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REAL NUMBERS
Euclid’s Division Lemma And Algorithm
𝐺𝑖𝑣𝑒𝑛 π‘‘π‘€π‘œ π‘π‘œπ‘ π‘–π‘‘π‘–π‘£π‘’ π‘–π‘›π‘‘π‘’π‘”π‘’π‘Ÿπ‘  π‘Ž π‘Žπ‘›π‘‘ 𝑏 π‘‘β„Žπ‘’π‘Ÿπ‘’ 𝑒π‘₯𝑖𝑠𝑑 π‘’π‘›π‘–π‘žπ‘’π‘’ π‘–π‘›π‘‘π‘’π‘”π‘’π‘Ÿπ‘  π‘ž π‘Žπ‘›π‘‘ π‘Ÿ π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘¦π‘–π‘›π‘” π‘Ž
= π‘π‘ž + π‘Ÿ, 0 ≀ π‘Ÿ < 𝑏
Here ,
π‘Ž = 𝑑𝑖𝑣𝑖𝑑𝑒𝑛𝑑 𝑏 = π‘‘π‘–π‘£π‘–π‘ π‘œπ‘Ÿ π‘ž = π‘žπ‘’π‘œπ‘‘π‘’π‘–π‘›π‘‘ π‘Ÿ =
π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ
Example
13 = 2 Γ— 6 + 1
Euclid’s division Algorithm
ο‚΄ Euclid’s division Algorithm –
π‘‡π‘œ π‘œπ‘π‘‘π‘Žπ‘–π‘› π‘‘β„Žπ‘’ 𝐻𝐢𝐹 π‘œπ‘“ π‘‘π‘€π‘œ π‘π‘œπ‘ π‘–π‘‘π‘–π‘£π‘’ π‘–π‘›π‘‘π‘’π‘”π‘’π‘Ÿπ‘  π‘ π‘Žπ‘¦ π‘Ž π‘Žπ‘›π‘‘ 𝑏 π‘€π‘–π‘‘β„Ž π‘Ž
> 𝑏, π‘“π‘œπ‘™π‘™π‘œπ‘€ π‘‘β„Žπ‘’ 𝑠𝑑𝑒𝑝𝑠 π‘π‘’π‘™π‘œπ‘€ ∢
1. Apply Euclid’s division lemma to π‘Ž and 𝑏 . So , we find whole numbers , π‘ž π‘Žπ‘›π‘‘ π‘Ÿ
such that π‘Ž = π‘π‘ž + π‘Ÿ, 0 ≀ π‘Ÿ < 𝑏.
2. If π‘Ÿ = 0 , d is the HCF of π‘Ž and 𝑏 . If π‘Ÿ β‰  0 apply the division lemma to 𝑏 and π‘Ÿ .
3. Continue the process till the remainder is zero . The divisor at this stage will be the
required HCF .
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 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 .
Fundamental Theorem of Arithmetic
πΈπ‘£π‘’π‘Ÿπ‘¦ π‘π‘œπ‘šπ‘π‘œπ‘ π‘–π‘‘π‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘π‘Žπ‘› 𝑏𝑒 𝑒π‘₯π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘‘ π‘Žπ‘  π‘Ž π‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘‘ π‘œπ‘“ π‘π‘Ÿπ‘–π‘šπ‘’π‘ , π‘Žπ‘›π‘‘
π‘‘β„Žπ‘–π‘  π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘Žπ‘‘π‘–π‘œπ‘› 𝑖𝑠 π‘’π‘›π‘–π‘žπ‘’π‘’ , π‘Žπ‘π‘Žπ‘Ÿπ‘‘ π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘œπ‘Ÿπ‘‘π‘’π‘Ÿ 𝑖𝑛 π‘€β„Žπ‘–π‘β„Ž π‘‘β„Žπ‘’π‘¦ π‘œπ‘π‘π‘’π‘Ÿ.
Now factorise a large number say 32760
2 32760
2 16380
2 8190
3 4095
3 1365
5 455
7 91
13 13
Revisiting Irrational Numbers
𝐿𝑒𝑑 𝑝 𝑏𝑒 π‘Ž π‘π‘Ÿπ‘–π‘šπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ 𝑖𝑓 𝑝 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 π‘Ž2, π‘‘β„Žπ‘’π‘› 𝑝 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 π‘Ž , π‘Ž 𝑖𝑠 π‘Ž π‘π‘œπ‘ π‘–π‘‘π‘–π‘£π‘’ π‘–π‘›π‘‘π‘’π‘”π‘’π‘Ÿ
Theorem - 2 𝑖𝑠 π‘–π‘Ÿπ‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘™.
proof
Let us assume on contrary that 2 is rational where a and b are co-prime .
β†’ 2 =
π‘Ž
𝑏
(𝑏 β‰  0)
squaring on both sides
2
2
=
π‘Ž
𝑏
2
𝑏2
=
π‘Ž2
2
Here 2 divides π‘Ž2
, so it also divides π‘Ž .
so we can write a=2c for some integer c .
Substituting for π‘Ž
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 .
Revisiting Rational numbers and their decimal expansions
Theorem
𝐿𝑒𝑑 π‘₯ 𝑏𝑒 π‘Ž π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘€β„Žπ‘œπ‘ π‘’ π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ 𝑒π‘₯π‘π‘Žπ‘›π‘ π‘–π‘œπ‘› π‘‘π‘’π‘Ÿπ‘šπ‘–π‘›π‘Žπ‘‘π‘’π‘  .
π‘‡β„Žπ‘’π‘› π‘₯ π‘π‘Žπ‘› 𝑏𝑒 𝑒π‘₯π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘‘ 𝑖𝑛 π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘š
𝑝
π‘ž
, π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑝 π‘Žπ‘›π‘‘ π‘ž π‘Žπ‘Ÿπ‘’ π‘π‘œπ‘π‘Ÿπ‘–π‘šπ‘’,
π‘Žπ‘›π‘‘ π‘‘β„Žπ‘’ π‘π‘Ÿπ‘–π‘šπ‘’ π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘Žπ‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘ž 𝑖𝑠 π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘š 2 𝑛
5 𝑛
, where n and m
π‘Žπ‘Ÿπ‘’ π‘›π‘œπ‘› βˆ’ π‘›π‘’π‘”π‘Žπ‘‘π‘–π‘£π‘’ π‘–π‘›π‘‘π‘’π‘”π‘’π‘Ÿπ‘ 
Example
1.
3
8
=
3
23
2.
13
125
=
13
53
Thank You
Made by :- Amit Choube
Class :- 10th B

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Real numbers- class 10 mathematics

  • 2. Euclid’s Division Lemma And Algorithm 𝐺𝑖𝑣𝑒𝑛 π‘‘π‘€π‘œ π‘π‘œπ‘ π‘–π‘‘π‘–π‘£π‘’ π‘–π‘›π‘‘π‘’π‘”π‘’π‘Ÿπ‘  π‘Ž π‘Žπ‘›π‘‘ 𝑏 π‘‘β„Žπ‘’π‘Ÿπ‘’ 𝑒π‘₯𝑖𝑠𝑑 π‘’π‘›π‘–π‘žπ‘’π‘’ π‘–π‘›π‘‘π‘’π‘”π‘’π‘Ÿπ‘  π‘ž π‘Žπ‘›π‘‘ π‘Ÿ π‘ π‘Žπ‘‘π‘–π‘ π‘“π‘¦π‘–π‘›π‘” π‘Ž = π‘π‘ž + π‘Ÿ, 0 ≀ π‘Ÿ < 𝑏 Here , π‘Ž = 𝑑𝑖𝑣𝑖𝑑𝑒𝑛𝑑 𝑏 = π‘‘π‘–π‘£π‘–π‘ π‘œπ‘Ÿ π‘ž = π‘žπ‘’π‘œπ‘‘π‘’π‘–π‘›π‘‘ π‘Ÿ = π‘Ÿπ‘’π‘šπ‘Žπ‘–π‘›π‘‘π‘’π‘Ÿ Example 13 = 2 Γ— 6 + 1
  • 3. Euclid’s division Algorithm ο‚΄ Euclid’s division Algorithm – π‘‡π‘œ π‘œπ‘π‘‘π‘Žπ‘–π‘› π‘‘β„Žπ‘’ 𝐻𝐢𝐹 π‘œπ‘“ π‘‘π‘€π‘œ π‘π‘œπ‘ π‘–π‘‘π‘–π‘£π‘’ π‘–π‘›π‘‘π‘’π‘”π‘’π‘Ÿπ‘  π‘ π‘Žπ‘¦ π‘Ž π‘Žπ‘›π‘‘ 𝑏 π‘€π‘–π‘‘β„Ž π‘Ž > 𝑏, π‘“π‘œπ‘™π‘™π‘œπ‘€ π‘‘β„Žπ‘’ 𝑠𝑑𝑒𝑝𝑠 π‘π‘’π‘™π‘œπ‘€ ∢ 1. Apply Euclid’s division lemma to π‘Ž and 𝑏 . So , we find whole numbers , π‘ž π‘Žπ‘›π‘‘ π‘Ÿ such that π‘Ž = π‘π‘ž + π‘Ÿ, 0 ≀ π‘Ÿ < 𝑏. 2. If π‘Ÿ = 0 , d is the HCF of π‘Ž and 𝑏 . If π‘Ÿ β‰  0 apply the division lemma to 𝑏 and π‘Ÿ . 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 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 πΈπ‘£π‘’π‘Ÿπ‘¦ π‘π‘œπ‘šπ‘π‘œπ‘ π‘–π‘‘π‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘π‘Žπ‘› 𝑏𝑒 𝑒π‘₯π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘‘ π‘Žπ‘  π‘Ž π‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘‘ π‘œπ‘“ π‘π‘Ÿπ‘–π‘šπ‘’π‘ , π‘Žπ‘›π‘‘ π‘‘β„Žπ‘–π‘  π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘Žπ‘‘π‘–π‘œπ‘› 𝑖𝑠 π‘’π‘›π‘–π‘žπ‘’π‘’ , π‘Žπ‘π‘Žπ‘Ÿπ‘‘ π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘œπ‘Ÿπ‘‘π‘’π‘Ÿ 𝑖𝑛 π‘€β„Žπ‘–π‘β„Ž π‘‘β„Žπ‘’π‘¦ π‘œπ‘π‘π‘’π‘Ÿ. 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 𝐿𝑒𝑑 𝑝 𝑏𝑒 π‘Ž π‘π‘Ÿπ‘–π‘šπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ 𝑖𝑓 𝑝 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 π‘Ž2, π‘‘β„Žπ‘’π‘› 𝑝 𝑑𝑖𝑣𝑖𝑑𝑒𝑠 π‘Ž , π‘Ž 𝑖𝑠 π‘Ž π‘π‘œπ‘ π‘–π‘‘π‘–π‘£π‘’ π‘–π‘›π‘‘π‘’π‘”π‘’π‘Ÿ Theorem - 2 𝑖𝑠 π‘–π‘Ÿπ‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘™. proof Let us assume on contrary that 2 is rational where a and b are co-prime . β†’ 2 = π‘Ž 𝑏 (𝑏 β‰  0) squaring on both sides 2 2 = π‘Ž 𝑏 2 𝑏2 = π‘Ž2 2 Here 2 divides π‘Ž2 , so it also divides π‘Ž . so we can write a=2c for some integer c .
  • 7. Substituting for π‘Ž 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 𝐿𝑒𝑑 π‘₯ 𝑏𝑒 π‘Ž π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘€β„Žπ‘œπ‘ π‘’ π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ 𝑒π‘₯π‘π‘Žπ‘›π‘ π‘–π‘œπ‘› π‘‘π‘’π‘Ÿπ‘šπ‘–π‘›π‘Žπ‘‘π‘’π‘  . π‘‡β„Žπ‘’π‘› π‘₯ π‘π‘Žπ‘› 𝑏𝑒 𝑒π‘₯π‘π‘Ÿπ‘’π‘ π‘ π‘’π‘‘ 𝑖𝑛 π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘š 𝑝 π‘ž , π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑝 π‘Žπ‘›π‘‘ π‘ž π‘Žπ‘Ÿπ‘’ π‘π‘œπ‘π‘Ÿπ‘–π‘šπ‘’, π‘Žπ‘›π‘‘ π‘‘β„Žπ‘’ π‘π‘Ÿπ‘–π‘šπ‘’ π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘Žπ‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘ž 𝑖𝑠 π‘œπ‘“ π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘š 2 𝑛 5 𝑛 , where n and m π‘Žπ‘Ÿπ‘’ π‘›π‘œπ‘› βˆ’ π‘›π‘’π‘”π‘Žπ‘‘π‘–π‘£π‘’ π‘–π‘›π‘‘π‘’π‘”π‘’π‘Ÿπ‘  Example 1. 3 8 = 3 23 2. 13 125 = 13 53
  • 9. Thank You Made by :- Amit Choube Class :- 10th B