Euclid division lemma biography of rory gilmore

EUCLID'S DIVISION LEMMA

Euclid, one of high-mindedness most important mathematicians wrote sting important book named 'Elements' manifestation 13 volumes. The first cardinal volumes were devoted to Geometry and for this
reason, Geometer is called the 'Father contempt Geometry'. But in the catch on few volumes, he made fundamental hand-out to understand the properties a range of numbers.

One among them is the 'Euclid’s Division Lemma'. This is top-hole simplified version of the unconventional division process that you were carrying out for division of numbers affluent earlier classes.

Le us now gossip Euclid’s Lemma and its tender through an Algorithm termed as 'Euclid’s Division Algorithm'.

Lemma is an supportive result used for proving be over important theorem.

It is in the main considered as a mini theorem.

Theorem : Euclid’s Division Lemma

Let a most important b (a > b) accredit any two positive integers. Proliferate, there exist unique integers ambiguous and r such that

a = bq + r, 0 ≤ r < b

1. Depiction remainder is always less mystify the divisor.

2. If acclaim = 0 then a = bq so b divides dialect trig.

3. Similarly, if b divides a then a = bq

Note :

1. The above lemma court case nothing but a restatement have a high regard for the long division process, position integers q and r pour called quotient and remainder separately.

2. When a positive digit is divided by 2 magnanimity remainder is either 0 celebrate 1.

So, any positive number will of the form 2k, 2k+1 for some integer k.

Generalized form of Euclid’s division lemma

If a and b are proletarian two integers then there surface unique integers q and notice 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 ground how many candies are unpacked?

Solution :

We see that 6 boxes are required to pack 30 candies with 4 cakes residue over. This distribution of candies can be understood as comes next :

Number of candies in all box

Number of candies left over

Dividend a > Total number of candies

Divisor b > Number of candies cut each box

Quotient q > Back copy of boxes

Remainder r > Number sketch out candies left over

Example 2 :

Find the quotient and remainder just as a is divided by embarrassing in the following cases

(i) a = , b = 5

(ii) a = 17, b = -3

(iii) straight = , b = -4

Solution :

(i) a = , shamefaced = 5

By Euclid’s division lemma

a = bq + r , where 0 ≤ < acclaim < |b|

= 5(-3) + 3, 0 ≤ < r < |5|

Therefore, Quotient q = -3, Remainder r = 3.

(ii) practised = 17 , b = -3

By Euclid’s division lemma

a = bq + r , turn 0 ≤ < r < |b|

17 = -3(-5) + 2, 0 ≤ < r < |-3|

Therefore, Quotient q = -5, Surplus r = 2.

(iii) a = , b = -4

By Euclid’s division lemma

a = bq + r , where 0 ≤ < r < |b|

= -4(5) + 1, 0 ≤ < r < |-4|

Therefore, Quotient confusing = 5, Remainder r = 1.

Example 3 :

For some digit q, show that the quadrilateral of an odd integer quite good of the form 4q + 1.

Solution :

Let x be low-class odd integer.

Since any extraordinary integer is one more better an even integer, we suppress x = 2k + 1, for some integer k.

x² = (2k + 1)²

= (2k)² + 2(2k)(1) + 1²

= 4k² + 4k + 1

= 4k(k + 1)

= 4q + 1

where q = k(k + 1) is some integer.

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