Question

geometrical e verify the Euclid Lemma / algorithm

Answer 1

Dear Student,
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Theorem
If a and b are integers and b>0, then there exist unique q and r(mathematically: ∃!q,r) such that:

a=bq+r and 0≤r<b

Proof
Define q=⌊ab⌋ and r=a−bq. Clearly a=bq+r. We must show that 0≤r<b: we know (do we? maybe it requires a proof...) that 

ab−1<⌊ab⌋≤ab

 If we multiply all members of the inequality by −b, which is a negative number, we get:

b−a>−b⌊ab⌋≥−a

If we now add a everywhere, and plug in ⌊ab⌋=q we get:

b>a−bq≥0

But we previously defined r=a−bqwhich implies that 0≤r<b

We therefore showed the "∃" part of the thesis, while we still miss the uniqueness (!): therefore, let us assume that there exists different values of q and r, so that

a=bq1+r1 and 0≤r1<b
and
a=bq2+r2 and 0≤r2<b

We must prove that r1=r2 and q1=q2: we can suppose that r1≠r2, let's say that r2>r1. Then

(bq1+r1)−(bq2+r2)=a−a=0
⇔b(q1−q2)+(r1−r2)=0
⇔b(q1−q2)=(r2−r1)

By definition of divisibility, b|r2−r1 and therefore b≤r2−r1, but since we assumed that b>r2>r1≥0 we have a contradiction, and this tells us that r1=r2. 

Finally, we also have that b(q1−q2)=0, and since we know that b≠0, we also conclude that q1=q2, and this completes the proof

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