Chapter 1 of the Class 10 NCERT Mathematics textbook focuses on the "Real Numbers" a fundamental concept in mathematics. Exercise 1.1 in this chapter introduces students to the basic properties and operations of the real numbers including their classification and operations. Understanding real numbers is crucial as they form the foundation for the more advanced topics in mathematics including algebra, calculus, and number theory. This exercise helps build a solid understanding of the number system and its applications in the various mathematical contexts.
Real Numbers
The Real numbers encompass all the numbers that can be found on the number line including both the rational and irrational numbers. They are essential in mathematics for representing quantities solving equations and modeling real-world scenarios. Real numbers are classified into various subsets such as natural numbers, whole numbers, integers, rational numbers, and irrational numbers each with unique properties and uses.
Exercise 1.1 Solutions
This exercise focuses on applying Euclid's division lemma to various problems, including finding the Highest Common Factor (HCF) and demonstrating the properties of integers. Below are the solutions to the questions in this exercise:
Question 1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Solution:
i)135 and 225
We need to apply Euclid's algo, for finding HCF, now First Step to take Divisor ,we always take Divisor as smaller than Dividend.
Divisor=135
Dividend=225
Quotient=225/135=1(quotient is an integer value)
Euclid's Div. Algo: 225=135*1+90
Since 90!=0 that means We have to again apply division lemma for remainder:90,Now Remainder Will become Divisor and Divisor
will become Dividend.
Euclid's Div. Algo: 135=90*1+45
Again we Need to apply Division lemma because Yet remainder is not equal to 0 (45!=0) and just similar to prev. ways we will get
Divisor =45 and Dividend=90
Euclid's Div. Algo:90=45*2+0
Here Remainder=0 that means We need to stop here,
When Remainder = 0 Then HCF = Divisor => HCF(225,135)=HCF(135,90)
=HCF(90,45)
=45.
ii)196 and 38220
Divisor=196
Dividend=38220
Quotient=38220/196=195
Euclid's Div. Algo: 38220=196*195+0
Remainder=0,We don't need to apply Division lemma further,
HCF(38220,196)=196
iii) 867 and 255
Divisor=255
Dividend=867
Quotient=867/225=3
Euclid's Div. algo: 867=255*3+102
Remainder = 102 (!=0) That means Again we need to apply Division lemma method, Divisor=102,Dividend=255,Quotient=255/102=2
Euclid's Div. algo: 255=102*2+51
Again Remainder =51(!=0), We need to apply Division lemma method again, Divisor=51,Dividend=102,Quotient=102/51=2
Euclid'S Div. algo:102=51*2+0
Remainder =0,We need to stop here ,
HCF(867,255)=HCF(255,102)
= HCF(102,51)
= 51
Question 2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
**We know that any odd no integer is not divisible by 2,
When we divide 6q by 2 then it is perfectly divisible by 2 and give result 3q without any remainder,=>6q is even number
let take any positive integer x and y=6
as Per Euclid's Div algo x=6q+r, q>=0 & 0<=r<6
if r=0 => x=6q, we earlier concluded that 6q is divisible by 2 =>x=6q is positive even integer
if r=2, 4 =>x=6q+2, x=6q+4 also An positive even integer because 6q is even as well as 2,4 also even and even divided by even gives
even
if r=1
Dividing 6q+1 by 2:
Dividend=6q+1,Divisor=2
Quotient = (6q+1)/2 =3q
as per Euclid's div algo 6q+1=3q*2+1
As We Got Remainder = 1 after dividing by 2 that means 6q+1 is odd number, As q>=0 that means 6q+1 is positive odd integer
if r=3
Dividing 6q+3 by 2:
Dividend=6q+3,Divisor=2
Quotient = (6q+3)/2 =3q+1
as per Euclid's div algo 6q+3=(3q+1)*2+1
As We Got Remainder = 1 after dividing by 2 that means 6q+3 is odd number, As q>=0 that means 6q+3 is positive odd integer
if r=5
Dividing 6q+5 by 2:
Dividend=6q+5,Divisor=2
Quotient = (6q+5)/2 =3q+2
as per Euclid's div algo 6q+5=(3q+2)*2+1
As We Got Remainder = 1 after dividing by 2 that means 6q+5 is odd number, As q>=0 that means 6q+5 is positive odd integer
Overall Conclusion : Any positive integer x can be in form 6q+1,6q+3 or 6q+5 if it is odd otherwise it is in form 6q, 6q+2, 6q+4
Question 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution:
Given,
Total number of army contingent members=616
Total number of army band members=32
Given two groups are to march in same number of column then maximum number of columns will be Highest Common Factor between two
groups ie HCF(616,32)
Finding HCF(616,32)
---------------------
Divisor=32,Dividend=616,quotient=616/32=19
Euclid's Div. Algo: 616=32*19+8
Remainder = 8(!=0), again Apply Division lemma
Divisor=8,Dividend=32,Quotient=32/8=4
Euclid's Div. Algo: 32=8*4+0
Remainder = 0,We have to stop here
HCF(616,32)=8
Therefore, Maximum Number of Columns Will be 8 in which Both groups can march
Question 4.Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Solution:
Let x be any positive integer, And y=3,
As per Euclid's Div algo. x=3q+r;q>=0 && 0 <= r < 3 => r=(0,1,2)
If r=0: x=3q
Squaring both sides: x2 =3*3*q2
let assume m1=3q2 ,Here m1 will be any positive integer because q>=0.
=>x2=3m1
if r=1: x=3q+1
Squaring both sides:x2 =(3q+1)2
=>x2 =3(3q2 + 2q)+1
let assume m2=3q2+2q
=> x2 = 3m2 + 1 ,Here m2 will be any positive integer because q>=0.
if r=2 : x=3q+2
Squaring both sides:x2 =(3q+2)2
=>x2 =3(3q2 + 4q+1)+1
let assume m3=3q2+4q+1
=> x2 = 3m3 + 1 ,Here m3 will be any positive integer because q>=0.
Since m1,m2,m3 are positive integer ,therefore we can Conclude that x2=3m or 3m+1 ,Where m is some integer.
This proved that the square of any positive integer(x) is either of the form 3m or 3m + 1 for some integer m
Question 5.Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let x be any positive integer, and y=3
As per Euclid's Div algo. x=3q+r; q>=0 && 0 <= r < 3 => r=(0,1,2)
if r=0: x=3q
Cubing both sides: x3 =9*3*q3
let assume m1=3q3 ,Here m1 will be any positive integer because q>=0.
=>x2=9m1
if r=1: x=3q+1
Cubing both sides: x3 =(3q+1)3
=>x3 =9(3q3+3q2+q)+1
Let assume m2=3q3+3q2+q
=> x3 = 9m2 + 1 ,Here m2 will be any positive integer because q>=0.
if r=2 : x=3q+2
Cubing both sides: x3 =(3q+2)3
=>x3 =9(3q3 + 6q2+4q)+8
let assume m3=3q3+6q2+4q
=> x3 = 9m3 + 8 ,Here m3 will be any positive integer because q>=0.
Since m1,m2,m3 are positive integer ,therefore we can Conclude that x3=9m,9m+1 or 3m+8, where m is some integer.
This proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Related Articles:
- Real Numbers
- What is a real number system in mathematics?
- Understanding Rational and Irrational Numbers
- Number Systems: An Overview
- Introduction to Algebraic Expressions
Summary
Exercise 1.1 of NCERT Class 10 Chapter 1 - Real Numbers focuses on Euclid's division lemma and its applications. It introduces students to the concept that for any positive integers a and b, there exist unique integers q (quotient) and r (remainder) such that a = bq + r, where 0 ≤ r < b. This lemma forms the basis for understanding divisibility, the division algorithm, and finding the HCF of two positive integers. The exercise helps students apply this concept to solve various problems involving division and remainders.