- Chapter 1- Rational and Irrational Numbers
- Chapter 2- Compound Interest [Without Using Formula
- Chapter 3- Compound Interest [Using Formula
- Chapter 4- Expansions
- Chapter 5- Factorisation
- Chapter 6- Simultaneous Equations
- Chapter 7- Indices [exponents]
- Chapter 8- Logarithms
- Chapter 9- Triangles [Congruency in Triangles]
- Chapter 10- Isosceles Triangle
- Chapter 11- Inequalities
- Chapter 12- Mid-Point and Its Converse
- Chapter 13- Pythagoras Theorem
- Chapter 14- Rectilinear Figures
- Chapter 15- Construction of Polygons
- Chapter 16- Area Theorems
- Chapter 17- Circles
- Chapter 18- Statistics
- Chapter 19- Mean and Median
- Chapter 20- Area and Perimeter of Plane Figures
- Chapter 21- Solids
- Chapter 22- Trigonometrical Ratios
- Chapter 23- Trigonometrical Ratios of Standard Angles
- Chapter 24- Solution Of Right Triangles
- Chapter 25- Complementary angles
- Chapter 26- Co-ordinate Geometry
- Chapter 27- Graphical Solution
- Chapter 28- Distance Formula
Selina Solutions Class 9 Mathematics Solutions for Exercise 1(B) in Chapter 1 - Chapter 1- Rational and Irrational Numbers
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Prove that each of the following numbers is irrational:
(i) √𝟑 + √𝟐
(ii) 𝟑 − √𝟐
(iii) √𝟓 − 𝟐
(i) √𝟑 + √𝟐
Let √𝟑 + √𝟐 be a rational number.
⇒√𝟑 + √𝟐 = x
Squaring on both the sides, we get
\begin{array}{l} (\sqrt{3}+\sqrt{2})^{2}=x^{2} \\ \Rightarrow 3+2+2 \times \sqrt{3} \times \sqrt{2}=x^{2} \\ \Rightarrow x^{2}-5=2 \sqrt{6} \\ \Rightarrow \sqrt{6}=\frac{x^{2}-5}{2} \end{array}
Here, x is a rational number.
⇒ x² is a rational number.
⇒ x² - 5 is a rational number.
\frac{x^{2}-5}{2}is a rational number.
But is an irrational number.
is an irrational number.\sqrt{6}
x2- 5 is an irrational number.
⇒ x2 is an irrational number.
⇒ x is an irrational number.
Rational and Irrational Numbers
But we have assume that x is a rational number.
∴ we arrive at a contradiction.
So, our assumption that √𝟑 + √𝟐 is a rational number is wrong.
∴ √𝟑 + √𝟐 is an irrational number.
(ii) 𝟑 − √𝟐
Let 𝟑 − √𝟐 be a rational number.
⇒ 𝟑 − √𝟐 = x
Squaring on both the sides, we get
\begin{array}{l} (3-\sqrt{2})^{2}=x^{2} \\ \Rightarrow 9+2-2 \times 3 \times \sqrt{2}=x^{2} \\ \Rightarrow 11-x^{2}=6 \sqrt{2} \\ \Rightarrow \sqrt{2}=\frac{11-x^{2}}{6} \end{array}
Here, x is a rational number.
⇒ x2 is a rational number.
⇒ 11 - x2 is a rational number.
\frac{11-x^{2}}{6} is also a rational number.
\Rightarrow \sqrt{2}=\frac{11-x^{2}}{6} is a rational number.
But \sqrt{2} is an irrational number.
\Rightarrow \sqrt{2}=\frac{11-x^{2}}{6} is an irrational number.
⇒ 11 - x2 is an irrational number.
⇒ x2 is an irrational number.
⇒ x is an irrational number.
But we have assume that x is a rational number.
∴ we arrive at a contradiction.
So, our assumption that 𝟑 − √𝟐 is a rational number is wrong.
∴ 𝟑 − √𝟐 is an irrational number.
(iii) √𝟓 − 𝟐
Let √𝟓 − 𝟐 be a rational number.
⇒√𝟓 − 𝟐 = x
Squaring on both the sides, we get
$(\sqrt{5}-2)^{2}=x^{2}$ $\Rightarrow 5+4-2 \times 2 \times \sqrt{5}=x^{2}$ $\Rightarrow 9-x^{2}=4 \sqrt{5}$ $\Rightarrow \sqrt{5}=\frac{9-x^{2}}{4}$
Here, x is a rational number.
⇒ x2 is a rational number.
⇒ 9 - x2 is a rational number.
⇒ is also a rational number.
is a rational number.
But is an irrational number.
is an irrational number.
⇒ 9 - x2 is an irrational number.
⇒ x2 is an irrational number.
⇒ x is an irrational number.
But we have assume that x is a rational number.
∴ we arrive at a contradiction.
So, our assumption that √𝟓 − 𝟐 is a rational number is wrong.
∴ √𝟓 − 𝟐 is an irrational number.
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