 I'm Zor. Welcome to Unizor Education. I will present problem number three for mathematical induction. Here's the problem. Let's assume you have a certain number of lines on the plane and we will assume that there are no parallel lines and there are no lines which cross no more than two lines are crossing in any particular point. So I am excluding this case when there are parallel lines and I'm excluding case when three lines are crossing in exactly the same point. So this is not happening. We have only non-parallel lines which are with crossings of no more than two lines in any point. The question is how many cross points we have in case of n lines? Well in case of one line obviously we have no crossing. In case of two lines obviously we have only one crossing right? Two lines are crossing in only one point. No parallels. In case of three lines so this is n and this is cross. In case of three lines as you see here we have three crossings. Well let's just go on in case of four. Well four actually does something like this right? It's not parallel to anything which means it's crossing anything. I added one to three points so I have six. So these are my sequences. It looks like the formula should be very easy and I'll just give it to you and you will try to prove it. The formula is n times n minus one divided by two. Let's check it. For one it's zero because this is zero. For two this is two this is one divided by two it's one. For three three times two divided by two it's three and for four four times three twelve divided by two six. All right so that seems to be like with right formula. Okay try to prove it using the method of mathematical induction. Press the pause button and do it yourself and I will try to prove it here. So the formula is n lines no parallels and cross two lines only. No three lines are crossing in the same point. And the formula which I would like to prove is that the number of crossing points is this. All right let's go through the steps. Step number one check. Well we have already checked for one two and three formula actually holds so the check is fine. Let's say for n is equal to one for one line. The formula is zero no crossings because there's only one line within that. Okay assume what do we assume that n lines of that type cross in let's look this way k lines I don't want to use n anymore. Let's assume that n is equal to k. So we have k lines are crossing in k times k minus one divided by two points. And three we have to check that the formula is correct for n is equal to k plus one. So let's just draw another line. We have k lines. We draw another line. As I said no parallel lines which means this new line will cross each and every one of those old lines. Now all these crossings will be new because no old cross points should be crossed again because I said no more than two lines are crossing to the same point. So whenever we draw the next line the k plus first line it crosses all the old lines which means it introduces new k points of crossing. So when I have n equals to k plus one I actually add k points crossing points. Well let's see if our formula will will be held in this particular case. So we used to have we assume that k lines by themselves are crossing in k times k plus minus one divided by two lines. I introduced k new points. And how the formula will look? Well that's easy. It's k times k minus one plus two k divided by two equals k. Let's factor out k and what will be? Will be k minus one plus two which is k plus one divided by two. Now let me just rewrite this formula slightly differently. Since I need space I will wipe out this. Instead of k I will use k plus one minus one. That's the same thing, right? So I assume that my formula for n equals to k looks like this and for n equals to k plus one we derive that the formula looks like this. But as you see this is exactly the same formula, right? k plus one this is instead of n we put k plus one. So the formula looks exactly the same way. It's number n whatever the n is in this case in this case is k plus one in this case is k. Same number n minus one. So the formula is exactly the same. So for n is equal to k it's this one and for n is equal to k plus one formula has exactly the same form and shape. So the formula holds and that actually proves the whole point. So if you have n lines on the plane they cross under these conditions obviously no parallels and no more than two lines are crossing at the same point. So they cross each other in n times n minus one divided by two points. And now here is an interesting thing. If you remember looked at the lecture which we started mathematical induction with we started with a summation example which young Carl Gauss when he was seven years old he added up numbers from one to one hundred. And obviously what happens with lines on the plane is exactly the same as what happens with numbers if you are adding numbers from one to n. Why? Because each new crossing is introducing the next number in this summation. So if you have one line you have zero crossing. If you have two lines it introduces crossing with old lines. If you have three lines it introduces new two crossings with old two lines. If you have four lines it introduces three new crossings with three new lines. So as you see this is exactly the same thing. So if you have n that's n minus one actually. So the number of crossings if you have n lines is a sum from this to this. And that will be exactly the same formula. Okay that basically concludes this problem number three. Thank you very much.