 the principle of mathematical induction now the principle of mathematical induction provides us a very powerful technique to prove several mathematical results the basic idea is that suppose we have a statement or a more precisely a predicate which depends on in a positive integer value n and we expect it to be true or false for all n greater than or equal to certain fixed positive integer then what we can do is to prove this statement for n equal to that fixed integer and assume that the statement is true for some positive integer k which is greater than or equal to that fixed integer and assuming that it is true for k prove that the statement is true for k plus 1 now let me write formally let pn be a statement which for all positive integer n may be either true false to prove that pn is true for all integers n greater than or equal to 1 it is sufficient to prove 1 p1 is true for all k greater than or equal to 1 pk implies pk plus 1 now thus we see that if we start from 1 and suppose p1 is true and suppose for all k greater than or equal to 1 pk implies pk plus 1 this means that if pk is true then pk plus 1 is true now if we can prove these two facts then since p1 is true and pk implies pk plus 1 therefore p2 is also true since p2 is true p3 is also true so we will start a chain like this that p1 is true implying p2 is true which in turn implies p3 is true and so on and this is all for the two facts that we have already proved that p1 is true and for k greater than or equal to 1 pk implies pk plus 1 now what we observe over here is that this number 1 is very specific and we can relax the situation a little more so it may so happen that some statements may not be true for 1 2 3 up to a fixed positive integer but after that the statement may be true for all the other integers greater than that specific integer in order to bring this slight generalization into our framework we state this whole principle in a slightly different way we state that to prove pn is true for all integers n greater than or equal to n0 where n0 is a fixed positive integer determined previously it suffices to prove that 1 pn0 is true to for all k greater than or equal to n0 pk implies pk plus 1 so this is our slight generalization of the principle of mathematical induction that we stated in the beginning of the lecture now we move on to the required steps in a proof which uses the principle of mathematical induction a proof by using the principle of mathematical induction has the following steps one the first step is called the basis of induction basis of induction in the basis of induction we have to show that pn0 is true if we are unable to show this then we cannot start induction because it will be meaningless induction hypothesis the induction hypothesis is a hypothesis that pk the statement that we get by putting the value of n equal to k is true so we assume that pk is true three inductive step show that pk plus 1 is true on the basis of the induction hypothesis if we can successfully complete these three steps then we will have a proof by using mathematical induction now let us look at one example now we are considering the sum of first n positive integers let us write Sn equal to 1 plus 2 plus 3 plus and so on up to n now suppose we have a conjecture that Sn is equal to n to n plus 1 divided by 2 and suppose we want to prove this first of all let us see that for small values of n this formula works so for example if I put n equal to 1 then s1 equal to 1 which is equal to 1 into 1 plus 1 divided by 2 n equal to 2 s2 equal to 1 plus 2 which is equal to 3 and if you look at the formula it should be 2 2 plus 1 divided by 2 this is also equal to 3 therefore we see that at least for n equal to 1 and 2 Sn equal to n plus n by 2 sorry n plus n plus 1 by 2 works so we can form the basis of induction 1 basis of induction for n equal to 1 1 equal to s1 equal to 1 into 1 plus 1 divided by 2 which is equal to 1 so s1 equal to 1 1 plus 1 divided by 2 is true or in other words Sn equal to n into n plus 1 divided by 2 is true for n equal to 1 now we move on to the induction hypothesis and induction hypothesis we have to choose we say that suppose for k greater than or equal to 1 sk equal to sk into k plus 1 divided by 2 is true this is my induction hypothesis now we come to the third point which is the inductive step what do we do here we now start from sk plus 1 and write sk plus 1 as it is defined this is 1 plus 2 plus 3 plus and so on plus k plus k plus 1 now here we observe that we can always sum this first k entries by using the formula that we have already assumed therefore I can write this is equal to 1 plus 2 plus 3 plus and so on up to k plus k plus 1 this is equal to k into k plus 1 divided by 2 plus k plus 1 and naturally we will sum this expression to get to k into k plus 1 plus 2 times k plus 1 which is equal to 2 in the denominator and on the numerator k plus 1 into k plus 2 the numerator can be written as 2 k plus 1 and then k plus 1 plus 1 thus we see that the formula that we wrote over here holds for k plus 1 if we assume that it hold if that we assume that it holds for k so this means very strictly speaking sk equal to k into k plus 1 by 2 implies sk plus 1 equal to k plus 1 into k plus 1 plus 1 divided by 2 thus this formula sn equal to n into n plus 1 divided by 2 true is going to be true for all positive integers greater than or equal to 1 this is because we know that it is true for 1 and we know that if it is true for k it is going to be true for k plus 1 therefore since is true for 1 is going to be true for 2 since is true for true then it will be going to be true for 3 and so on and it will cover the set of positive integers now let us look at another example where a problem is solved by using mathematical induction the problem is as follows find and prove a formula for the sum of the first n cubes that is 1 cube plus 2 cube plus 3 cube plus and so on up to n cube well this is of course a difficult problem in the sense that nobody has given me a formula if a formula is given I can quickly check whether I what I can do by using mathematical induction for the formula we need some imagination and some something which cannot be really quantified but let us check by experiment what happens so 1 cube equal to 1 equal to 1 square 1 cube plus 2 cube equal to 9 which is equal to 3 square 1 cube plus 2 cube plus 3 cube equal to 36 which is equal to 6 square 1 cube plus 2 cube plus 3 cube plus 4 cube is equal to 100 which is equal to 10 square now if we go on in this way we will find that whenever we are taking sum of cubes up to n it is becoming a perfect square we can check few more few more terms then somehow we can argue of course without any possible concrete proof that probably whatever the sum maybe it is a perfect square but perfect square of what that we do not know again if we just sum up to n terms we will see that 1 equal to 1 1 plus 2 equal to 3 1 plus 2 plus 3 equal to 6 1 plus 2 plus 3 plus 4 equal to 10 surprisingly we see that these sum of cubes is it looks like it is as if square of the sum of the usual sub now of course this is not a proof but this may lead us to a conjecture like this well this is just a conjecture now what we can see that it is a very neat conjecture and it is worth checking by using mathematical induction whether this is indeed true for that we will start again from basis of induction here we see that 1 cube is indeed 1 into 1 plus 1 divided by 2 whole square equal to 1 thus we have proved the basis of induction and now we come to induction hypothesis now induction hypothesis we assume that assume that sk equal to k into k plus 1 divided by 2 whole square yeah that is it for a k greater than or equal to 1 3 now we have inductive step again like before we consider k plus 1 so if I have sk plus 1 and write explicitly the sum I will get 1 cube plus 2 cube plus 3 cube plus and so on up to k cube plus k plus 1 cube again I observe that the first k terms are are essentially sk and therefore since I have assumed sk equal to k plus 1 k into k plus 1 divided by 2 whole square I can write this as k square k plus 1 whole square by 2 square plus k plus 1 whole cube and of course I can simplify this expression as putting denominator 2 square in the denominator and on the numerator we have k square k plus 1 whole square plus 2 square k plus 1 whole cube this gives us 2 square and here we will have k plus 1 whole square k square plus 4 k plus 4 now this gives me 2 square k plus 1 whole square and this is k plus 2 whole square thus by doing again a small manipulation we get k plus 1 and k plus 1 within bracket plus 1 whole square so the expression k plus 1 whole square into k plus 2 whole square divided by 2 square is equal to k plus 1 into k plus 1 within bracket plus 1 divided by 2 and the whole expression is squared and we see that this conforms exactly with the formula that I predicted that is sn equal to n plus 1 n into n plus 1 divided by 2 whole square therefore we can conclude that the conjecture is true now we move on to one more application of mathematical induction and of course there are several applications of mathematical induction and here we will use this mathematical induction to prove something related to logic specifically de morgan's laws that we have studied in previous lectures let us go to the example prove that n is greater than or equal to 2 then the generalized de morgan's law that is not of p1 and p2 and so on and pn by conditional not of p1 or not of p2 or not of pn is true now we go on to the solution now here we see that it is somewhat meaningless to start from n equal to 1 it is from n equal to 2 because the de morgan's law the one that we have already studied involves two propositions so for n equal to 2 we have not of p1 and p2 by conditional not of p1 or not of p2 is true in fact we can use our previous knowledge to write that these two statements not of p1 intersection p2 and not of p1 sorry not intersection here and not of p1 or not of p2 they are equivalent because if the by conditional is always true that is a tautology then these two propositions are equivalent therefore we write that p1 and not of that is equivalent to not of p1 or not of p2 so this essentially forms the induction hypothesis I am sorry this essentially forms the basis of induction so I write the statement as this that I have got a statement pn which is not of p1 p2 and pn equivalent to not of p1 or not of p2 or not of pn first step basis of induction p2 is true which is essentially due to de morgan's law now the second step is induction hypothesis the induction hypothesis states that pk is true this means that not of p1 and so on up to pk is equivalent to not of p1 and so on up to not of pk now we go to inductive step so we start with pk plus one which is not of p1 and up to pk and then and pk plus one and now we see that this is equivalent to not of p1 and pk we can put a bracket enclosing p1 up to pk that is because after all we know that and is associative and then we put and pk plus one and once we have this we see that we have one proposition p1 and up to and pk and then another one is pk plus one and not of that and we can use de morgan's law for the original de morgan's law the reason is that we have proved p2 is true therefore we will write not of p1 and so on up to pk or not of pk plus one and now we use the induction hypothesis to write that not of p1 and up to pk is not of p1 or not of p2 and not of pk we can of course put enclose this whole expression by bracket and then it is not of pk plus one and now we know that in this case we can remove brackets so we can write not of p1 or not of p2 or and so on up to not of pk or not of pk plus one thus we see that I have to write equivalences over here so thus we see that pk plus one is true so pk implies pk plus one for k greater than or equal to 2 thus we see that pn is true for all n greater than or equal to 2 this completes our proof here we see that we have used a slightly different technique in that we are not only depending on the truth of pk but we are depending on truth of p p2 which that is what we have proved in the basis of induction step by this we end today's lecture thank you.