Euclid's division algorithm

PPT on Euclid's division algorithm for class 10 maths.
For more contact me.

1. Euclid’s Division Lemma Made by- Shubham Kumar Class-10 A Roll.-31

2. a = b × q + r 0 ≤ r < b Euclid’s Division Lemma The general equation can be represented as: Where: a = Dividend, b= Divisor, q = Quotient, r = Remainder

3. Using Euclid’s Division Lemma show that any positive integer is of the form 4q, 4q + 1, 4q+2 or 4q+3 where “q” is some integer The general form can be explained as: TERM DIVISOR QUOTIENT REMAINDER 4q 4 q 4q+1 4 q 1 4q+2 4 q 2 4q+3 4 q 3 QUESTION Euclid’s Division Lemma

4. SOLUTION Let ‘a’ be any positive integer On dividing “a” by 4 , let “q” be the quotient and “r” be the remainder Then, by Euclid’s Division Lemma, we have: ( a = b × q + r ) a = 4q + r , where 0 ≤ r < 4  a = 4 q + r , where r = 0, 1, 2 or 3 a = 4q ( when r = 0) a = 4q + 1 ( when r = 1) a = 4q + 2 ( when r = 2) a = 4q + 3 ( when r = 3)  A ny positive integer is of the form 4q, 4q + 1, 4q+2 or 4q+3 Euclid’s Division Lemma

5. Show that any positive even integer is of the form 6q or 6q + 2 or 6q + 4 and that any positive odd integer is of the form 6q+1 or 6q+3 or 6q+ 5 where “q” is some integer Thumb Rules Even Number × Any number = Even N umber Any number × Even Number + Odd number = Odd Number TERM DIVISOR QUOTIENT REMAINDER 6q 6 q 6q+1 6 q 1 6q+2 6 q 2 6q+3 6 q 3 6q+4 6 q 4 6q+5 6 q 5 QUESTION Euclid’s Division Lemma

6. Let “a” be any positive integer On dividing “a” by 6 , let “q” be the quotient and “r” be the remainder. Then , by Euclid’s Division L emma , we have a = 6q + r , where 0 ≤ r < 6 a = 6q + r , where “r” = 0, 1, 2 , 3 , 4, or 5 Positive even integers Positive odd integers a = 6q (where r = 0) a = 6q + 1 ( where r = 1) a = 6q + 2 ( where r = 2) a = 6q + 3 ( where r = 3) a = 6q + 4 ( where r = 4) a = 6q + 5 ( where r = 5) SOLUTION  Any positive even integer is of the form 6q or 6q + 2 or 6q + 4 and that any positive odd integer is of the form 6q+1 or 6q+3 or 6q+ 5 where “q” is some integer. Euclid’s Division Lemma

7. Case 1 Case 2 If “m” is Even If “m” is Odd when m = 2q when m= 2q + 1 m 2 = (2q) 2 m 2 = (2q + 1) 2 m 2 = 4q 2 m 2 = 4q 2 + 4q + 1 m 2 = 4q(q+0) m 2 = 4q(q+1) + 1 m 2 = 4p {  p= q(q+0) } m 2 = 4p +1 {  p = q(q+1 )}   QUESTION Show that the square of any positive integer is of the form 4p or 4p+1 for some integer “p” SOLUTION Let ‘ m’ be the positive integer  The square of any positive integer is of the form 4p or 4p + 1 for some integer “p ”. Euclid’s Division Lemma

8. On dividing a by 8 , let “q” be the quotient and “r” be the remainder Then , by Euclid’s Division Lemma, we have a = 8q + r , where 0 ≤ r < 8 (r = 0,1,2,3,4,5,6 or 7)  W e want only positive odd integers, we will consider only the odd values of ‘r’ which are a= 8q +1, a= 8q+3, a= 8q + 5 or a= 8q + 7 QUESTION Show that the square of any positive odd integer is of the form 8m + 1 for some integer “m” SOLUTION Let ‘a’ be any positive odd integer Euclid’s Division Lemma

9. Case 1 when a = 8q + 1 a 2 = (8q + 1) 2 a 2 = 64 q 2 +16 q + 1 a 2 = 8q(8q+2 ) + 1 a 2 = 8m + 1 {m= q(8q+2)} Case 2 when a = 8q + 3 a 2 = (8q + 3) 2 a 2 = 64 q 2 +48 q + 9 a 2 = 64 q 2 +48 q + 8 + 1 a 2 = 8(8q 2 +6q + 1 ) + 1 a 2 = 8m + 1 {m = ( 8q 2 + 6+1)} Case 3 when a = 8q + 5 a 2 = (8q + 5) 2 a 2 = 64 q 2 +80 q + 25 a 2 = 64 q 2 +80 q + 24 + 1 a 2 = 8(8 q 2 +10 q + 3) + 1 a 2 = 8m + 1 {m = (8q 2 +10q+1)} Case 4 when a = 8q + 7 a 2 = (8q + 7) 2 a 2 = 64 q 2 +112 q + 49 a 2 = 64 q 2 +112 q + 48 + 1 a 2 = 8(8 q 2 +14 q + 6) + 1 a 2 = 8m + 1 {m = ( 8q 2 + 14q+6)}  The square of any positive odd integer is of the form 8m + 1 for some integer “m” Euclid’s Division Lemma

10. Arranging the terms in the given question as per the Euclid’s Division Lemma 2. Identification of the Dividend, Divisor, Quotient and Remainder . 3. Any positive integer can be represented as ‘2q’ or ‘2q+1’. 4. Even number × Any Number = Even number . 5. Even number × Any Number + Odd number = Odd number . a = b × q + r REVISION General equation Euclid’s Division Lemma