Arrange 2 3 , 3 4 and 4 3 in descending order. Laws of Exponents for Real Numbers Laws of Exponents for Real Numbers\text { Solve the following equations for } x \text { : }Solve the following equations for $x$ :Solve the following equations for $x$ :

We are given to arrange the numbers 2 3 , 3 4 and 4 3 in descending order. The given numbers are: 2 3 , 3 4 and 4 3 i..e, 2 1 3 , 3 1 4 , 4 1 3 Now, the L C M of 3 and 4 = 12 , so write the each number with exponent 1 12 Therefore, 2 3 = 2 1 3 = 2 4 1 12 = ( 16 ) 1 12 3 4 = 3 1 4 = 3 3 1 12 = ( 27 ) 1 12 And 4 3 = 4 1 3 = 4 4 1 12 = ( 256 ) 1 12 As 256 > 27 > 16 ⇒ 4 3 > 3 4 > 2 3 Hence, the given numbers can be written in descending order as: 4 3 , 3 4 , 2 3 .

Find a rational number between 2 3 and 3 5 . Laws of Exponents for Real Numbers Laws of Exponents for Real Numbers\text { Solve the following equations for } x \text { : }Solve the following equations for $x$ :Solve the following equations for $x$ :

We are given to find the rational number between 2 3 and 3 5 We can use Mean or Average Method to find the rational number between any two given rational numbers. A rational number between 2 3 and 3 5 = 2 3 + 3 5 2 = 10 + 9 15 2 = 19 30 Hence, the rational number between 2 3 and 3 5 is 19 30

Find the irrational number between 3 5 and 2 3 . Laws of Exponents for Real Numbers Laws of Exponents for Real Numbers\text { Solve the following equations for } x \text { : }Solve the following equations for $x$ :Solve the following equations for $x$ :

We are given to find the irrational number between 3 5 and 2 3 The decimal expansions of given numbers are: 3 5 = 0 . 6 and 2 3 = 0 . 666 . . . = 0 . 6 ¯ To find an irrational number between 3 5 and 2 3 , we find a decimal number which is non-terminating non-recurring, terminating and lying between them i..e, between 0 . 6 and 0 . 6 ¯ . There are infinitely such numbers. One such number is 0 . 62101101110 …

Find a rational number between 1 and 2 . Laws of Exponents for Real Numbers Laws of Exponents for Real Numbers\text { Solve the following equations for } x \text { : }Solve the following equations for $x$ :Solve the following equations for $x$ :

We have to find the rational number between 1 and 2 . Consider the squares of 1 and 2 . 1 2 = 1 and 2 2 = 2 For any two rational numbers, their average or mid-point is rational. Proceeding similarly, we can generate infinitely many rational numbers. Take any rational number between 1 and 2 , which is perfect square of a rational number. One such number is 1 . 21 , and 1 . 21 = 1 . 1 2 i.e. 1 . 21 = 1 . 1 . As 1 < 1 . 21 < 2 ⇒ 1 < 1 . 21 < 2 ⇒ 1 < 1 . 1 < 2 Hence, 1 . 1 is a rational number between 1 and 2 .

Find an irrational number between 2 and 3 . Laws of Exponents for Real Numbers Laws of Exponents for Real Numbers\text { Solve the following equations for } x \text { : }Solve the following equations for $x$ :Solve the following equations for $x$ :

We have to write one irrational number between 2 and 3 To find an irrational number between 2 and 3 , we find a decimal number which is non-terminating non-recurring, terminating and lying between them i..e, between 2 and 3 . There are infinitely such numbers. One such number is 1 . 5010010001 …

How many rational numbers lie between 0 . 01 and 0 . 1 ? Laws of Exponents for Real Numbers Laws of Exponents for Real Numbers\text { Solve the following equations for } x \text { : }Solve the following equations for $x$ :Solve the following equations for $x$ :

Any time you take two rational numbers and find their average i.e., add them, then divide the sum by 2 , you get another rational number. Furthermore, if those two rational numbers are different from each other, this average will lie halfway between them. Now imagine finding the average of 0 . 01 and 0 . 1 . That gives you a rational number between 0 . 01 and 0 . 1 Next, repeat the process, but this time use 0 . 01 and the average from the last step. Averaging them gives another rational number between 0 . 01 and the first average, so it is between 0 . 01 and 0 . 1 . You can do this forever and you will keep getting a new rational number that is between the ones you averaged, so it is between 0 . 01 and 0 . 1 Hence, there are infinitely many rational numbers between 0 . 01 and 0 . 1 .

How many irrational numbers lie between 0 . 01 and 0 . 1 ? Laws of Exponents for Real Numbers Laws of Exponents for Real Numbers\text { Solve the following equations for } x \text { : }Solve the following equations for $x$ :Solve the following equations for $x$ :

We know that a rational number can be expressed in the form p q such that p , q are integers and q ≠ 0 . An irrational numbers is a number that cannot be expressed in the form p q . We also know the inequality of square roots; a < b ⇒ a < b We are asked in the question to find the irrational numbers between 0 . 01 and 0 . 1 . We know how to convert decimals into fractions. We write the number without decimal point as numerator and put same number of zeros as the number of digits after the decimal point. So the fraction form the given numbers are 0 . 1 = 1 10 and 0 . 01 = 1 100 We take the square of obtained factions. So we have; ( 0 . 1 ) 2 = 1 10 2 = 1 10 × 10 = 1 100 ( 0 . 01 ) 2 = 1 100 2 = 1 100 × 100 = 1 10000

Without actual division, determine whether the decimal expansion of the rational number 51 84 is terminating or non-terminating recurring. Laws of Exponents for Real Numbers Laws of Exponents for Real Numbers\text { Solve the following equations for } x \text { : }Solve the following equations for $x$ :Solve the following equations for $x$ :

First, we note that the given rational number 51 84 is in the lowest terms. Its denominator = 84 = 4 × 21 = 2 2 × 7 × 3 , which contains a prime factor other than 2 and 5 . As we know, The decimal expansion of a rational number p q is non-terminating recurring if q has a prime factor other than 2 or 5 . Therefore, the given rational number 51 84 is a non-terminating recurring decimal.

E M B I B E

CBSE Syllabus Standard Mathematics for Class IX>Chapter 1 - Number System>Exercise 1.5

M L Aggarwal Solutions for Chapter: Number System, Exercise 5: Exercise 1.5

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Number System, Exercise 5: Exercise 1.5

Attempt the practice questions on Chapter 1: Number System, Exercise 5: Exercise 1.5 with hints and solutions to strengthen your understanding. CBSE Syllabus Standard Mathematics for Class IX solutions are prepared by Experienced Embibe Experts.