 discussion on mathematical induction in the last lecture we have studied mathematical induction and some problems related to mathematical induction now we will start by giving another alternative and interestingly equivalent version of the principle of mathematical induction which is called the strong mathematical induction at this point it is worth mentioning that the original mathematical induction that we studied in the last lecture and the strong mathematical induction that we are discussing now are equivalent but sometimes depending on the problem the first version is more useful and sometimes the second one that is the one which we are going to study now becomes more useful let us first see what is meant by strong mathematical induction let Pn be a statement for each integer n may be either true or false then is true for all positive integers there is an integer q greater than or equal to 1 such that 1 p1 p2 pq are all true when k is greater than or equal to q the assumption that p i is true for all integers 1 less than or equal to i less than or equal to k implies that pk plus 1 is true so here we see that our hypothesis and assumptions are slightly different than the previous one what we prove here what we have to start prove in the beginning is that there is a positive integer q which is greater than or equal to 1 for which p1 p2 and pq are all true that is we have to show that for a positive integer any positive integer less than that will satisfy Pn then our assumption is that we take a q sorry we take a k greater than or equal to q and assume that for all i between 1 to k p i is true and then we have to prove that this implies that pk plus 1 is true if we can do this then we claim that the statement Pn is true for all n greater than or equal to 1 and this is what we mean by the principle of strong mathematical induction as I have already said that the two principles are exactly same they have the same power but we use either of them depending on the problem under consideration now let us check the steps of the proof or steps of a proof involving strong mathematical induction the steps of a proof involving strong mathematical induction so first like before we have basis of induction basis of induction show that p1 p2 pq are all true strong inductive hypothesis assume that p i is true for all integers i such that 1 less than or equal to i less than or equal to k where k is greater than or equal to q now if you compare the previous version there we assume that pk is true for some k greater than or equal to n0 and we prove that n0 pn0 is true and then we went on to prove that pk plus 1 is true in this case we assume that for all i between 1 to k where k is greater than or equal to q p i is true and then we go on to prove this in the inductive step show that pk plus 1 is true on the basis the strong inductive hypothesis now before we go on to discuss some examples where strong mathematical induction is useful we will quickly have a look at recursions and how we define recursions let n be the set of non-negative integers a function from the set of non-negative integers is defined recursively if the value of f at 0 is given for each positive integer n the value fn is defined in terms of the values of fk where 0 less or equal to k less or equal to n now we look at some recurrence relations and the most probably the most famous recurrence relation that is the Fibonacci sequence f0 is defined as 1 which is equal to f1 and fn plus 1 is fn plus fn minus 1 for all n greater than or equal to 1 this is how we define Fibonacci sequence recursively so let us start calculating the elements of the sequence so as we see for 0 f0 equal to 1 then for 1 f1 equal to 1 then for 2 this value is f0 plus f1 which gives me 2 then for 3 this value is f2 plus f1 that is 2 plus 1 which is equal to 3 for 4 this is f3 plus f2 which is equal to 2 plus 3 equal to 5 and so on now it is known that the nth Fibonacci number that is fn is given by 1 over root of 5 then 1 plus square root of 5 over 2 whole raise to the power n plus 1 minus 1 minus square root of 5 over 2 whole raise to the power n plus 1 for all n greater than or equal to 0 and we would like to prove this relation by using strong mathematical induction proof let us put n equal to 0 if we do that then f0 is 1 by root 5 and here it will be 1 plus square root of 5 divided by 2 minus 1 minus square root of 5 divided by 2 this is equal to 1 for n equal to 1 we have f1 equal to 1 by square root of 5 and this is 1 by square root of 5 over 2 and square minus 1 minus square root of 5 by 2 square now this is equal to 1 by square root of 5 and 1 plus square root of 5 plus 1 minus square root of 5 whole divided by 2 into 1 plus square root of 5 minus 1 plus square root of 5 by 2 yes and therefore here it will get cancelled and this factor is equal to 1 whereas this factor is equal to square root of 5 and this gives me 1 so we have checked that our formula given here works for n equal to 0 and n equal to 1 now we go for the induction hypothesis which in this case becomes strong induction hypothesis for n greater than or equal to 1 assume that fk equal to 1 by square root of 5 into within bracket 1 plus square root of 5 divided by 2 raise to the power k plus 1 minus 1 minus square root of 5 divided by 2 raise to the power k plus 1 yes for each integer k for each integer k where 0 less than or equal to k less than or equal to n now once this is my induction hypothesis I go for the induction inductive step inductive step we start with fn plus 1 the question is why fn plus 1 and not fk plus 1 so we go to the induction hypothesis and we see that here we have assumed that I have an n and for all k from 0 to n this formula works and I am now checking for n plus 1 if I am successful in proving that this formula works for fn plus 1 that means for all k between 0 to n plus 1 the formula works and therefore we can start checking for n plus 2 but that is of course that will work by using the same argument because we have we are able to choose n any positive integer greater than 1 and fix it so we now have fn plus 1 and if we use the basic recursion formula we know that fn plus 1 is equal to fn plus fn minus 1 and since in both these cases the subscript n and n minus 1 are less or equal to n we can write by using the induction hypothesis that 1 by root 5 1 plus root 5 divided by 2 n plus 1 minus 1 minus root 5 divided by 2 n plus 1 plus 1 by root 5 1 plus root 5 raise to the power n minus 1 minus root 5 by 2 raise to the power n now for convenience we replace 1 plus root 5 upon 2 by a and 1 minus root 5 divided by 2 by b and the expression will become a to the power n plus 1 minus b to the power n plus 1 plus a to the power n minus b to the power n which in turn becomes 1 by root 5 a to the power n a plus 1 minus b to the power n b plus 1 now if we start with a plus 1 we will see that a plus 1 is equal to 1 plus root 5 by 2 plus 1 which is of course equal to 1 plus root 5 by 2 and 1 plus root 5 plus 2 which after simplification becomes 3 plus root 5 by 2 and by a sudden streak of imagination if we are able to check a square that is 1 plus root 5 upon 2 whole square which is equal to 4 in the denominator and the numerator 1 plus root 5 whole square which gives us 4 1 plus 5 plus 2 root 5 which brings us to 6 plus root 5 divided by 4 which is 3 plus root 5 divided by 2 I said that we need imagination somehow we have to guess the right thing and to to realize that what we get for a plus 1 is same thing as a square so this is of course a plus 1 similarly we will see that b square equal to 1 plus b and if we replace these two expressions in the calculation of f n plus 1 then I get 1 by root 5 equal to a a to the power n into a square so that gives me a n plus 2 and minus b n plus 2 and which is which shows me that the formula works for n plus 1 and therefore our proof is complete from this we can say thus the formula of f n works for f n plus 1 hence our proof is complete thus we have seen one example using the principle of strong mathematical induction now let us look at another problem which uses the principle of mathematical induction and it is a problem related to postage stamps suppose that we have stamps of different denominations rupees 3 and rupees 5 to show that it is possible to make up any postage of rupees 8 or more by using these two stamps to show that it is possible to make up any postage of rupees 8 or more by using the stamps of these denominations only now to solve this problem we start from the beginning if we have a postage of rupees 8 of course 8 can be written as 3 plus 5 now let us check 9 9 can be written as 3 plus 3 plus 3 10 10 can be written as 5 plus 5 now if we come to 11 11 can be written as 5 plus 3 plus 3 so we see that at least for first few consecutive possible postages we can make up those postages by using the denominations of rupees 3 and 5 now the question is that true then we can say that let us use mathematical induction we have already proved the basis of induction so basis of induction rupees 8 postage can be made up of the denominations rupees 3 rupees 8 and no rupees 3 this is rupees 3 and rupees 5 of course because 8 equal to 3 plus 5 now we go to the induction step no we go towards induction hypothesis assume the result to be true for a postage of rupees k where k is greater than or equal to 8 now we come to induction step in the inductive step we have two cases case one to make up the postage k at least 1 rupees 5 stamp is required what do we do then replace the rupees 5 stamp by two stamps of rupees 3 if we do that then we see that 5 gets replaced by 6 and therefore the total postage will be k plus 1 the postage will be k plus 1 now we come to case 2 now in the case 2 all the stamps required to make up the postage of k is of denomination 3 so there is no 5 denomination 5 stamp so in this case all the stamps are denomination rupees 3 now what we realize over here is that this postage cannot be 8 so it can be 9 in which case it is 3 plus 3 plus 3 9 and if it is something more than 9 then also it will have at least three stamps of denomination 3 in this case there must be three stamps of denomination 3 rupees 3 now our strategy will be to replace these three stamps by two stamps of denomination 5 so replace them that is all the three stamps by two stamps of denomination 5 rupees 5 to obtain a postage of rupees k plus 1 now we see that this is exactly what we wanted we wanted to show that if we have a postage of k and if we can make up a postage of k by using stamps of denomination 3 and 5 whatever the case may be I can make up a postage of rupees k plus 1 once we have proved that and we know that we can make up postage of 8 therefore by mathematical induction we know that we have got the complete proof the last problem in this lecture on mathematical induction is involving chess boards now let us consider an 8 by 8 chess board well this is an 8 by 8 chess board and suppose we take out one block from it suppress one block or square from it then we will call this a defective 8 by 8 chess board defective 8 by 8 chess board now there is another object that we would like to introduce which is called a triomino it looks like this the question that we are asking is that can I can I cover this 8 by 8 chess board by using triominoes of this type let me correct myself my question is that can I cover a defective 8 by 8 chess board by triominoes or in general is it possible to tile an n by 2 to the power n by 2 to the power n by n defective chess board by triominoes question is it possible to tile 2 to the power n by 2 to the power n defective chess board by using triominoes now we will try to use mathematical induction suppose we take n equal to 1 then I get a 2 by 2 chess board and I know that I can tile it by triomino suppose because of being defective this shaded square is excluded then whatever is left is a triomino it does not matter which square I leave out if the square is this then I can place the triomino like this or if the square is like this then I can place the triomino like this or if the square is over here which is removed then the triomino will be kept like this so I can I can put any triomino one triomino and by tiling I mean that I want to cover the chess board by triominoes and distinct triominoes must not intersect that is they must not overlap that is something that we have to be careful and for n equal to 1 I say that I have only single triomino so I can do this without any overlap now the question is that what about a 4 by 4 triomino sorry what about a 4 by 4 chess board defective chess board so n equal to 2 so I have got a situation where I have got 4 by 4 so this is 2 square by 2 square chess board and let us suppose that I put a defect over here now we see over here is that a 4 by 4 chess board is made up of 4 2 by 2 chess boards 1 here this is 2 2 and this is 2 2 2 2 2 so we have got 2 by 2 chess boards over here 4 2 by 2 chess boards are giving me 1 4 by 4 chess board and if we want to go in general we will see that 2 to the power k plus 1 by 2 to the power k plus 1 chess board is made up of 4 chess boards each of which are 2 to the power k by 2 to the power k all right now we know that there is a defect so the defect in the 2 to the power k plus 1 by 2 to the power k plus 1 chess board has to lie in one of these smaller chess board suppose it is here then what we can do is that we can put a triomino around the center here making the remaining 3 chess boards defective now let us assume that 2 to the power k 2 to the power k defective chess boards can be tiled by triomino then this one can be tiled this one can be tiled this one can be tiled this one can be tiled and of course this one can be tiled because this is defective from the beginning and therefore the whole chess board can be tiled so if we assume that tiling is possible for k we have proved that tiling is possible for k plus 1 assuming that the tiling is possible for k we have that is to be more precise 2 to the power k by 2 to the power k defective chess board let us write more precisely so 2 to the power k by 2 to the power k defective chess board we have proved that it is possible for 2 to the power k plus 1 by 2 to the power k plus 1 defective chess board and we have already seen that tiling works for n equal to 1 therefore we know that if we have a 2 to the power n by 2 to the power n defective chess board no matter how large n is we can tile it and of course in that case we can tile an 8 by 8 chess board and that is all for today thank you