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3. Prove that, given any n € Nand a € R,there exists 6 such that la ~al < 1...

Question

3. Prove that, given any n € Nand a € R,there exists 6 such that la ~al < 1

3. Prove that, given any n € Nand a € R,there exists 6 such that la ~al < 1



Answers

Prove each statement by mathematical induction. See Examples $1-3$. $$\text { If } 0<a<1, \text { then } a^{n}<a^{n-1}$$

They want us to prove that 60 factor of in Q plus three and squared plus two in for all values off in. Okay, so let's go ahead and see what we can do with this. So we're gonna do this by induction, So let's go ahead and call this just our statement yet. Now we need to check our base case, which is going to be in. Is he gonna one Since we're doing this for all natural numbers in and that's going to be so, it will just plug it one someone cute plus three times one squared, plus two times one. So that all adds up to six. And being a factor six or having a factor of six is just a fancy way of saying gets weaken divided by six. So six is divisible by six. So it's one so checks up all right now Or are induction hypothesis we will need to do the all the link. Well, we want to say Piquet is a statement. So six actor Oh que cute plus three k squared plus two k And then we want to show that PK plus one It is also true, which is the statement. Six factor of K plus one cute plus three K plus one squared plus, uh, two K plus one. All right, now, let's go ahead and start the proof. So remember, we always want tell the reader what kind of proof were doing because there's multiple ways to prove things. So by way, uh, induction, assume p K is true. Now we want to start with this, Okay? Plus one cute times, three K plus one squared, plus two times K plus one. So we somehow want to get this k cubed lost three K squared, plus two k to come out from this and just looking at it, I really don't see a good way to do that. And if we were to factor out a K plus one it doesn't look like would help either, because we just have, like, a K squared term in the middle. So, uh, that doesn't look like it's Ah, good way to try to go about this. So why don't we just expand everything? So let's see. So if we were to expand this, we should get que que plus three k squared plus three K plus one. If we were to expand this here That should give us plus three k squared plus six K plus three and then that last term is just gonna be two k plus two. So let's add everything. Go. Um, we only have one case square term, so that would just be K squared. We have three k squared three k squared. So that's going to give us six k squared. We have three K plus six K plus two case, that's 11 K And unless we have one plus two plus three, which is six. Okay, so we somehow want to get this cake plus three k squared, plus two K to come up. So I'm gonna write that down right here. Que cute plus three k squared plus two K. Well, that shoes pulls some of these terms out, and right over here, we're gonna have K cubed plus three K squared plus two K. So if we do that, then so that case squared or kick, it becomes just zero. This K squared should become plus three case. Where and then over here, this 11 K is going to become, um, nine K. And in the six just days there All right, so we have this now, so we know for sure This portion in blue is visible by six by our induction hypothesis by we don't know. Sorry. Let me just say this is PK appear, but we don't know if this term here is divisible by six. So one of the tricks we tried in some of the other problems is to just keep pulling out mawr terms of it until we get this to be something that's all divisible by six. But you might know. So we try to do that again. So we would have negative k cubed here, and then we would have, um, no k squared. And then we'd have that nine k, but it would be seven k. Then here would be plus six still. And then this would be two times que que plus three k squared plus two K like this. And then we would say, Okay, well, this still isn't anything that's divisible by six, and we just keep repeating. And since this nine here isn't an even number will never be able to get anything that's divisible by six. If we keep subtracting because I'll always be odd. So we might need to try something else or this one. So what do we do? Some scratch work over here on the side or this So scratch? Well, we can factor three outs. Let's try that. So we have three times k squared plus three K plus two. And then this quadratic is also factor. What would be K plus one? Okay, plus two. Okay, well, if something is divisible by six, that means it's divisible by two and three. So we already know this is gonna be divisible by three since we're multiplying by three. So what we need to ask ourselves is this expression visible by two. And if you think about it for a little, you should be able to agree with me that this will be divisible by two since K plus one in K plus two are consecutive integers. And since it's consecutive integers, one of them has to be even is if you think about it. So if we start counting so 01234567 and so on. It alternates between even and odd. So one of these has to be even and thus the definition of even is that it's divisible by two. So that means this is divisible by six. So let's just go ahead and rewrite this real fast looks k cubed plus three casework plus two K And then over here, this is going to be plus three times K plus one times K plus two. Yeah. So this is the statement we're working with now. Well, by induction hypothesis, So by our induction hypothesis, ih que que plus three k squared plus two K is divisible I six and now to show that three K plus one times K plus two is divisible, we're pretty much is going to explain the same thing we did over here about this being three times some even number so so and three times K plus one times K plus two is divisible by six cents. It is too visible by three and it must be divisible by two since K plus one and K plus two or consecutive integers and one must be either. All right, so what we just show is now or I should before we do that. So since both of these are true so since both terms are divisible by six, so must be there some and so we pretty much shown that PK plus one is also a true statement of it being divisible by six. So let's just go ahead and wrap this upper class by saying so since P K plus one is true, assuming P K is true, then peon is true for all in elements of the natural numbers. And then once you do this, you could put your proof box and a smiley face because you're glad you're done with it.

When we want to show that in Cuba minus and is divisible by six for all end in the positive vintages. Um, so we want to check the base case, which is in equals one. Since we're talking with the positive managers, then one cubed minus one is equal to reserve, which is divisible by six. Uh so n equals one is true. We assumed this is true up to n equals K that would tell us k cubed minus K is divisible by six. Then for n equals K plus one, that would mean K plus one cubed minus K plus one. And we want to try to ah show that this thing is also visible by six. We'll have to expand it. So I'm gonna take out a common factor of K plus one left with K plus one squared minus one, and that works out to K plus one times k squared plus two K. So this is easy to distribute. This is K cubed, uh, plus three K squared plus two K and ah, we want to find a way to use that inductive steps. So I want cake cubed minus k somewhere. So I will write it as K cubed minus K plus three k squared which would change my last term too. Three K three k j f k. That creates the negative two k in the previous step. So this is still equivalent. Um, the que cubed minus K here can be written as six A. Because it is divisible by six. So that's six times some integer. We don't know what it is plus three k squared plus three k where a ah is an interview. So that means I can take it a common factor of three on. Then I would have to a plus free Ansari plus K squared plus k left over. Ah, that only shows that is divisible by three. I need to show that it's divisible by six. So I need to find another factor of two that can be pulled out. This first part is great. I know that is divisible by two. So I just need to show that this thing back here is divisible by two. So let's consider case Grid plus K on the side here K squared plus que is the same as K Times K plus one. So let's choose K to be odd. If Kay is odd, then I would have an odd number. Que Plus one would then be an even number because they're consecutive numbers, odd times and even is going to be. Even if Kay is even to begin with, then it would be even and K plus one would be odd again, even times AAD is still even so it turns out that k squared plus que is always going to be even, which means I can write, ah three times to a case great plus que is always even so I can write that as to be because if it's even, it's always divisible by two where B is an integer and if I write it like this, you can tell that I could pull out another two, which is six times a plus b and yeah, hey, plus B r ever variables that we've introduced, but we don't care what their actual values are. We've just shown that there is a six in front, which means que plus one to the power of three miles. K plus one is divisible by six, therefore and cubed minus end is divisible by six for all end in the positive into choose

All right. So I'm this problem. Uh, the things I say might sound redundant, but I'm gonna do my best to explain what's coming on here. Uh, and what we're gonna try and show is that any time that you take a positive exponent here is on is the basis between zero and one. You're still gonna end up with some number between zero and one. Um, so I would just start with the base case, which is? Ah, saying that And is one Well, is this again? I mentioned this is redundant because we know that Ah, a is between zero and one. Well, if you take a to the first, power is just gonna equal a well. We already know A is between zero and one. So this statement is true. So then if we do the inductive step so now that we know the base case is true, weaken, do the inductive step. That's for anise equals K plus one. So basically what it boils down to is ah so money put box. We know this statement is true. All right, so now if we were to change a to the power to be a to the K plus one power. Um, basically, what it boils down to is this is equal to a to the K times. Uh, so I didn't want to put that times right there times a to the first. Um, so basically, what this boils down to is because we know eight of the first is between zero and one. If we add another thing to it, that's between zero and one. We know that eight of the K is true times by another. Maybe I should underline that this is between zero and one, and this is between zero and one that if we multiply them together, it's still between zero and one. So I'm trying my best to explain this kind of in depth of you multiply a decimal byte decimal. You just get another decimal between zero and one. Um, so this is really all the work I would show to get the right answer. Your instructor might be expecting a little bit more, but I think this is sufficient. It's explaining pretty well that, um exactly what we wanted to show

Hello. So here are given sequence is the end fruit of eight to the end plus B to the end. So if you take the limit as N goes to infinity of the end. Fruit of a de emphasis to the end. While we have an indeterminate form which is going to be infinity to the zero. Can then apply low petals role. Um And we take again the limit of well this is going to give us the limit of um age of the fsb to the end to the one over. And which is going to give us the limit as N approaches infinity of we get here um A A to the end times Ln of A plus B to the end times Ln of be all over A. To the N. B to the end. Which is going to give us here A A over B to the N. Times Ln of A plus Ln of B. All divided by A over B. To the N. Um plus one. And then um taking the limit here, this is gonna gonna just convert to us to the natural log of BE. So therefore yes, we take the limit um As N. Goes to infinity of the natural log of eight to the end to be to the end to the one over end is going to be equal um and converged to be.


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