The document explains Euclid's division lemma and algorithm for finding the highest common factor (HCF) of two positive integers. It illustrates the process using the example of finding the HCF of 12576 and 4052, ultimately determining that the HCF is 4. Additionally, it discusses the fundamental theorem of arithmetic regarding prime factorization and touches on irrational and rational numbers with examples.

1. REAL NUMBERS

2. Euclid’s Division Lemma And Algorithm
𝐺𝑖𝑣𝑒𝑛 𝑡𝑤𝑜 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑎 𝑎𝑛𝑑 𝑏 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡 𝑢𝑛𝑖𝑞𝑢𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑞 𝑎𝑛𝑑 𝑟 𝑠𝑎𝑡𝑖𝑠𝑓𝑦𝑖𝑛𝑔 𝑎
= 𝑏𝑞 + 𝑟, 0 ≤ 𝑟 < 𝑏
Here ,
𝑎 = 𝑑𝑖𝑣𝑖𝑑𝑒𝑛𝑑 𝑏 = 𝑑𝑖𝑣𝑖𝑠𝑜𝑟 𝑞 = 𝑞𝑢𝑜𝑡𝑒𝑖𝑛𝑡 𝑟 =
𝑟𝑒𝑚𝑎𝑖𝑛𝑑𝑒𝑟
Example
13 = 2 × 6 + 1

3. Euclid’s division Algorithm
 Euclid’s division Algorithm –
𝑇𝑜 𝑜𝑏𝑡𝑎𝑖𝑛 𝑡ℎ𝑒 𝐻𝐶𝐹 𝑜𝑓 𝑡𝑤𝑜 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠 𝑠𝑎𝑦 𝑎 𝑎𝑛𝑑 𝑏 𝑤𝑖𝑡ℎ 𝑎
> 𝑏, 𝑓𝑜𝑙𝑙𝑜𝑤 𝑡ℎ𝑒 𝑠𝑡𝑒𝑝𝑠 𝑏𝑒𝑙𝑜𝑤 ∶
1. Apply Euclid’s division lemma to 𝑎 and 𝑏 . So , we find whole numbers , 𝑞 𝑎𝑛𝑑 𝑟
such that 𝑎 = 𝑏𝑞 + 𝑟, 0 ≤ 𝑟 < 𝑏.
2. If 𝑟 = 0 , d is the HCF of 𝑎 and 𝑏 . If 𝑟 ≠ 0 apply the division lemma to 𝑏 and 𝑟 .
3. Continue the process till the remainder is zero . The divisor at this stage will be the
required HCF .

4. Example :- Using Euclid’s division algorithm find the HCF of 12576 and 4052
.
Since 12576 > 4052 we apply the division lemma to 12576 and 4052 to get
12576 = 4052 × 3 + 420
Since the remainder 420 ≠ 0 , we apply the division lemma to 4052 and 420 to get
4052 = 420 × 9 + 272
We consider the new divisor 420 and new remainder 272 apply the division lemma to get
420 = 272 × 1 + 148
Now we continue this process till remainder is zero .
272 = 148 × 1 + 124
148 = 124 × 1 + 24
124 = 24 × 5 + 4
24 = 4 × 6 + 0
The remainder has now become 0 , so our procedure stops . Since the divisor at this stage is 4 ,
the HCF of 12576 and 4052 is 4 .

5. Fundamental Theorem of Arithmetic
𝐸𝑣𝑒𝑟𝑦 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑐𝑎𝑛 𝑏𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑎𝑠 𝑎 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑝𝑟𝑖𝑚𝑒𝑠, 𝑎𝑛𝑑
𝑡ℎ𝑖𝑠 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑠𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑢𝑛𝑖𝑞𝑢𝑒 , 𝑎𝑝𝑎𝑟𝑡 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑜𝑟𝑑𝑒𝑟 𝑖𝑛 𝑤ℎ𝑖𝑐ℎ 𝑡ℎ𝑒𝑦 𝑜𝑐𝑐𝑢𝑟.
Now factorise a large number say 32760
2 32760
2 16380
2 8190
3 4095
3 1365
5 455
7 91
13 13

6. Revisiting Irrational Numbers
𝐿𝑒𝑡 𝑝 𝑏𝑒 𝑎 𝑝𝑟𝑖𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑓 𝑝 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎2, 𝑡ℎ𝑒𝑛 𝑝 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 𝑎 , 𝑎 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟
Theorem - 2 𝑖𝑠 𝑖𝑟𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙.
proof
Let us assume on contrary that 2 is rational where a and b are co-prime .
→ 2 =
𝑎
𝑏
(𝑏 ≠ 0)
squaring on both sides
2
2
=
𝑎
𝑏
2
𝑏2
=
𝑎2
2
Here 2 divides 𝑎2
, so it also divides 𝑎 .
so we can write a=2c for some integer c .

7. Substituting for 𝑎
we get
2𝑏2
= 4c2
𝑏2
= 2c2
𝑐2
=
𝑏2
2
Here 2 divides 𝑏2 , so it also divides 𝑏 .
This creates a contradiction that a and b have no common factors other than 1 .
This contradiction has arisen because of our wrong assumption .
So we conclude that 2 is a irrational number .

8. Revisiting Rational numbers and their decimal expansions
Theorem
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 𝑐𝑜𝑝𝑟𝑖𝑚𝑒,
𝑎𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑠𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑞 𝑖𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 2 𝑛
5 𝑛
, where n and m
𝑎𝑟𝑒 𝑛𝑜𝑛 − 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠
Example
1.
3
8
=
3
23
2.
13
125
=
13
53

9. Thank You
Made by :- Amit Choube
Class :- 10th B