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Fax initial proof

this is the beginner's guide to mathematical proofs and today I'm going to go through five of the most common sorts of proofs so the first one is the direct proof so let's prove that if X is even then x squared is even so I like to think of it this way we start with what we know which is if X is even so that's at the top down the bottom the last line is where we want to go we want the last line to be therefore x squared is even and in the middle a logical flow so each statement in the middle must follow logically from the previous statements and anything else that we know that's obvious so we start with what we know if X is even well that means then that x equals 2 times our for some integer R because that's what it means for a number to be even it's twice another integer so that then tells us that x squared is equal to R or squared which is equal to 4 R squared which is equal twice or 2 times 2 R squared so this means that X is equal to 2 times s for the integer s equals to our squared inside therefore x squared is even because as we said before that's what it means through a number to be even it's twice or 2 times an integer okay now let's move on to the contrapositive and this works in basically swapping around the the two parts of the the proof so generally we want to prove something like if a then B and the contrapositive says that that's the same as proving if not B and not a we can see this graphically here's a situation where we have a is this set here and B is the biggest circle and if something is in a then you can see that it's in B but the other way to look at it is to say this is the area that's not B and if something's not B then it's not that it's in the area that's not a which is the area here so let's prove that if x squared is even then X is even and we're going to do that using a contrapositive and so that tells us that we start by swapping around the two parts of the proof and putting not in front of both so what we're going to show is that if X is not even then x squared is not even so let's do the proof now in this in similar to the way we did the direct proof what do we know we know that if they're weeks well we start with if X is not even and we know we're going to end up with down the bottom therefore x squared is not even so let's go back at the top if X is not even in x equals 2 times our plus 1 for some integer R after all that's what it means for a number to be odd it's twice or two times in each J plus one so this means that x squared is equal to 2 our plus 1 all squared which is equal to 4 R squared plus 4 R plus 1 which is equal to 2 times 2r squared plus 2r plus 1 so this means that x squared is actually equal to 2 times s plus 1 for the integer s equals 2 our squared plus 2r so if x squared is equal to 2 times twice s plus 1 then clearly we can conclude therefore that x squared is not even and that would come that would finish off the proof so that's the contrapositive next we'll go on to the proof by cases but before we do that I just mentioned if you like this video I've got quite a number of other videos in the made easy playlist and they're suitable for late high school and early university students where I go through it some typical exam or test questions on my youtube channel I got lots of other videos on mathematics which you might find interesting as well okay let's do now proof by cases and what we're going to do is prove this we're going to say let n be an integer prove that the floor of n over 2 plus the floor of n plus 1 over 2 equals in and I just remind you that the floor just means around down to the next lowest integer so the floor of 5 on 2 is 2 because 5 on 2 is 2 and 1/2 from the floor of 2 and 1/2 is 2 so that's an example so how do we go about proving this statement about the sum of these two floors equaling your end well we split it up in two cases the first case is that we're going to look at anything even the second case n is odd and of course n must be either even or odd so in this case two cases is sufficient to complete the proof so let's do that case 1 assume that n is even that is N equals 2 times n for some integer m and then what have we got we've got the floor of n over 2 plus the floor of n plus 1 over 2 is equal to the floor of 2m on - plus the floor of 2m plus 1 on 2 of justthere substituting in the ends that I have because I know that N equals 2 N and that equals the floor of n plus the floor of n plus 1/2 which is equal to M plus M which is equal to 2m which is equal to n so in the case where n is even we've proven what we need to prove now let's do the case two let's assume that n is odd that is N equals 2 times n plus 1 for some integer n that's what it means for an integer to be odd and then we go back to the left hand side of what we're trying to prove so we've got the floor of n on 2 plus the floor of n plus 1 on 2 and now we have to feed in N equals 2 n plus 1 so we get here the floor of 2n plus 1 on 2 plus the floor of 2n plus 2 on 2 and I'm just going to be careful here that's the floor of n plus 1/2 plus the floor of n plus 1 and that equals M plus n plus 1 which is equal to 2n plus 1 and 2 n plus 1 is equal to n so we've shown that it's true in the case of n being odd so we've done N equals odd we've done N equals even so we've done everything and that is sufficient and that's the proof by cases okay now let's go on to proof by contradiction one of the interesting ways to prove something so we're going to prove that the log of 7 to the base 10 is irrational so the idea with the contradiction is we start with this line here suppose that log 7 to the base 10 is not irrational so I'm starting with what I'm the opposite of what I'm trying to prove then we're going to have a whole lot of logical statements that follow from the previous statements and anything else that we know that's obvious and at the end we're going to get a contradiction and because we get a contradiction something that can't be true we know all the logical statements are true the only thing that the only thing that can't be true is our first statement in this case that Locke 7 to the base 10 is not irrational so that will then prove that log 7 to the base 10 is irrational so let's see how this works suppose that like 7 to the base 10 is not irrational so that means the log 10 to the 7 years rational so that so that means that it equals P divided by Q where P and Q are integers what we're saying is in this case if what 7 to the base 10 is rational then it's just a fraction which I've just shown there with P and Q so that means because of the definition what log is that means 10 to the power of P divided by Q equals 7 and if we now take the cutes power of both sides we get 10 to the P equals 7 to the power Q now the left-hand side must be even because 10 to the 10 to the power of anything is even and the right-hand side must be odd because 7 to the power of anything must be an odd number so we're saying that the left-hand side is even the right hand side is odd this is impossible so this is the contradiction that we have so once again every statement that I've made everything flows logically the only thing that we can conclude is our first statement is not correct and so therefore log 7 to the base 10 is sorry is not rational it is irrational okay we've got one more left I want to do and that's proof by induction now proof by induction you can think of this way imagine you were telling someone how to climb up this ladder the first thing is you could say there's an initial step and that is get onto the first rung of the ladder that's the initial step and then you could say okay if you're on the first step go to the second step if you're on the second step go to the third if you're on the third gather 450 on the fourth one fifth and etc until you get to the top now we can make it we can abbreviate that a little bit the first step is to say get on the first rung of the ladder and the second step we can say is if you are on a rung of a ladder go to the next rung of the ladder and of course that can be repeated and that will be enough to ensure that you get to the top of the ladder so that's the concept of induction so let's see how this might work let's prove that five divides 11 to the power n minus six for any positive integer n so the way you generally do these is you start off with a proposition that could be either true or false so let's let TN be the proposition that 5 divides 11 to the power n minus 6 now if you recall what we did with the ladder we're going to do the same sort of thing so the initial step is that we say what happens when N equals 1 the lowest value that we're interested in so 5 does in fact divide 11 to the power 1 minus 6 because 5 divides 5 so that means that P 1 is true so we've established something we've established that P 1 is true now for the induction step let's suppose that PK is true for some K that's equal to 1 2 3 4 100 whatever so that means that 5 divides 11 to the K minus 6 then now let's have a think about 11 to the K + 1 minus 6 well 11 to the K + 1 minus 6 is equal to 11 times 11 to the K minus 6 and then turns out that you have to add 60 now by assumption we've said that 5 divides 11 to the K minus X so that means that 5 divides 11 times 11 to the K minus X and also 5 divides 60 so if we look up here what we're saying is 5 divides both parts of the right hand side so 5 must divide the left hand side so therefore 5 divides 11 to the K + 1 minus 6 so that means that P to the K + sot k plus 1 is true so what we said is if TK is true then PK plus 1 is true and actually that's all you need to do because if you think about it we prove the T 1 is true and then we said if P 1 is true then P 2 is true but if P 2 is true then P 3 is true if T 3 is true then P 4 is 3 etc so what we can conclude is that 5 divides 11 to the N minus 6 for N equals 1 2 3 4 etc and we'd normally put by induction to explain how we did it so that's five proofs that'll help you get started with mathematical proofs if you're doing a course like discrete maps that requires you to specifically do proofs or either you know you just need to use proofs in some of your other mathematical studies so I hope you found it useful