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THEORETICAL PROBLEMS= (1-4) Answer each of the following statements use any proof technique that You would like to use) For every nonnegative real number and natura...

Question

THEORETICAL PROBLEMS= (1-4) Answer each of the following statements use any proof technique that You would like to use) For every nonnegative real number and natural number n. (1 +x)" > 1+nx_ Use the well-ordering principle to show that there is no natural number x such that 0 Hint: Do not use induction O the well-ordering proof technique, use the well- ordering principle directly: Prove that if 23 + 2x2 0. then 2x + 5 <11. Prove that the limit of the following sequence is lilln-+o &q

THEORETICAL PROBLEMS= (1-4) Answer each of the following statements use any proof technique that You would like to use) For every nonnegative real number and natural number n. (1 +x)" > 1+nx_ Use the well-ordering principle to show that there is no natural number x such that 0 Hint: Do not use induction O the well-ordering proof technique, use the well- ordering principle directly: Prove that if 23 + 2x2 0. then 2x + 5 <11. Prove that the limit of the following sequence is lilln-+o "+3 n3+1 = 0 Let a. b, and c be integers. Prove that if c2 + 62 22 , then at least one of & and b is even Prove that for all natural numbers n > 1. 2" < 2n+1 2"-1 _ -1. Prove that 10n+1 + 3.4"-1 + 5 is divisible by 9 for all natural numbers n 2 1. Let a, b, and be integers Prove that if a k bc; then *b. Prove that for all natural numbers n, 1 + {+}+f+3+ +321+3 Hint: Unlike previous formulae for sums you ve had to prove; the number of additional terms at each step as n increases is more than Let A, B, and € be sets. Prove that (4 B) U (A C) = A iff AnBnc = 0. Let a, b, and c be odd integers and a real solution to az2 _ br +e= 0. Prove that ~is irrational Let a, b. and be positive integers. Prove that if ab 2 c then 2 Vc o1 b 2 Vc.



Answers

In this chapter we began with the natural numbers and gradually built up to the real numbers. A completely rigorous discussion of this process requires a little book in itself (see Part $V$ ). No one has ever figured out how to get to the real numbers without going through this process, but if we do accept the real numbers as given, then the natural numbers can be defined as the real numbers of the form $1,1+1,1+1+1,$ etc. The whole point of this problem is to show that there is a rigorous mathematical way of saying "etc." (a) A set $A$ of real numbers is called inductive if (1) 1 is in $A$ (2) $k+1$ is in $A$ whenever $k$ is in $A$ Prove that
(i) $\quad \mathbf{R}$ is inductive. (ii) The set of positive real numbers is inductive. (iii) The set of positive real numbers unequal to $\frac{1}{2}$ is inductive. (iv) The set of positive real numbers unequal to 5 is not inductive. (v) If $A$ and $B$ are inductive, then the set $C$ of real numbers which in both $A$ and $B$ is also inductive.
(b) A real number $n$ will be called a natural number if $n$ is in every inductive set. (i) Prove that 1 is a natural number. (ii) Prove that $k+1$ is a natural number if $k$ is a natural number.

So it does seem likely that ruin is irrational wherever the natural number is not the square of another natural number. Um So what we want to do for this problem is we want to turn this argument into a rigorous proof by complete induction. So we're going to have the fact that route and we're gonna let this be such that and is not a perfect square. So what it means for end not to be a perfect square is it means that for all N, there cannot exist a K. Such that K squared um equals. And yeah, with this in mind, we're able to show by induction that this is going to be an irrational number.

High in this question. We are asked to prove using induction the formula, the inequality regarding it's some and this formula. So we one end to start with to like, at least two. And here we go. We let this whole state men BP au vin the basic step and equal to two we have on the left hand side of some off one and 1/4 on the right hands. I in the formal Abu have to minus Uh huh. And this will become so five over full, less than tree over too. Wishes correct, Right? So the basic step is true. Next for the inactives that we supposed us that men is true for some in and we want to show that the in plus one is true s bill. And one way to do this is to start with P au vin here. So we have this inequality and IBU ad this term on both sides so that the some on the left become the sum we want in the in. Plus one statement What's left is to ling like is to show that this right hand side is less than the farm lobby. One right and How do I dump from there to there? Well, we let's look at this. I claim that that this inequality is true. Why? Because let's see, this term 10 n plus one square is less than one away in plus one, right? Because in this positive integer and one over in plus one is if I multiply in on, like, above and below, so it doesn't change anything. I can break this into this to turn one over in minus 10 n plus one. And this this whole lie off a human man, the a bold claim. True, because we can just move one over in to the left side, off the inequality and that that's make us this inequality. And so we use this Indy in the previous that that'd be dumb from that to there. And oh, in all this showed up the statement when off p and plus one is true, right? And so we have proven the inactive state as well. And so with this two step, we have proven the Steadman fall. Any positive integer in greater than one on That is it. Thank you

Are they giving a problem? We uh the result we have has a more important generalization as the arithmetic mean and the geometric mean. So it's good that we define these first the arithmetic mean Is going to be the more common mean that we're used to also known as the average. So it's going to be a one plus A two all the way to A. N. And then this is divided by. And but then we also have the geometric mean G seven Mhm. Which is equal to the square root of a one times all the way through A. M. When we do that, we take the and through of it. So that's going to be our final answer.

Hi. This question We are us to prove that this formula for the some off into just square. Ah, correct. And it is true for any positive integer in. And we are going to use induction to do this. First we let this whole statement, uh, name it, pee off in, and we're going to do this into step right. Basic step and inductive step. So first bet, X step. We just need to show that it is true for the first positive in HAGIA which is one year the Steadman at one woo. How Levin's I asked just a some off one terms, right? And the right hands Ibuki put one into substitute in for one and we gonna get that this is six or six wishes one. So the statement is true for people to sorry for an equal to one so basics that is clear. Next is inductive step. So in this step, we want to cool that eve. The statement is true for any end for any positive integer in it will forth the next state man that is pee in plus one to be true as well. So P o n is what we can use is our like a dot thought that we can use p in ple as one of these. The result would be one. So we want to modify peel in somehow to arrive at p N plus one. So let's do this. So first we started p Elvin and we have this ass fact because inductive hepatitis is next. I put ab the the next term that we want in plus one square on both sigh So on the right hands I we we just add this term into, like, added with fraction. So we time six to it so that we can much everything together and have 106 year and the middle the middle term rule. You will be able to separate it into this too. And you can see that this is in fact, the the formula to be one for win. So is a formula for p r n plus one, right? Because the last one two in plus t is in fact, two times in plus one and then plus one again. So this is the correct form. A lot of the one we have shown up the some off into just acquire from one to in plus one equal this formula. So we have prevented the inductive Is that and we clear on both step now means that by meant induction It is true fall any positive Fantasia This is a kiss because any any positive integer supposed k that we Well I want to know if the statement is true P okay must be true because it is not true p o came in this one the one before you also I do X. This also will not be true all right Because we just proved that if even if the one in front of it is true that it must be true If we assume that for some k the statement is not true than the one before it also must not be true But then we can have this sequins Uh false. That meant back until we get we hit p 01 right? Because positive in Hagia no matter how high it is Eventually this sequence we hit Pier one and it will force p off one to be false wish is not the case. We just show in basic step so this whole acumen can happen. So any k Any kid that we looked at the statement must be true. And that is it for this question. Thank you.


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