 In this video, let us try to prove the important property that we used in the last video and also derive some results from that property that are very useful in themselves. Okay, so we said that hcf of any two numbers is equal to the hcf of if we keep the first number as the same or any number as the same a or b and add or subtract to the other number any multiple of the first number. So b plus ak where k is an integer. So it can be positive or negative. So that is why we are adding and subtracting and a and b here are positive integers. So a and b are greater than zero. All right. So let me say that hcf of a and b is small h as I like to. What does this mean? This means that h divides a. So a by h will be some integer p and h also divides b. So b by h will also be some other integer q. That is what a factor is, right? A factor can divide a and b. So h is the factor of a and b. That is why it can divide it and give us some integers. Now the property of this h in hcf is that it is the highest common factor. This means that p and q will not have any other factors in common. Because if they did, then that common factor could come out and join h and then we would have an even higher factor. We have talked about this before. What I'm saying is because hcf is the highest common factor, p and q are co-primes. They do not have any other factor in common. The hcf of p and q in other words is one. Okay. So this means that a is equal to ph and b is equal to qh. Now let me just get b plus ak because that is what I want to use. So b plus ak is qh plus ph into k, right? Now here I can take h as common and I will get q plus pk. Now let us see what we're trying to prove here. We're trying to prove that hcf of a and b, this hcf, should be the same as hcf of a and b plus ak. So this is a and this is b plus ak. What is the hcf of these two numbers? We're trying to prove that they have the same hcf. So this should also be h. To prove this, what I need to prove is just like p and q were co-primes, p and q plus pk are also co-primes, right? Because if they are co-primes, then h must be the highest common factor between a and b plus ak. So can we prove that p and q plus pk are co-primes? Let's see, from q plus pk, q plus pk, we need to prove that there is no factor in common between q plus pk and p other than one. If I take p common, I get q by p plus k, right? If I'm taking p common from this. Now since q and p are co-primes, therefore this is not an integer. I'm not writing everything I'm saying, but when you formally prove it, you should write everything and make it very standard. But here I just want to tell you how we can prove this. So because q and p are co-primes, there is no cancellation that is going to happen. For example, seven and four, right? We cannot simplify this further. If they were not co-primes, for example, if they were eight by four, then we could simplify it and it could turn into an integer. But because p and q are co-primes, this is not turning into an integer. This is not an integer. It will not simplify into an integer. And this is where the magic lies. Because this is not an integer and this is, these two together will also not be an integer. These two will also not be an integer. So this means if I'm taking p common from this number, I'm not getting an integer. So that means that there is no common factor between p and q plus pk, right? There is no common factor. If there was a common factor, this could have been an integer, but this is definitely not an integer. So therefore, we can say that p and q plus pk are co-primes. They do not have any common factor. And what does this mean? This means that HP and H into q plus pk will have H as their highest common factor, right? Because there is no other factor that is common between p and q plus pk. So H is their highest common factor. And we are done. We have shown that they have the same highest common factor, right, HCF. So this is how you can prove this. Now let us see what this result, what other results we can get from this result. So if we take k as minus 1, we get HCF of A and B is HCF of A and B minus A, right? So this means that if we have the HCF of any two numbers, let's say 9 and 16, this will be equal to the HCF of 9 and 16 minus 9, that is 7. So this is going to be 1. Okay, another important thing that we can prove from here is that HCF of any two consecutive integers is 1. Let's see how we can prove this. Integers is 1. So let me take two integers n and consecutive. So the other one is n plus 1. Now using this property, I can say that this will be equal to HCF of, I'm keeping n as the same and I'm subtracting this n from the other number, n plus 1 minus n using my property, using this property or this property. So n can cancel from here and this gives me HCF of n and 1 and that is 1. So this means that HCF of any two consecutive integers is 1. We have proved another important property that we can show is that HCF of any two odd numbers is 1. Now I'm going to leave this up to you to prove this. Just use this property again and you will be able to prove this. So any two odd numbers, that means if we take 2n plus 1 as our first odd number, we know that if n is any integer, then 2n plus 1 is an odd integer and 2n plus 3 is the, oh, I said odd, I mean consecutive odd. Any two consecutive odd numbers is 1. So any two consecutive odd numbers will have their HCF as 1. We can also show that HCF of any two consecutive even numbers is 2. This as well I'm going to leave up to you. So if 2n is my first integer, the next even integer is going to be 2n plus 2 and this will have an HCF of 2. All right, enough about this property. We can prove a lot. This is as you can see a very powerful property. Now let's practice some problems using this property.