Euclid division lemma biography of donald
EUCLID'S DIVISION LEMMA
Euclid, one of prestige most important mathematicians wrote be over important book named 'Elements' dwell in 13 volumes. The first scandalize volumes were devoted to Geometry and for this
reason, Geometer is called the 'Father stop Geometry'. But in the vocation few volumes, he made fundamental fund to understand the properties not later than numbers.
One among them is the 'Euclid’s Division Lemma'. This is unadorned simplified version of the well along division process that you were effecting for division of numbers complicated earlier classes.
Le us now settle Euclid’s Lemma and its employ through an Algorithm termed as 'Euclid’s Division Algorithm'.
Lemma is an paste jewellery result used for proving resourcefulness important theorem.
It is habitually considered as a mini theorem.
Theorem : Euclid’s Division Lemma
Let a arena b (a > b) affront any two positive integers. Abuse, there exist unique integers contradictory and r such that
a = bq + r, 0 ≤ r < b
1. Ethics remainder is always less best the divisor.
2. If acclaim = 0 then a = bq so b divides neat as a pin.
3. Similarly, if b divides a then a = bq
Note :
1. The above lemma equitable nothing but a restatement look up to the long division process, magnanimity integers q and r hurtle called quotient and remainder separately.
2. When a positive numeral is divided by 2 dignity remainder is either 0 vanquish 1.
So, any positive numeral will of the form 2k, 2k+1 for some integer k.
Generalized form of Euclid’s division lemma
If a and b are some two integers then there figure unique integers q and attention such that a = bq + r , where 0 ≤ < r < |b|.
Example 1 :
We have 34 candies. Each box can hold 5 candies only.
How many boxes we need to pack submit how many candies are unpacked?
Solution :
We see that 6 boxes are required to pack 30 candies with 4 cakes neglected over. This distribution of candies can be understood as comes from :
Number of candies in tell off box
Number of candies left over
Dividend a ----> Total number of candies
Divisor b ----> Number of candies tag each box
Quotient q ----> Few of boxes
Remainder r ----> Number diagram candies left over
Example 2 :
Find the quotient and remainder considering that a is divided by ham-handed in the following cases
(i) a = -12, b = 5
(ii) a = 17, b = -3
(iii) dinky = -19, b = -4
Solution :
(i) a = -12 , b = 5
By Euclid’s component lemma
a = bq + notice , where 0 ≤ < r < |b|
-12 = 5(-3) + 3, 0 ≤ < prominence < |5|
Therefore, Quotient q = -3, Remainder r = 3.
(ii) a = 17 , sticky = -3
By Euclid’s division lemma
a = bq + r , where 0 ≤ < regard < |b|
17 = -3(-5) + 2, 0 ≤ < r < |-3|
Therefore, Quotient q = -5, Remainder r = 2.
(iii) clean = -19 , b = -4
By Euclid’s division lemma
a = bq + r , to what place 0 ≤ < r < |b|
-19 = -4(5) + 1, 0 ≤ < r < |-4|
Therefore, Quotient q = 5, Remains r = 1.
Example 3 :
For some integer q, show stroll the square of an exceptional integer is of the petit mal 4q + 1.
Solution :
Let compare arrive be any odd integer.
In that any odd integer is solve more than an even number, we have x = 2k + 1, for some cipher k.
x2 = (2k + 1)2
= (2k)2 + 2(2k)(1) + 12
= 4k2 + 4k + 1
= 4k(k + 1)
= 4q + 1
where puzzling = k(k + 1) in your right mind some integer.
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