Every student has a fear of Mathematics. Students feared learning theorems, formulas, and some of them forgot the basic concept but now, no more fear of the word MATHS! This article provides you with basic concepts of class 10th lesson 1, Real Numbers, and some examples to help you.
Euclid’s Division Lemma
An algorithm is a series of well-defined steps that give a procedure for solving a problem. A lemma is a proven statement used for proving another statement. Euclid’s division algorithm is a technique to compute the highest common factor of two given positive integers.
To obtain the HCF of two positive integers, say c and d, with c>d, follow the steps below:
- Step1. Apply Euclid s division lemma to c and d. So, we find the whole numbers q and r such that c=dq+r, 0≤r<d.
- Step2. If r=0, d is the HCF of c and d. If r is not equal to 0, apply the division lemma to d and r.
- Step3. Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
The Fundamental Theorem Of Arithmetic
Every compost number can be expressed as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
Rational and Irrational Numbers
A number s is called rational if it can be written as p,q.
Where p and q are integers and q is not equal to zero.
A number s is called irrational if it can be written as p, q where p and q are integers and qis not equal to zero.
The irrationality of Square Roots of 2, 3, and 5.
Let p be a prime number. If p divides the square of a where a is a positive integer.
√2, √3 are irrational.
Decimals expansions of Rational Numbers
Let x be a rational number whose decimal expansion terminates,. Then we can express x in the form of pq., where p and q are coprime, and the prime factorization of q is of the form 2 power m multiplied 5 power n where m and n are non-negative integers.
Let x= pq be a rational number, such that the prime factorization of q is not of the form 2 power m multiplied 5 power m where m and n are non-negative integers. Then x has a decimal expansion which is non-terminating repeating (recurring).
NCERT Solutions Of Class 10th Real Numbers
Q1. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Sol.
By Euclid’s algorithm,
a = 6q + r, and r = 0, 1, 2, 3, 4, 5
Hence, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.
Hence, these numbers are odd numbers.
Q2. 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?
Sol.
Euclid’s algorithm
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8.
Q3. Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Solution:
(i) Given numbers are 135 and 225.
On applying Euclid’s division algorithm, we have
225 = 135 x 1 + 90
Since the remainder 90 ≠ 0, so again, we apply Euclid’s division algorithm to 135 and 90 to get
135 = 90 x 1 + 45
Since the remainder 45 ≠ 0, so again we apply Euclid’s division algorithm to 90 and 45, to get
90 = 45 x 2 + 0
The remainder has now become zero, so we stop.
∵ At the last stage, the divisor is 45
∴ The HCF of 135 and 225 is 45.
Alternatively:
(i) By Euclid’s Division Algorithm, we have
225 = 135 x 1 + 90 135
= 90 x 1 + 45 90
= 45 x 2 + 0
∴ HCF (135, 225) = 45.
(ii) Given numbers are 196 and 38220
On applying Euclid’s division algorithm, we have
38220 = 196 x 195 + 0
Since we get the remainder zero in the first step, so we stop.
∵ At the above stage, the divisor is 196
∴ The HCF of 196 and 38220 is 196.
Alternatively:
(ii) By Euclid’s Division Algorithm, we have
38220 = 196 x 195 + 0
196 = 196 x 1 + 0
∴ HCF (38220, 196) = 196.
Given numbers are 867 and 255
On applying Euclid’s division algorithm, we have
867 = 255 x 3 + 102
Since the remainder 102 ≠ 0, so again we apply Euclid’s division algorithm to 255 and 102. to get
255 = 102 x 2 + 51
Since the remainder 51 ≠ 0, so again we apply Euclid’s division algorithm to 102 and 51, to get
102 = 51 x 2 + 0
We find the remainder is 0 and the divisor is 51
∴ The HCF of 867 and 255 is 51.
Alternatively:
(iii) 867 and 255
Step 1: Since 867 > 255,
apply Euclid’s division lemma, to a =867 and b=255 to find q and r
such that 867 = 255q + r, 0 ≤ r<255
On dividing 867 by 255 we get quotient as 3 and remainder as 102
i.e 867 = 255 x 3 + 102
Step 2: Since remainder 102 ≠ 0,
we apply the division lemma to a=255 and b= 102 to find whole numbers q and r
such that 255 = 102q + r where 0 ≤ r<102
On dividing 255 by 102 we get quotient as 2 and remainder as 51
i.e 255 = 102 x 2 + 51
Q4. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Let ‘a’ be any positive integer and b = 6.
∴ By Euclid’s division algorithm, we have
a = bq + r, 0 ≤ r ≤ b
a = 6q + r, 0 ≤ r ≤ b [ ∵ b = 6] where q ≥ 0 and r = 0,1, 2, 3, 4,5
Now, ‘a’ may be of the form of 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5
If ‘a’ is of the form 6q, 6q + 2, 6q + 4 then ‘a’ is an even.
In the above, we can see clearly that the numbers of the form 6q, 6q + 2, 6q + 4 are having the factor 2.
∴ The numbers of the form 6q, 6q + 2, 6q + 4 are even.
If ‘a’ is of the form 6q +1,6q +3, 6q + 5 then ‘a’ is an odd.
As if
∵ We know that the number of the form 2k + 1 is odd.
∴ The numbers of the form 6q + 1, 6q + 3, 6q + 5 are odd.
Alternatively:
Let a be any odd positive integer we need to prove that a is of the form 6q + 1 , or 6q + 3 , or 6q + 5 , where q is some integer. Since a is an integer consider b = 6 another integer applying Euclid’s division lemma
we get a = 6q + r for some integer q ≤ 0, and r = 0, 1, 2, 3, 4, 5 since
0 ≤ r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
However since a is odd so a cannot take the values 6q, 6q+2 and 6q+4
(since all these are divisible by 2)
Also, 6q + 1 = 2 x 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are odd numbers.
Therefore, any odd integer can be expressed is of the form
6q + 1, or 6q + 3, or 6q + 5 where q is some integer.
Q5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
Let a be a positive integer, q be the quotient and r be the remainder.
Dividing a by 3 using Euclid’s Division Algorithm, we have,
a = 3q + r, where 0 ≤ r < 3
Putting r = 0, 1 and 2, we get:
a = 3q, a = 3q + 1 and a = 3q + 2
If a = 3q, then a3 = 27q3 = 9(3q3) = 9m. (Assuming m = 3q3.)
If a = 3q + 1, then
a3 = (3q + l)3 = 27q3 + 9q(3q + 1) + 1 = 9(3q3 + 3q2 + q) + 1 = 9m + 1, (Assuming m = 3q3 + 3q2 + q)
If a = 3q + 2, then a3 = (3q + 2)3
= 27q3 + 18q(3q + 2) + (2)3
= 9(3q3 + 6q2 + 4q) + 8
= 9m + 8, (Assuming m – 3q3 + 6q2 + 4q)
Hence, a3 is of the form 9m, 9m + 1 or 9m + 8.
Conclusion
We have discussed the short notes of class 10th Maths Real Numbers along with the ncert solutions to some questions to help the students to understand the concepts and get the gist of the lesson. Hope you all find it useful.
