3. Euclidβs division Algorithm
ο΄ Euclidβs division Algorithm β
ππ πππ‘πππ π‘βπ π»πΆπΉ ππ π‘π€π πππ ππ‘ππ£π πππ‘πππππ π ππ¦ π πππ π π€ππ‘β π
> π, ππππππ€ π‘βπ π π‘πππ πππππ€ βΆ
1. Apply Euclidβs division lemma to π and π . So , we find whole numbers , π πππ π
such that π = ππ + π, 0 β€ π < π.
2. If π = 0 , d is the HCF of π and π . If π β 0 apply the division lemma to π and π .
3. Continue the process till the remainder is zero . The divisor at this stage will be the
required HCF .
4. Example :- Using Euclidβs 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 Γ 3 + 420
Since the remainder 420 β 0 , we apply the division lemma to 4052 and 420 to get
4052 = 420 Γ 9 + 272
We consider the new divisor 420 and new remainder 272 apply the division lemma to get
420 = 272 Γ 1 + 148
Now we continue this process till remainder is zero .
272 = 148 Γ 1 + 124
148 = 124 Γ 1 + 24
124 = 24 Γ 5 + 4
24 = 4 Γ 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
πΈπ£πππ¦ ππππππ ππ‘π ππ’ππππ πππ ππ ππ₯ππππ π ππ ππ π πππππ’ππ‘ ππ ππππππ , πππ
π‘βππ ππππ‘ππππ ππ‘πππ ππ π’ππππ’π , πππππ‘ ππππ π‘βπ πππππ ππ π€βππβ π‘βππ¦ ππππ’π.
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
πΏππ‘ π ππ π πππππ ππ’ππππ ππ π πππ£ππππ π2, π‘βππ π πππ£ππππ π , π ππ π πππ ππ‘ππ£π πππ‘ππππ
Theorem - 2 ππ πππππ‘πππππ.
proof
Let us assume on contrary that 2 is rational where a and b are co-prime .
β 2 =
π
π
(π β 0)
squaring on both sides
2
2
=
π
π
2
π2
=
π2
2
Here 2 divides π2
, so it also divides π .
so we can write a=2c for some integer c .
7. Substituting for π
we get
2π2
= 4c2
π2
= 2c2
π2
=
π2
2
Here 2 divides π2 , so it also divides π .
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 .