Euclid's division lemma  

In this case main concern that is taken into consideration is that there are 2 variables that are to be considered, namely a and b.

As per the theorem, the main aspect that is considered is that b has to be a positive non zero positive integers. The equation that are performed in a manner as per a=bq+r. This formula is basically performed for proper establishing the equation. The main aspect that is performed for performing the critical evaluation process. 0≤r<b.

The proof that is performed as per analysis+ of the data analysis for performing the proving part of the Euclid’s division lemma requires processing of the q=⌊a/b⌋. The equation that is considered for defining r=a-bq. Clearly it can be stated that a=bq+r.  From the above two equations it can be stated that proper assessment of this lemma related performance, it is an important aspect that is to be considered  for process8ng of the entire equation is that 0≤r<b0≤r

a/b−1<⌊a/b⌋≤a/b, this is the equation that is considered for performing the inequality of the members that are required fir performing data set management and hence wise analysis of numeric data will be performed. It is also stated that processing r=a-bq. The section also helps in better understanding of the numeric value of r. It is stated that performing analysis of the data that are collected and proper understanding of equation is performed. It is also stated that the entire equation is multiplying the entire equation with –b will be performed. This leads to the fact that the equation will be getting performed in a manner that the outcome of the derivation will be b−a>−b⌊a/b⌋≥−a. After per the equation formation, the main aspect that is considered in this case is that a/b is interchanged with q. This is the main reason that performing management of the equation will be having a proper aspect of differentiation in between q and r. the equations that are considered for the experiment are a=bq1+r1 and 0≤r1r1r2>r1. It is also stated that the resultant that will be received will be

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

As per the definition of divisibility, it can be stated that if the condition is b|r2−r1, then the resultant will be b≤r2−r1. As per assumed it can be stated that the b>r2>r1≥0. This will lead to the fact that regular processing will be including the fact that r1=r2. As per the completion of the equation, it can be stated that b(q1−q2)=0b(q1−q2)=0. This is estimated that b will not be a non zero value. After commencing of the project it can be stated that q1 will be equal to q2.

To start off with the fact that existence of q and r is to be proved. This will ensure that the values of q and r will be unique in nature. This uniqueness in the project will be having better assessment of the recalling of the term q. This is one of the main aspect that b will be a positive number. Hence keeping the result to be a non negative assessing will be including the fact that subtraction process will be performed. Hence performing the attempt of processing of q and r’s existence will be making the entire process will become a non negative section.

In case the value of b is considered to be 27, the main aspect that is concerned is that subtraction process will be done in regards to the multiple of 5. The result that will be achieved will be 8,−3,2,7,12,17,22,27,32,37,… and so on. This reduction process will be performed in a manner that smallest elements will be performed. The hint of the tool that is considered in this process is performed as per recalling the performing non empty sets. The main aspect includes s=b-ka. In this case the section the main aspect that will be considered is that b-ka has to be a non negative number.

It can be considered that b≥0. This is the main reason that functioning of the k=b can be considered. It can be stated that b−ka=b−ba=b(1−a) will be performed. The factor that a being greater than 0 can be performed. hence this equation is considered to be proven: b−ka≥0. Again it is considered that b=qa+r with 0≤r<a, then those values of q and r are unique.