB^2-OB^2=2^2-1^2=3\quad\Longrightarrow\quad OA=\sqrt3\text{ cm}\]

Question 7

Solved

Construct a line segment of length \(\sqrt8\) cm.

View step-by-step solution＋

1. Draw a straight line \(XY\) and mark \(O\).
2. At \(O\), draw \(OB\perp XY\) with \(OB=1\) cm.
3. With centre \(B\) and radius \(3\) cm, cut the line at \(A\), so \(AB=3\) cm.
4. Join \(OA\).

\[OA^2=AB^2-OB^2=3^2-1^2=8\quad\Longrightarrow\quad OA=\sqrt8\text{ cm}\]

Question 8(i)

Solved

Show that \(x^3+\dfrac1{x^3}=52\), if \(x=2+\sqrt3\).

View step-by-step solution＋

\[\frac1x=2-\sqrt3,\qquad x+\frac1x=4\]

\[x^3+\frac1{x^3}=\left(x+\frac1x\right)^3-3\left(x+\frac1x\right)=4^3-3(4)=52\]

Question 8(ii)

Solved

Show that \(x^2+\dfrac1{x^2}=34\), if \(x=3+2\sqrt2\).

View step-by-step solution＋

\[\frac1x=3-2\sqrt2,\qquad x+\frac1x=6\]

\[x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2=36-2=34\]

Question 8(iii)

Solved

Show that

\[ \frac{3\sqrt2-2\sqrt3}{3\sqrt2+2\sqrt3}+\frac{2\sqrt3}{\sqrt3-\sqrt2}=11 \]

View step-by-step solution＋

Factor \(\sqrt6\) from the first numerator and denominator:

\[ \frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}+\frac{2\sqrt3}{\sqrt3-\sqrt2} \]

\[ = (\sqrt3-\sqrt2)^2+2\sqrt3(\sqrt3+\sqrt2) \]

\[=5-2\sqrt6+6+2\sqrt6=11\]

Question 9

Solved

Show that \(x\) is irrational if (i) \(x^2=6\), (ii) \(x^2=0.009\), and (iii) \(x^2=27\).

View step-by-step solution＋

1. \(x=\pm\sqrt6\), irrational because \(6\) is not a perfect square.
2. \(x=\pm\sqrt{0.009}=\pm\dfrac{3\sqrt{10}}{100}\), irrational.
3. \(x=\pm\sqrt{27}=\pm3\sqrt3\), irrational.

Question 10

Solved

Show that \(x\) is rational if (i) \(x^2=16\), (ii) \(x^2=0.0004\), and (iii) \(x^2=1\dfrac79\).

View step-by-step solution＋

1. \(x=\pm4\), integers and hence rational.
2. \(x=\pm0.02\), terminating decimals and hence rational.
3. \(1\dfrac79=\dfrac{16}{9}\), so \(x=\pm\dfrac43\), rational.

Question 11

Solved

Find

\[ \frac1{1+\sqrt2}+\frac1{\sqrt2+\sqrt3}+\cdots+\frac1{\sqrt{2025}+\sqrt{2026}} \]

View step-by-step solution＋

For a general term,

\[\frac1{\sqrt n+\sqrt{n+1}}=\sqrt{n+1}-\sqrt n\]

Therefore, the series telescopes:

\[(\sqrt2-1)+(\sqrt3-\sqrt2)+\cdots+(\sqrt{2026}-\sqrt{2025})\]

\[=\sqrt{2026}-1\]

Complete Solutions

Case-Study Based Questions

Application-oriented questions on rationality and rationalizing factors.

Case-Study Question 1

Solved

Real numbers include rational and irrational numbers. A rational number can be written as \(p/q\), where \(p,q\) are integers and \(q\ne0\); an irrational number cannot.

1. Is the difference between a rational number and an irrational number always rational?
2. Is \(0.5555\ldots\) rational?
3. Is \(0.5050050005\ldots\) rational?
4. Is the product of \(3-\sqrt7\) and its rationalizing factor rational?

View step-by-step solution＋

1. No. Rational minus irrational is irrational; for example, \(2-\sqrt3\).
2. Yes. \(0.5555\ldots=0.\overline5=5/9\).
3. No. The decimal is non-terminating and non-recurring, so it is irrational.
4. Yes. The rationalizing factor is \(3+\sqrt7\), and \((3-\sqrt7)(3+\sqrt7)=9-7=2\), which is rational.

Case-Study Question 2

Solved

Under the Taruner Swapna Scheme of the West Bengal Government, eligible backward or economically disadvantaged students may receive financial assistance and digital devices such as smartphones or tablets to support online learning and help bridge the digital divide.

Rohan, who lives in Purulia district, receives a tablet and studies the number system through e-learning. He is particularly interested in the irrational number \(7+4\sqrt3\).

1. Find its rationalizing factor.
2. Find its reciprocal. Are the reciprocal and rationalizing factor the same?
3. If \(x=7+4\sqrt3\), find \(x+1/x\).

View step-by-step solution＋

1. The rationalizing factor is the conjugate \(7-4\sqrt3\).
2. \[\frac1{7+4\sqrt3}=\frac{7-4\sqrt3}{49-48}=7-4\sqrt3\]Yes. They are the same because the product of the conjugates is \(1\).
3. \[x+\frac1x=(7+4\sqrt3)+(7-4\sqrt3)=14\]

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