 Hello friends welcome to this session in this session. We are going to prove Euclid's division lemma in the last few sessions We saw what is Euclid's division lemma. We saw some examples We also saw some practical cases where we could explain Euclid's division lemma. Now in this session Let us try and prove it. So first of all, let us go through what exactly is Euclid's division lemma so lemma is For given integers a and b greater than 0 both greater than 0 there exists unique integers q and r such that a equals bq plus r right, so for any two Integers a and b there exists another set of integers q and r and there must be unique such that a equals bq plus r and 0 is less than r less than b so Let's first take an example and understand what exactly this means So let's say I take 26 and 5, okay where 26 is my a and 5 is b So, you know, you know from our previous discussion a is called the dividend the dividend and b is the divisor divisor and q is the quotient quotient and R is remainder These are the four terms of four Things in this relationship. Now, let us take a as 26 and b is 5 So I can express this after dividing 26 by 5 I can write 26 equals 5 times 5 plus 1 Okay, so if you see this is my Dividend dividend dividend. This is my divisor divisor this 5 happens to be the quotient and This happens to be a remainder now clearly if you can see r is r. This is r here. So r is less than my b Okay, so This is what is given by Euclid's division lemma now Clearly let us let us take another example so that it becomes much clear. So another example could be let's say 57 is equal to 5 times 11 5 times 11 plus 2 So this happens to be a this happens to be b this happens to be q and This happens to be r Okay, now what are what we are going to do is let us first write all multiples of 5 So first of all, let us draw a line Okay, so let us draw a line. Let's say this is the line and And let's say let me just jot down all the all the multiples of 5 So let's say this is 0. This is 5. This is 10 This is 15 then 20 then 25 then 30 35 and Let's say the line was a little longer Line a little longer. So 35 40 45 50 and 55 and 60 So if you see where does 57 lie so clearly 57 lies somewhere in the middle of 55 and 60 Okay, and for that matter any number you take if let's say let's let us fix b first b is fixed and b equals 5 Okay, now let us take any value of a let us take 37 Let us take a equals 37. Where will it lie? It will lie here between 35 and 40, isn't it? Let us take a as 25, where does it lie? It lies exactly on 25 here. Let us take a as 29 so if you see a 29 will lie somewhere here somewhere here. So why am I doing this if you see any number any number when divided by 5 either Lies on the multiple of 5 or it lies between any two consecutive multiples of 5 so if you see in case of 37 it was lying between 35 and 40 two consecutive multiples of 5 and In case of 29 it was lying between 25 and 30 now if you see can I not write this as and here here are negative numbers as well So there is no restriction on The sign of a and b so it can be any any two integers Now if you see what is what is what are these multiples of five? How can we express them so it is nothing but five times minus two This is five times minus one then five times zero Then let me change a little you know take another color so that it becomes differentiated. This is five into one This is five into two. This is five into three Five into four five into five and so on and so forth. Why am I doing this? So if you see 29 lies between so if you see 29 lies between five into five and Five into six so any any number any dividend for that matter any value of a can be expressed as some multiple of five and And the the dividend will lie between some multiple of five and its immediate Neighbor, isn't it? Now what was 37 like 37 was if you see this was five into five into seven and it is five into eight Isn't it and 25 was Was five into five since it is equal so hence this inequality is also valid and it was less than five into six So if you see generalizing can I not say any a any a lies between five into q and Five into q plus one If you see these were the q's in case of 29 five was the q in case of 37 Seven was the q quotient in case of 25 five was q isn't it so this is what I can say now What was five if you see what was five? five was my B this was be the The divisor is it it so hence I can easily write B times q is less than a is less than B times q Plus one Isn't it? Isn't it now if let me just write this inequality of fresh Here yeah So let's let's focus here. So what is it if you see from this analysis? We can say that B q divisor times quotient is less than equal to The dividend itself is less than is there is less than B times q plus one now in this inequality If I subtract a positive quantity across the inequality the inequality doesn't change. What does it mean? So I can very well subtract Bq from the entire inequality. So let's say Bq minus Bq Will be less than equal to a minus Bq and will be less than Bq I'll open Bq plus. So is it is Bq plus B and minus Bq is it? Okay, so if you simplify this what will happen? So I hope this this particular Statement is you know understood to all of you all of you now Hence if you see in this part it will it will become zero. So zero is less than a minus Bq a minus Bq and Is less than B? And it's less than B. Is it not is it not correct? So let us say let us say let us say a minus Bq is equal to r Is equal to r then Then what we can what so now if you see a a can be written as a can be written as Bq Bq plus a minus Bq Bq plus a minus Bq why because a Bq minus Bq gets cancelled so a remains so hence a can be expressed as this So hence a can be expressed as Bq plus and what was a minus Bq r? So r Isn't it? So hence we we've proved the first part. We've proved the first part that Bq as there is for every a and b integers There exists q and r such that This relation holds so we for every a and b you will get one q and one r And hence the relationship hold now the second part is to prove that they are unique What is that so prove that they are unique what is unique that means for? 29 when you express 29 as 5 into 5 Plus 4 there cannot be any other value of this value this 5 or this thing such which will give you 29 again So it is valid or let's say 29 can be expressed uniquely only by this If you have fixed the The divisor right if the divisor is 5 there's no way but this to express 29 you cannot have 5 into 7 and minus something to get you or anyways we are talking about positive integers So yes, you can have different values of q and r if you let negative integers also come in Why because you can see 5 into 7 is 35 minus 6 can also be Expressed at 29, but we are talking only about We are only talking about positive Values of q and r so hence if q and r are positive there cannot be any other way, but this to express 29 Okay, so how do we prove that so we got here that there exists q and r But how do we know that they are unique? Okay, so the best way in mathematics to prove such things is Proving by contradiction So let us say that It's not unique. Let us say that there is there is you know There exists another pair of q and r which will give you same value a and b. So let us let us try and understand that so Yes here. So if you see let us say that there exists. So let us assume that now we are doing what part b of the prove that q and r are unique Is it it? How do we prove that they are unique? So best is let's you know start with the contradicting hypothesis Let us say there exists two pair of q1 two pair of q q1 comma r1 and Let's say q2 comma r2 there exists two pairs such that a is equal to b times q1 plus r1 and B sorry and a can also be expressed as a can also be expressed as a equals b times q2 plus r2 that is what is meant by let's say if you are trying to prove that there is only one q and r and You have to prove it. Then what do we do? Let us assume that there are two and then contradict our hypothesis and then we'll conclude that This contradiction is happening because we are considering two different pairs of q and r So hence, you know, it cannot exist. So So what do we do now so we can equate both of them? So if you see if you let's say this is equation number one and This is equation number two Okay, so hence from one and two I can write bq1 plus r1 is equal to bq2 Plus r2 and reshuffling you can find out b times q1 minus q2 why take this Take this this thing in LHS and this thing in RHS. So you will get what? b times q1 minus q2 is equal to r2 minus r1 Am I not am I not correct? So hence, what will you what what is what is observed from here? So isn't it a integer? Q1 minus q2 is an integer. Why because if q1 is an integer q2 is an integer So q1 minus q2 has to be an integer from here. We see that b divides R2 minus r1 Yeah, so b times an integer is if you remember our first session on divisibility So what was divisibility? So divisibility was so a is equal to b times c So if b if this kind of a integral equation is there that means If a, b, c all are integers and we find this kind of a relationship, then we say that b divides a So if you see here b times q1 minus q2 so you can consume q1 minus q2 to bc So b times c is equal to r2 minus r1 That is a here So you can clearly say that b divides r2 minus r1 so hence b divides r2 minus r1, but If you see in our previous case we what did we find we found that r2 was less than b if you see a Previous case here. We just proved some time back that if you see This was zero my zero a minus bq is less than b and a minus bq was r So r has to be r was proved to be less than b So here coming back to our proof here if you see If you see r R both r1 and r2 will be less than b. So r2 is less than b R1 is also less than b then how can so clearly r2 minus r1 will be less than b So there is an integer which is less than b and you are saying that b divides that integer which is less than b how is that possible a for example, you're saying that 25 is Divisible by 50 Is not divisible by 50s, right? So let's say this is b which is greater than this value So in no case a greater number can divide a smaller number since r2 minus r1 is less than b so we we conclude We conclude what do we conclude on so we conclude on that b doesn't divide r2 minus r1 it cannot Yeah, either this is not happening or or or r2 r2 minus r1 If you from here this this particular thing is coming from where when we considered R2 is different from our or this can be possible only when R2 minus r1 equals 0. So when this becomes 0 then b divides this R2 minus r1 so hence the only possibilities r2 minus r1 is 0 from here I will get r1 is equal to r2 and the moment r1 equals to r2 you put here r1 equals to r2 So what you can do is you can deploy r1 equals to r2 back here in this equation So this will lead to what? b times q1 minus q2 Equals 0 so since b is not equal to 0 or device that cannot be 0 So only possibility is this value has to be 0. So hence q1 minus q2 is 0 So q1 equals q2. So hence it what do we see we see r1 equals to r2 and q1 equals to q2 So they are not distinct but same number so hence the second part of the proof is also done Right. So hence if there if there are two integers a and b there will exist There will exist what? q and r such that a equals a equals a equals bq plus r where r is less than 0 equal less than equal to 0 and less than b and q and r are Unique unique for a pair of a pair of a and b This is what is Euclid's division lemma and we could prove it here Thanks for watching this video