# Real Numbers - Class 10 : Notes

Notes for real numbers chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.

**(1) Euclidâ€™s Division Lemma:
**

**Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 â‰¤ r < b.**

*Theorem*:**(2) Euclidâ€™s division algorithm:Â ****To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below:
**Step 1: Apply Euclidâ€™s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 â‰¤ r < d.

Step 2: If r = 0, d is the HCF of c and d. If r â‰ 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.

** For Example:Â **Use Euclidâ€™s division algorithm to find the HCF of 135 and 225.

Step 1: Here 225 > 135, on applying the division lemma to 225 and 135, we get 225 = 135 x 1 + 90

Step 2: Since, remainder â‰ 0, we again apply division lemma to 135 and 90, we get 135 = 90 x 1 + 45

Step 3: Again, applying division lemma to 90 and 45, we get 90 = 45 x 2 + 0

The remainder has become zero. And since the divisor at this step is 45, the HCF of 135 and 225 is 45.

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

Let

*a*be any positive odd integer. And we apply division algorithm with

*a*and

*b*= 4.

As 0 â‰¤ r < 4, the possible remainders could be 0, 1, 2 and 3.

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

Now, since

*a*is odd, so

*a*cannot be 4q or 4q + 2 (as both are divisible by 2).

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

**(3) The Fundamental Theorem of Arithmetic:
**

**Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. In general, given a composite number x, we factorise it as x = p**

*Theorem*:_{1}p

_{2}... p

_{n}, where p

_{1}, p

_{2},..., p

_{n}are primes and written in ascending order, i.e., p

_{1}â‰¤ p

_{2}â‰¤ . . . â‰¤ p

_{n }. If we combine the same primes, we will get powers of primes.

** For Example:Â **The prime factors of 32760 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5 Ã— 7 Ã— 13 = 2

^{3}Ã— 3

^{2}Ã— 5 Ã— 7 Ã— 13.

** For Example:Â **Find the HCF and LCM of 96 and 404 by prime factorisation method.

The prime factorisation of 96 is 2

^{5}x 3. And that of 404 is 2

^{2}x 101.

Hence, HCF of 96 and 404 will be 2

^{2}= 4.

Now, LCM (96, 404) = (96 x 404)/(HCF (96, 404)) = (96 x 404)/4 = 9696.

** For Example:Â **Check whether 6

^{n}can end with the digit 0 for any natural number n.

If a number ends with digit 0, then, it must be divisible by 10 or in other words, it will be divisible by 2 and 5 as 10 = 2 x 5.

Now, prime factorisation of 6

^{n}= (2 x 3)

^{n}.

Here, 5 is not in the prime factorisation of 6

^{n}. Hence, for any value of n, 6

^{n}will not be divisible by 5.

Thus, 6

^{n}cannot end with the digit 0 for any natural number n.

**(4) Revisiting Irrational Numbers:
**

**Irrational Number :**are the numbers which cannot be written in

*p/q*form, where

*p*and

*q*are integers and

*q*â‰ 0.

**Let p be a prime number. If p divides a**

*Theorem 1*:^{2}, then p divides a, where a is a positive integer.

*Proof*:Â Suppose the prime factorisation of a is as follows:

(i) a = p

_{1}p

_{2}....p

_{n}, where p

_{1},p

_{2},....pn are primes.

(ii) On squaring both the sides, we get,

(iii) a

^{2}= (p

_{1}p

_{2}....p

_{n}) ( p

_{1}p

_{2}....p

_{n}) = p

_{1}

^{2}p

_{2}

^{2}....p

_{n}

^{2}.

(iv) It is given that p divides a

^{2}. Hence, we can say that p is one of the prime factors of a

^{2}as per the Fundamental Theorem of Arithmetic.

(v) However, as per the uniqueness part of the Fundamental Theorem of Arithmetic, we can deduce that the only prime factors of a

^{2 }are p

_{1}p

_{2}....p

_{n}. Thus, p is one of p

_{1}p

_{2}....p

_{n}.

Since, a = p

_{1}p

_{2}....p

_{n},p divides a.

**Â Â**

** Theorem 2:** âˆš2 is irrational.

*Proof*:Â We shall start by assuming âˆš2 as rational. In other words, we need to find integers x and y such that âˆš2 = x/y.

(i) Let x and y have a common factor other than 1, and so we can divide by that common factor and assume that x and y are coprime. So, yâˆš2 = x.

(ii) Squaring both side, we get, 2y

^{2}= x

^{2}.

(iii) Thus, 2 divides x

^{2}. and by theorem we can say that 2 divides x.

(iv) Hence, x = 2z for some integer z.

(v) Substituting x, we get, 2x

^{2}= 4z

^{2}e. y

^{2}= 4z

^{2}; which means y

^{2 }is divisible by 2, and so y will also be divisible by 2.

(vi) Now, from theorem, x and y will have 2 as a common factor. But, it is opposite to fact that x and y are co-prime.

(vii) Hence, we can conclude âˆš2 is irrational.

** For Example:Â **Prove that âˆš3 is irrational.

We shall start by assuming âˆš3 as rational. In other words, we need to find integers x and y such that âˆš3 = x/y.

Let x and y have a common factor other than 1, and so we can divide by that common factor and assume that x and y are coprime. So, yâˆš3 = x.

Squaring both side, we get, 3y

^{2}= x

^{2}.

Thus, x

^{2}is divisible by 3, and by theorem we can say that x is also divisible by 3.

Hence, x = 3z for some integer z.

Substituting a, we get, 3x

^{2}= 9z

^{2}i.e. y

^{2}= 3z

^{2}; which means y

^{2 }is divisible by 3, and so y will also be divisible by 3.

Now, from theorem, x and y will have 3 as a common factor. But, it is opposite to fact that x and y are co-prime.

Hence, we can conclude âˆš3 is irrational.

** For Example:Â **Prove that 6 + âˆš2 is irrational.

Let us assume 6 + âˆš2 to be rational.

Therefore, we must find two integers a, b (b â‰ 0) such that

6 + âˆš2 = a/b i.e. âˆš2 = a/b â€“ 6.

Since, a and b are integers, a/b â€“ 6 is also rational and hence âˆš2 must be rational.

Now, this contradicts the fact that âˆš2 is irrational.

Hence, 6 + âˆš2 is irrational.

**(5) Revisiting Rational Numbers and Their Decimal Expansions:
**

**Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form, p/q where p and q are co-prime, and the prime factorisation of q is of the form 2**

*Theorem 1*:^{n}5

^{m}, where n, m are non-negative integers.

**13/125 = 13/5**

*For Example*:Â^{3}= (13 x 2

^{3})/(2

^{3 }x 5

^{3}) = 104/10

^{3}= 0.104

** Theorem 2:** Let x = p/q be a rational number, such that the prime factorisation of q is of the form 2

^{n }5

^{m}, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

** Theorem 3:** Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2

^{n}5

^{m}, where n, m are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating (recurring).

**Without actually performing the long division, state whether 6/15 will have a terminating decimal expansion or a non-terminating repeating decimal expansion.**

*For Example*:ÂThe prime factorisation of 6/15 can be written as

6/15 = (2 x 3)/(3 x 5) = 2/5

Here, the denominator is of the form 5

^{n}.

Hence, decimal expansion of 6/15 is terminating.

** For Example:Â **Write down the decimal expansions of 17/8.

The decimal expansion of 17/8 is

** For Example:Â **The following real number has decimal expansions as given below. Decide whether it is rational or not. If it is rational, and of the form, p/q what can you say about the prime factors of q?

Here, as the decimal expansion is non-terminating recurring, the given number is a rational number of the form p/q.

Moreover, q is not of the form 2

^{n}5

^{m}, hence, prime factors of q will also have factors other than 2 or 5.

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