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CBSE NOTES CLASS 10 MATHEMATICS CHAPTER 1

Real Numbers

1. Summary Of Basic Concepts

2. Detailed Notes

CBSE NOTES CLASS 10 MATHEMATICS CHATER 1

Real Numbers

Summary Of Basic Concepts

Lemma: A lemma is a proven statement used for proving another statement.

Algorithm: An algorithm is a series of well defined steps which give a procedure for solving a type of problem.

Theorem 1.1: Euclid’s Division Lemma

Given positive integers a and b, there exist unique integers q and r satisfying

a = bq + r, 0 ≤ r < b.

Euclid’s Division Algorithm

To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below:

Step 1: Apply Euclid’s division lemma, to c and d to find whole numbers, q and r such that

c = dq + r, 0 ≤ r < d.

Step 2: If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.

Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

Theorem 1.2: Fundamental Theorem of Arithmatic):

Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

Or

The prime factorization of a natural number is unique, except for the order of its factors.

In general, given a composite number x, we factorize it as x = p1p2 ... pn, where p1, p2,..., pn are primes and written in ascending order, i.e., p 1 ≤ p2 ≤ . . . ≤ pn. If we combine the same primes, we will get powers of primes.

Theorem 1.3: Let p be a prime number. If p divides a2, then p divides a, where ‘a’ is a positive integer.

Theorem 1.4: 2 is irrational. 3 etc. are also irrational

Theorem 1.5: Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form pq, where p and q are coprime, and the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers.

Theorem 1.6: Let x = pq be a rational number, such that the prime factorization of q is of the form 2n5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

Theorem 1.7: Let x = pq be a rational number, such that the prime factorization of q is not of the form 2n5m, where n, m are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating (recurring).

To determine after how many decimal places, the decimal expansion will terminate

Any number of the form

x =  p2m5n(m>n) 

can be written as

x =  p2m5n ×5m-n5m-n 

 x =  p×5m-n2m5m 

Hence, the number p×5m-n is divided by a multiple of 10 with m trailing zeroes. By looking at the number we can find the decimal places in the result.

CBSE NOTES CLASS 10 MATHEMATICS CHATER 1

Real Numbers

Detailed Chapter Notes

Lemma: A lemma is a proven statement used for proving another statement.

Algorithm: An algorithm is a series of well defined steps which give a procedure for solving a type of problem.

Theorem 1.1: Euclid’s Division Lemma

Given positive integers a and b, there exist unique integers q and r satisfying

a = bq + r, 0 ≤ r < b.

Take examples

Euclid’s Division Algorithm

To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below:

Step 1: Apply Euclid’s division lemma, to c and d to find whole numbers, q and r such that

c = dq + r, 0 ≤ r < d.

Step 2: If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.

Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

Example 1: Use Euclid’s algorithm to find the HCF of 4052 and 12576.

[NCERT]

Solution:

Step 1: Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get 12576 = 4052 × 3 + 420

Step 2: Since the remainder 420 ≠ 0, we apply the division lemma to 4052 and 420, to get

4052 = 420 × 9 + 272

420 = 272 × 1 + 148

272 = 148 × 1 + 124

148 = 124 × 1 + 24

124 = 24 × 5 + 4

24 = 4 × 6 + 0

Step 3: The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.

We see that HCF (24, 4) = HCF (124, 24) = HCF (148, 124) = HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052) = 4.

Example 2: Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer.

[NCERT]

Solution: Let a be any positive integer and b = 2. Then, by Euclid’s algorithm,

a = 2q + r, 0 ≤ r < 2, for some integer q ≥ 0, and r = 0 or r = 1.

So,

a = 2q or 2q + 1.

If a is of the form 2q, then a is an even integer. Also, a positive integer can be either even or odd. Therefore, any positive odd integer is of the form 2q + 1.

Example 3: Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

[NCERT]

Solution:

Let us start with taking a = 4q + r, where a is a positive integer.

Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3.

That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.

However, since a is odd, it cannot be 4q or 4q + 2 (since they are both divisible by 2).

Therefore, any odd integer is of the form 4q + 1 or 4q + 3.

Example 4: A sweet seller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number of that can be placed in each stack for this purpose?

[NCERT]

Solution: To find solution of such problems we need to find HCF of two figures.

420 = 130 × 3 + 30

130 = 30 × 4 + 10

30 = 10 × 3 + 0

So, the HCF of 420 and 130 is 10.

Therefore, the sweet seller can make stacks of 10 for both kinds of barfi.

Theorem 1.2: Fundamental Theorem of Arithmatic):

Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

Or

The prime factorization of a natural number is unique, except for the order of its factors.

In general, given a composite number x, we factorize it as x = p1p2 ... pn, where p1, p2,..., pn are primes and written in ascending order, i.e., p 1 ≤ p2 ≤ . . . ≤ pn. If we combine the same primes, we will get powers of primes. For example,

32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13

= 23 × 32 × 5 × 7×13

Theorem 1.3: Let p be a prime number. If p divides a2, then p divides a, where ‘a’ is a positive integer.

Theorem 1.4: 2 is irrational.

Proof (by contradiction):

Let us assume, to the contrary, that 2 is rational.

So, we can find integers r and s (≠ 0) such that

2=rs

Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get 2  = ab, where a and b are coprimes.

So, b 2  = a.

Squaring both sides and rearranging, we get,

2b2 = a2.

Therefore, 2 divides a2

⇒ 2 divides a.

So, we can write a = 2c for some integer c.

Substituting for a, we get.

2b2 = 4c2, that is, b2 = 2c2.

This means that 2 divides b2, and so 2 divides b.

Therefore, a and b have at least 2 as a common factor.

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that 2 is rational.

So, we conclude that 2 is irrational.

Start with assumption that 2 + 3 is rational, then,

2 + 3 =ab , for some a & b such tha a and b are coprimes

 2b +3  b=a 

3 =a-2bb

Since both a and are integers, a-2b and b are integers, 

Hence we can write, 3 =rs, where r and s are coprimes.

That means 3   is rational. But this contradicts the fact that 3   is irrational.

Hence our assumption (that 2 + 3 is rational), was wrong.

Therefore, we conclude that 2 + 3 is irrational.

Theorem 1.5: Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form pq, where p and q are coprime, and the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers.

Theorem 1.6: Let x = pq be a rational number, such that the prime factorization of q is of the form 2n5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

Theorem 1.7: Let x = pq be a rational number, such that the prime factorization of q is not of the form 2n5m, where n, m are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating (recurring).

To determine after how many decimal places, the decimal expansion will terminate

Any number of the form

x =  p2m5n(m>n) 

can be written as

x =  p2m5n ×5m-n5m-n 

 x =  p×5m-n2m5m 

Hence, the number p×5m-n is divided by a multiple of 10 with m trailing zeroes. By looking at the number we can find the decimal places in the result.

Example: After how many decimal places the decimal expansion of the rational number 3323×5 will terminate ?

Solution:

The given number can be represented as

3323×5=3323×5×53-153-1

=33×5223×53= 33×251000

=8251000=0.825

Hence the number will terminate after 3 decimal places.