Properties of Rational Number –

• Expressed in decimal as P/q (q ≠ 0) and if q is of form 2n5m (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 = 22×5×7

HCF and LCM by prime factorisation method

Let any two number be 6 & 20

6 = 2 × 3

20 = 22 × 5

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

LCM (6, 20) =22 × 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,

3b2 = a2         So, a2 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.

• Pi (π) is the 16th 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 = π2 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

1. Show that 5 + 2√7 is an irrational number, where √7 is given to be an   irrational number                                                                                    

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 12n can end with the digit 0 for any natural number n.   12n = (2 × 2 × 3)n                                                                                     

Solution:

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

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

12n = (2 × 2 × 3)n

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

⸫ it is not a multiple of 5.

Hence, 12n 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.       

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

a2 = 9m2………………………………..   (1)

a2 = 9m2 + 6m +1 ………………….   (2)

a2 = 9m2 + 12m + 4 ……………….   (3)

From (1), a2 = 3(3m2) = 3q                         ; Where q = 3m2

From (2), a2 = 3(3m2 + 2m) +1 = 3q + 1   ; Where q = (3m2 + 2m)

From (3), a2 = (3m + 2)2 = 3q                              ; Where q = (3m + 2)2

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

Solution:

Prime factorisation

2048 = 211

960 = 26 × 3 × 5

HCF (2048, 960) = 26 = 64

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

Solution:

Prime factorisation

404 = 22 × 101

96 = 25 × 3

HCF (404, 96) = 22 = 4

LCM (404, 96) = 25 × 3 × 101 = 9696

HCM × LCM = Product of the two given numbers

4 × 9696 = 404 × 96

38784 = 38784

Hence, verified

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