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REAL NUMBERS
Euclidโ€™s Division Lemma And Algorithm
๐บ๐‘–๐‘ฃ๐‘’๐‘› ๐‘ก๐‘ค๐‘œ ๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ๐‘  ๐‘Ž ๐‘Ž๐‘›๐‘‘ ๐‘ ๐‘กโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘’๐‘ฅ๐‘–๐‘ ๐‘ก ๐‘ข๐‘›๐‘–๐‘ž๐‘ข๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ๐‘  ๐‘ž ๐‘Ž๐‘›๐‘‘ ๐‘Ÿ ๐‘ ๐‘Ž๐‘ก๐‘–๐‘ ๐‘“๐‘ฆ๐‘–๐‘›๐‘” ๐‘Ž
= ๐‘๐‘ž + ๐‘Ÿ, 0 โ‰ค ๐‘Ÿ < ๐‘
Here ,
๐‘Ž = ๐‘‘๐‘–๐‘ฃ๐‘–๐‘‘๐‘’๐‘›๐‘‘ ๐‘ = ๐‘‘๐‘–๐‘ฃ๐‘–๐‘ ๐‘œ๐‘Ÿ ๐‘ž = ๐‘ž๐‘ข๐‘œ๐‘ก๐‘’๐‘–๐‘›๐‘ก ๐‘Ÿ =
๐‘Ÿ๐‘’๐‘š๐‘Ž๐‘–๐‘›๐‘‘๐‘’๐‘Ÿ
Example
13 = 2 ร— 6 + 1
Euclidโ€™s division Algorithm
๏‚ด Euclidโ€™s division Algorithm โ€“
๐‘‡๐‘œ ๐‘œ๐‘๐‘ก๐‘Ž๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐ป๐ถ๐น ๐‘œ๐‘“ ๐‘ก๐‘ค๐‘œ ๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ๐‘  ๐‘ ๐‘Ž๐‘ฆ ๐‘Ž ๐‘Ž๐‘›๐‘‘ ๐‘ ๐‘ค๐‘–๐‘กโ„Ž ๐‘Ž
> ๐‘, ๐‘“๐‘œ๐‘™๐‘™๐‘œ๐‘ค ๐‘กโ„Ž๐‘’ ๐‘ ๐‘ก๐‘’๐‘๐‘  ๐‘๐‘’๐‘™๐‘œ๐‘ค โˆถ
1. Apply Euclidโ€™s division lemma to ๐‘Ž and ๐‘ . So , we find whole numbers , ๐‘ž ๐‘Ž๐‘›๐‘‘ ๐‘Ÿ
such that ๐‘Ž = ๐‘๐‘ž + ๐‘Ÿ, 0 โ‰ค ๐‘Ÿ < ๐‘.
2. If ๐‘Ÿ = 0 , d is the HCF of ๐‘Ž and ๐‘ . If ๐‘Ÿ โ‰  0 apply the division lemma to ๐‘ and ๐‘Ÿ .
3. Continue the process till the remainder is zero . The divisor at this stage will be the
required HCF .
Example :- Using Euclidโ€™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 ร— 3 + 420
Since the remainder 420 โ‰  0 , we apply the division lemma to 4052 and 420 to get
4052 = 420 ร— 9 + 272
We consider the new divisor 420 and new remainder 272 apply the division lemma to get
420 = 272 ร— 1 + 148
Now we continue this process till remainder is zero .
272 = 148 ร— 1 + 124
148 = 124 ร— 1 + 24
124 = 24 ร— 5 + 4
24 = 4 ร— 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
๐ธ๐‘ฃ๐‘’๐‘Ÿ๐‘ฆ ๐‘๐‘œ๐‘š๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘’ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘๐‘Ž๐‘› ๐‘๐‘’ ๐‘’๐‘ฅ๐‘๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘’๐‘‘ ๐‘Ž๐‘  ๐‘Ž ๐‘๐‘Ÿ๐‘œ๐‘‘๐‘ข๐‘๐‘ก ๐‘œ๐‘“ ๐‘๐‘Ÿ๐‘–๐‘š๐‘’๐‘ , ๐‘Ž๐‘›๐‘‘
๐‘กโ„Ž๐‘–๐‘  ๐‘“๐‘Ž๐‘๐‘ก๐‘œ๐‘Ÿ๐‘–๐‘ ๐‘Ž๐‘ก๐‘–๐‘œ๐‘› ๐‘–๐‘  ๐‘ข๐‘›๐‘–๐‘ž๐‘ข๐‘’ , ๐‘Ž๐‘๐‘Ž๐‘Ÿ๐‘ก ๐‘“๐‘Ÿ๐‘œ๐‘š ๐‘กโ„Ž๐‘’ ๐‘œ๐‘Ÿ๐‘‘๐‘’๐‘Ÿ ๐‘–๐‘› ๐‘คโ„Ž๐‘–๐‘โ„Ž ๐‘กโ„Ž๐‘’๐‘ฆ ๐‘œ๐‘๐‘๐‘ข๐‘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
๐ฟ๐‘’๐‘ก ๐‘ ๐‘๐‘’ ๐‘Ž ๐‘๐‘Ÿ๐‘–๐‘š๐‘’ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘–๐‘“ ๐‘ ๐‘‘๐‘–๐‘ฃ๐‘–๐‘‘๐‘’๐‘  ๐‘Ž2, ๐‘กโ„Ž๐‘’๐‘› ๐‘ ๐‘‘๐‘–๐‘ฃ๐‘–๐‘‘๐‘’๐‘  ๐‘Ž , ๐‘Ž ๐‘–๐‘  ๐‘Ž ๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ
Theorem - 2 ๐‘–๐‘  ๐‘–๐‘Ÿ๐‘Ÿ๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™.
proof
Let us assume on contrary that 2 is rational where a and b are co-prime .
โ†’ 2 =
๐‘Ž
๐‘
(๐‘ โ‰  0)
squaring on both sides
2
2
=
๐‘Ž
๐‘
2
๐‘2
=
๐‘Ž2
2
Here 2 divides ๐‘Ž2
, so it also divides ๐‘Ž .
so we can write a=2c for some integer c .
Substituting for ๐‘Ž
we get
2๐‘2
= 4c2
๐‘2
= 2c2
๐‘2
=
๐‘2
2
Here 2 divides ๐‘2 , so it also divides ๐‘ .
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
๐ฟ๐‘’๐‘ก ๐‘ฅ ๐‘๐‘’ ๐‘Ž ๐‘Ÿ๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘คโ„Ž๐‘œ๐‘ ๐‘’ ๐‘‘๐‘’๐‘๐‘–๐‘š๐‘Ž๐‘™ ๐‘’๐‘ฅ๐‘๐‘Ž๐‘›๐‘ ๐‘–๐‘œ๐‘› ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘–๐‘›๐‘Ž๐‘ก๐‘’๐‘  .
๐‘‡โ„Ž๐‘’๐‘› ๐‘ฅ ๐‘๐‘Ž๐‘› ๐‘๐‘’ ๐‘’๐‘ฅ๐‘๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘’๐‘‘ ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘“๐‘œ๐‘Ÿ๐‘š
๐‘
๐‘ž
, ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘ ๐‘Ž๐‘›๐‘‘ ๐‘ž ๐‘Ž๐‘Ÿ๐‘’ ๐‘๐‘œ๐‘๐‘Ÿ๐‘–๐‘š๐‘’,
๐‘Ž๐‘›๐‘‘ ๐‘กโ„Ž๐‘’ ๐‘๐‘Ÿ๐‘–๐‘š๐‘’ ๐‘“๐‘Ž๐‘๐‘ก๐‘œ๐‘Ÿ๐‘–๐‘ ๐‘Ž๐‘ก๐‘–๐‘œ๐‘› ๐‘œ๐‘“ ๐‘ž ๐‘–๐‘  ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘“๐‘œ๐‘Ÿ๐‘š 2 ๐‘›
5 ๐‘›
, where n and m
๐‘Ž๐‘Ÿ๐‘’ ๐‘›๐‘œ๐‘› โˆ’ ๐‘›๐‘’๐‘”๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ๐‘ 
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. Euclidโ€™s Division Lemma And Algorithm ๐บ๐‘–๐‘ฃ๐‘’๐‘› ๐‘ก๐‘ค๐‘œ ๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ๐‘  ๐‘Ž ๐‘Ž๐‘›๐‘‘ ๐‘ ๐‘กโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘’๐‘ฅ๐‘–๐‘ ๐‘ก ๐‘ข๐‘›๐‘–๐‘ž๐‘ข๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ๐‘  ๐‘ž ๐‘Ž๐‘›๐‘‘ ๐‘Ÿ ๐‘ ๐‘Ž๐‘ก๐‘–๐‘ ๐‘“๐‘ฆ๐‘–๐‘›๐‘” ๐‘Ž = ๐‘๐‘ž + ๐‘Ÿ, 0 โ‰ค ๐‘Ÿ < ๐‘ Here , ๐‘Ž = ๐‘‘๐‘–๐‘ฃ๐‘–๐‘‘๐‘’๐‘›๐‘‘ ๐‘ = ๐‘‘๐‘–๐‘ฃ๐‘–๐‘ ๐‘œ๐‘Ÿ ๐‘ž = ๐‘ž๐‘ข๐‘œ๐‘ก๐‘’๐‘–๐‘›๐‘ก ๐‘Ÿ = ๐‘Ÿ๐‘’๐‘š๐‘Ž๐‘–๐‘›๐‘‘๐‘’๐‘Ÿ Example 13 = 2 ร— 6 + 1
  • 3. Euclidโ€™s division Algorithm ๏‚ด Euclidโ€™s division Algorithm โ€“ ๐‘‡๐‘œ ๐‘œ๐‘๐‘ก๐‘Ž๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐ป๐ถ๐น ๐‘œ๐‘“ ๐‘ก๐‘ค๐‘œ ๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ๐‘  ๐‘ ๐‘Ž๐‘ฆ ๐‘Ž ๐‘Ž๐‘›๐‘‘ ๐‘ ๐‘ค๐‘–๐‘กโ„Ž ๐‘Ž > ๐‘, ๐‘“๐‘œ๐‘™๐‘™๐‘œ๐‘ค ๐‘กโ„Ž๐‘’ ๐‘ ๐‘ก๐‘’๐‘๐‘  ๐‘๐‘’๐‘™๐‘œ๐‘ค โˆถ 1. Apply Euclidโ€™s division lemma to ๐‘Ž and ๐‘ . So , we find whole numbers , ๐‘ž ๐‘Ž๐‘›๐‘‘ ๐‘Ÿ such that ๐‘Ž = ๐‘๐‘ž + ๐‘Ÿ, 0 โ‰ค ๐‘Ÿ < ๐‘. 2. If ๐‘Ÿ = 0 , d is the HCF of ๐‘Ž and ๐‘ . If ๐‘Ÿ โ‰  0 apply the division lemma to ๐‘ and ๐‘Ÿ . 3. Continue the process till the remainder is zero . The divisor at this stage will be the required HCF .
  • 4. Example :- Using Euclidโ€™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 ร— 3 + 420 Since the remainder 420 โ‰  0 , we apply the division lemma to 4052 and 420 to get 4052 = 420 ร— 9 + 272 We consider the new divisor 420 and new remainder 272 apply the division lemma to get 420 = 272 ร— 1 + 148 Now we continue this process till remainder is zero . 272 = 148 ร— 1 + 124 148 = 124 ร— 1 + 24 124 = 24 ร— 5 + 4 24 = 4 ร— 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 ๐ธ๐‘ฃ๐‘’๐‘Ÿ๐‘ฆ ๐‘๐‘œ๐‘š๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘’ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘๐‘Ž๐‘› ๐‘๐‘’ ๐‘’๐‘ฅ๐‘๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘’๐‘‘ ๐‘Ž๐‘  ๐‘Ž ๐‘๐‘Ÿ๐‘œ๐‘‘๐‘ข๐‘๐‘ก ๐‘œ๐‘“ ๐‘๐‘Ÿ๐‘–๐‘š๐‘’๐‘ , ๐‘Ž๐‘›๐‘‘ ๐‘กโ„Ž๐‘–๐‘  ๐‘“๐‘Ž๐‘๐‘ก๐‘œ๐‘Ÿ๐‘–๐‘ ๐‘Ž๐‘ก๐‘–๐‘œ๐‘› ๐‘–๐‘  ๐‘ข๐‘›๐‘–๐‘ž๐‘ข๐‘’ , ๐‘Ž๐‘๐‘Ž๐‘Ÿ๐‘ก ๐‘“๐‘Ÿ๐‘œ๐‘š ๐‘กโ„Ž๐‘’ ๐‘œ๐‘Ÿ๐‘‘๐‘’๐‘Ÿ ๐‘–๐‘› ๐‘คโ„Ž๐‘–๐‘โ„Ž ๐‘กโ„Ž๐‘’๐‘ฆ ๐‘œ๐‘๐‘๐‘ข๐‘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 ๐ฟ๐‘’๐‘ก ๐‘ ๐‘๐‘’ ๐‘Ž ๐‘๐‘Ÿ๐‘–๐‘š๐‘’ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘–๐‘“ ๐‘ ๐‘‘๐‘–๐‘ฃ๐‘–๐‘‘๐‘’๐‘  ๐‘Ž2, ๐‘กโ„Ž๐‘’๐‘› ๐‘ ๐‘‘๐‘–๐‘ฃ๐‘–๐‘‘๐‘’๐‘  ๐‘Ž , ๐‘Ž ๐‘–๐‘  ๐‘Ž ๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ Theorem - 2 ๐‘–๐‘  ๐‘–๐‘Ÿ๐‘Ÿ๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™. proof Let us assume on contrary that 2 is rational where a and b are co-prime . โ†’ 2 = ๐‘Ž ๐‘ (๐‘ โ‰  0) squaring on both sides 2 2 = ๐‘Ž ๐‘ 2 ๐‘2 = ๐‘Ž2 2 Here 2 divides ๐‘Ž2 , so it also divides ๐‘Ž . so we can write a=2c for some integer c .
  • 7. Substituting for ๐‘Ž we get 2๐‘2 = 4c2 ๐‘2 = 2c2 ๐‘2 = ๐‘2 2 Here 2 divides ๐‘2 , so it also divides ๐‘ . 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 ๐ฟ๐‘’๐‘ก ๐‘ฅ ๐‘๐‘’ ๐‘Ž ๐‘Ÿ๐‘Ž๐‘ก๐‘–๐‘œ๐‘›๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘คโ„Ž๐‘œ๐‘ ๐‘’ ๐‘‘๐‘’๐‘๐‘–๐‘š๐‘Ž๐‘™ ๐‘’๐‘ฅ๐‘๐‘Ž๐‘›๐‘ ๐‘–๐‘œ๐‘› ๐‘ก๐‘’๐‘Ÿ๐‘š๐‘–๐‘›๐‘Ž๐‘ก๐‘’๐‘  . ๐‘‡โ„Ž๐‘’๐‘› ๐‘ฅ ๐‘๐‘Ž๐‘› ๐‘๐‘’ ๐‘’๐‘ฅ๐‘๐‘Ÿ๐‘’๐‘ ๐‘ ๐‘’๐‘‘ ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘“๐‘œ๐‘Ÿ๐‘š ๐‘ ๐‘ž , ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘ ๐‘Ž๐‘›๐‘‘ ๐‘ž ๐‘Ž๐‘Ÿ๐‘’ ๐‘๐‘œ๐‘๐‘Ÿ๐‘–๐‘š๐‘’, ๐‘Ž๐‘›๐‘‘ ๐‘กโ„Ž๐‘’ ๐‘๐‘Ÿ๐‘–๐‘š๐‘’ ๐‘“๐‘Ž๐‘๐‘ก๐‘œ๐‘Ÿ๐‘–๐‘ ๐‘Ž๐‘ก๐‘–๐‘œ๐‘› ๐‘œ๐‘“ ๐‘ž ๐‘–๐‘  ๐‘œ๐‘“ ๐‘กโ„Ž๐‘’ ๐‘“๐‘œ๐‘Ÿ๐‘š 2 ๐‘› 5 ๐‘› , where n and m ๐‘Ž๐‘Ÿ๐‘’ ๐‘›๐‘œ๐‘› โˆ’ ๐‘›๐‘’๐‘”๐‘Ž๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ๐‘  Example 1. 3 8 = 3 23 2. 13 125 = 13 53
  • 9. Thank You Made by :- Amit Choube Class :- 10th B