**Properties of Rational Number –**

- Expressed in decimal as P/q (q ≠ 0) and if q is of form 2
^{n}5^{m}(either ‘m’ or ‘n’ may be 0) then it is terminating decimal, otherwise, non-terminating repeating. - E.g.- Either terminating (7/2 = 3.5) or non-terminating repeating (1/3 = 0.3333_ _ _ or 1/7 = 0.142857…..)

Note : Irrational numbers are non-terminating and non-repeating.

**Euclid Division Lemma **

a = bq + r where, a – any +ve integer (known as dividend)

b – any +ve integer (known as divisor)

q – unique integer (known as quotient)

r – unique integer (known as remainder); 0 ≤ r < b

e.g.- let ‘a’ (dividend) = 16 and ‘b’ (divisor) = 3; then this can be written in Euclid division lemma form as

16 = 3 × 5 + 1 where 5 is quotient and 1 is remainder.

**Application of Euclid Division Lemma**

To compute the Highest Common Factor (HCF) of two positive integer.

e.g.- let two positive integer be 5 and 6.

Now applying Euclid Division Lemma, we get

Note: The last step will have the remainder 0 and its divisor will be the HCF.

To find properties of numbers

e.g.- Every even number is of the form 2q whereas every odd number is of the form 2q + 1

Let ‘a’ be any +ve integer as dividend and b = 2 as divisor

a = 2q + r ; q is quotient and r is remainder

since, 0 ≤ r < b and here b= 2

therefore, r = 0 or r = 1

Hence, a = 2q (even) or a = 2q +1 (odd)

Note : odd integer can be of the form 2q +1, 2q + 3, 2q +5 and so on.

**Fundamental theorem of arithmetic (Prime factorisation)**

- A composite number (product of atleast two numbers) can be expressed as a product of prime numbers.
- E.g.- 140 = 2
^{2}×5×7

**HCF and LCM** by prime factorisation method

Let any two number be 6 & 20

6 = 2 × 3

20 = 2^{2} × 5

HCF (6, 20) = 2 (product of common in both 6’s & 20’s prime factorisation)

LCM (6, 20) =2^{2} × 3 × 5=60 (product of the greatest power of each prime factor)

And also, the product of the two numbers is equal to the product of the HCF & LCM of that two numbers that is the above example can be written as:

**6 ****× ****20 = 2 (HCF) ****×**** 60 (LCM)**

**How to prove a number as irrational number**

1) To prove √3 as irrational

Method- Assume √3 as rational

Then √3 can be written in fraction, let write it as (a/b), where it is in its simplest form.

√3 = a/b b√3 = a……………..(1)

Squaring both the sides,

3b^{2} = a^{2} So, a^{2 }is divisible by 3

Which means that ‘a’ is also divisible by 3 or a multiple of 3

⸫ It can be written as; a = 3c and if it is so, then b is also divisible by 3 or a multiple of 3 from (1)

But this can not be possible as it contradicts the assumption (a/b is in its simplest form or ‘a’ & ‘b’ can have no common factor except 1)

⸫ √3 is not rational. Hence, √3 is an irrational number.

2) To prove 5-√3 as irrational

Method- Assume (5-√3) as rational, then

5-√3 = a/b √3 = 5 – (a/b)

√3 = (5b – a)/b

Here, √3 is irrational but 5 – (a/b) is rational

⸫ it contradicts the assumption. Hence, 5-√3 is an irrational.

**Some facts about irrational number**

- Pi (π) is the 16
^{th}letter of Greek alphabet. The value of π is estimated as the ratio of the circumference of a circle to its diameter. This can be calculated using any circle.

- Every irrational number has infinite decimal digits and that too non-repeating digits.

- The product of two same irrational numbers is not always a rational number.
- e.g.- π × π = π
^{2}is an irrational number

- Other than ‘π’, Euler’s number (e) and Golden ratio (Ψ) are also irrational numbers.

- Therefore we can say, irrational numbers are not absolute numbers.

**Previous years questions**

- Show that 5 + 2√7 is an irrational number, where √7 is given to be an irrational number [2020]

Solution:

Let assume 5 + 2√7 is an rational number

5 + 2√7 = a/b 2√7 = (a/b) – 5

√7 = (a – 5b)/2b

Here, √7 is an irrational number, but (a – 5b)/2b is rational

⸫ it contradicts the assumption. Hence, 5 + 2√7 is an irrational.

2. Check whether 12^{n} can end with the digit 0 for any natural number n. 12^{n} = (2 × 2 × 3)^{n} [2020]

Solution:

If 12^{n }ends with the digit 0, then 12^{n} has to be a multiple of 5.

Since, 5 is a prime number, it has to be there in the prime factorisation of 12^{n}.

12^{n} = (2 × 2 × 3)^{n}

As there is no 5 in the prime factorisation of 12^{n},

⸫ it is not a multiple of 5.

Hence, 12^{n} does not end with 0.

3. Use Euclid Division Lemma to show that the square of any positive integer is either of the form 3q or 3q + 1 for some integer q. [2020]

Solution:

Let ‘a’ be any +ve integer as dividend and b = 3 as divisor

a = 3m + n; m is the quotient and n is the remainder

0 ≤ n < 3

⸫ a is of the form 3m; 3m + 1; and 3m + 2

squaring each form of ‘a’ , we get

a^{2} = 9m^{2}……………………………….. (1)

a^{2} = 9m^{2} + 6m +1 …………………. (2)

a^{2} = 9m^{2} + 12m + 4 ………………. (3)

From (1), a^{2} = 3(3m^{2}) = 3q ; Where q = 3m^{2}

From (2), a^{2} = 3(3m^{2} + 2m) +1 = 3q + 1 ; Where q = (3m^{2} + 2m)

From (3), a^{2} = (3m + 2)^{2} = 3q^{ } ; Where q = (3m + 2)^{2}

4. Using Euclid’s Algorithm, find the HCF of 2048 and 960. [2019]

Solution:

Prime factorisation

2048 = 2^{11}

960 = 2^{6} × 3 × 5

HCF (2048, 960) = 2^{6} = 64

5. Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers. [2018]

Solution:

Prime factorisation

404 = 2^{2} × 101

96 = 2^{5} × 3

HCF (404, 96) = 2^{2} = 4

LCM (404, 96) = 2^{5} × 3 × 101 = 9696

HCM × LCM = Product of the two given numbers

4 × 9696 = 404 × 96

38784 = 38784

Hence, verified

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