P.S., I'm sorry if I ripped someone else's ideas off in your thread, I wouldn't know, because I didn't look through even close to everything; when I saw how much feedback there was for one problem I thought I would do better to keep myself from getting confused and just try it myself.
Apostol already proves in the section that any nonempty set bounded below has an infimum. Please Register or Log in to view the hidden image! Your proof is similar in direction to mine but longer. Here's mine: From Problem 2 there exists an integer u >= -x. Let S contain the negatives of all such integers u. S is bounded above by u and thus has a supremum n. (The fact that n is a supremum implies by definition that n<=x, although you went the long way and applied the definition to prove so). Since (n+1) is not an element of S, n+1>x. As for uniqueness, there is exactly one supremum of S and so exactly one integer satisfying the inequality. Seem good? Now if you could help me with #6.. I found the proof for 6 but it's rather.. unobvious?
I found a proof for 6 when looking around earlier for solutions to these exercises. It's actually a quiz from stanford in pdf format. Without the well ordering principle, it seems that you are expected to be very creative with this one.....I know there's a star by this section, but I think there should have been a much more significant disclaimer. I suppose it's assumed that if you are using the book, you are probably in an advanced class with very good students and a very good professor who can help you. Here's the link, look at number 6: http://math.stanford.edu/~vakil/014/quiz1sol.pdf Your proof seems fine, and is probably alot more similar to something youd find in a book because it outlines the important points and leaves the tedious parts for the reader to confirm. My problem, of course, is mainly with the theorems about irrationals. The only information we've been given this point is that they cannot be represented as the ratio of two integers. You can find basically the ENTIRE proof(s) leading up to the irrationality of the square root of 2 in the first few pages of rudin which you can look at on amazon. http://www.amazon.com/gp/sitbv3/rea...10?_encoding=UTF8&asin=007054235X#reader-link He also points out very casually the answer to my very first question in this thread, that "between any two reals there is another: if r < s, then r < (r+s)/2 < s"
http://wwwmaths.anu.edu.au/~john/Assets/Lecture Notes/B21H_97.pdf Hello alyosha. I did not sleep till 5 this morning when I found this little gem. On page 42 (31) of the pdf, you will find short proofs for the rational and irrational case. The book might come in handy. I'm going to play around a bit with generalizing the proof for irrationals since we're not supposed to 'know' sqrt(2) is irrational until later in the problem set. That's one of the differences between Apostol and Spivak. Spivak would probably have thrown in a time saving hint or provided (gasp) a solution to the proof in the back of the book. Spivak is more friendly to self learners, I suppose. A Proof that there exists at least one irrational z such that x < z < y x < y 0 < y - x Let i be a member of the set of positive irrational numbers. By the Archimedian property there exists a positive integer n such that ni(y-x) > 1 Let a be the smallest integer (by #5) such that nix < a then z = a/(ni) and z > x if z >=y then z>=x + (y-x) > x + 1/(ni) z > x + 1/(ni) (a-1)/(ni) > x, contradicting the fact that a is the smallest integer such that nix < a. I think the least obvious logical step is postulating the existence of a. I'm having trouble seeing how to formulate 10e though. 12 is a stunner as well but I think I understand the point.
Thank you so much for that pdf, southstar; its a goldmine. It goes all the way to manifolds; wow! I like your proof, and the only concern of mine is actually the existence of a positive irrational number. Obviously the square root of two is an example, but that problem is placed after this one. I'm also a little confused as to just what we are supposed to demonstrate in number 7. As for number 10e, I don't exactly know what your problem with it is, but it seems to be just a matter of showing that if a and b are both even, then the factor of 2 that they both share can be cancelled until one of them is odd. I haven't tried it yet though, I've just been looking at that pdf.
Apostol takes a few things for granted with the reals, it looks like he's just accepting the existence of irrational numbers at this point. This exercise would have been better placed follwing the proof sqrt(2) is irrational so you know you've got one, and it can be moved there without trouble (ex 10 and 11 don't depend on ex 9). Many options here. Try assuming you've got your rational as a/b where b is the smallest possible (in absolute value). If both a and b are even, find smaller. Esssentially that addition and multiplication of a (non-zero) rational and an irrational cannot be rational. You know the rationals are closed under these operations, but if you combine one with an irrational, you get an irrational. Try arguing by contradiction.
I was able to solve 7 by shmoe's method a short while after posting but I'm now more confused than ever about #12. It appears that he's asking us to disprove the least upper bound axiom but I have no clue how to show that the rationals do not satisfy that property. After looking around for a bit I thought the set S = {x | x is rational, x^2<2} would work because it had no least upper bound but indeed it does (a supremum does not need to be a member of the set). I can't figure out what else he could mean though.. Please Register or Log in to view the hidden image! He also claims in the problem that the Archimedian property was deduced for the real-number system but it seems clear that it was only deduced for the positive integers For #7: For a/b, if a and b are even, a = 2m and b = 2n. So a/b = m/n. This can be repeated a finite # of times until n and/or m is odd. What escapes me is how to justify this last statement because it seems so obvious.
It's supremum is not a rational number though. Confined to the rationals, your set S has no least upper bound, and hence the rational numbers does not have the least upper bound property. (compare with saying there is no multiplicative inverse for 2 in the integers, even though we know how to stick the integers into a larger set where 2 has an inverse) Look at the Archimedean property again, it's a statement about the real numbers. No matter how small a real number x>0 is, we can add x to itself enough times (i.e. multiply by some positive integer n) to make it larger than any given fixed real number y. This is why I suggested to start with a minimal b (by well ordering, it's equivalent to induction which you already have). such an m and n would contradict this choice of b.
For number 8, "Is the sum or product of two irrational numbers always irrational?" , the product part is taken care of easily with a specific case like the root of 2 multiplied by itself. Maybe it could be made more general simply by stating that if i is irrational, then so is 1/i, and i(1/i)=1. But what about the sums of irrational numbers? Also, the positive integer n in your irrational proof seems like it should be unecessary, but maybe it is with the tools we've been given so far. For a given real x>o, it should seem that there must be a positive irrational i that for an arbitrary real y xi>y, because if not, then there is no i that i > y/x implying that the set of positive irrationals is bounded above. But, sadly, I have no way to show that it isn't. And it seems like the well ordering principle is necessary to postulate the existence of your a, but of course that section comes after this one. So the question is, did we goof or did apostol?
That gives lots more examples, one is sufficient though. What can you add to sqrt(2) to get an integer?
I considered root 2 + -(root 2) to be a sufficient counterexample Please Register or Log in to view the hidden image!. shmoe's works too and is probably more along the point apostol was trying to get across (if he is thinking of the number (1 - root 2)). If this is addressed to me then I should point out that (as I mentioned in the proof) a is the integer n from #5. In other words, nix <= a < nix + 1 except if nix happens to be an integer of course, then a = nix + 1. im too tired now to determine whether this extra step is necessary though. if you haven't seen this by tomorrow i'll look it over. the n in the proof may seem superfluous for as shmoe pointed out, the archimedean property extends to the reals. let me see if i can churn out a sample proof before my head explodes: ... ok, that didn't work. that was my problem too with #12. the archimedean property provided only works for integers and although it may seem obvious that if a subset (the integers) is unbounded then the set itself (the reals) must be unbounded we don't actually have the tools to justify the assertion, except for the set of integers. i think we're just supposed to assume it though because, for example, apostol takes it for granted in theorem 1.28 that the supremum of a set of integers is itself an integer (this is not so obvious when you remember that a supremum is not necessarily a member of the set). maybe we're trying to out-think the man? Please Register or Log in to view the hidden image!
I guess my problem with using #5 is the logical demonstration that a is necessarily the smallest integer such that a > nix. It seems like if this were enough to show that it is the smallest, then this argument should be enough to prove the well ordering principle itself. It also seems like you necessarily must end up with x <= z < y.
It seems obvious to me that a is the smallest integer from the following facts: From 2 we know that there exists an integer u >= nix. Let S contain all such integers u. S is bounded below by u and thus has a minimum/infimum a. See? n-1 is not in S and so can be discounted and n+1 > n and so is not the smallest such integer. I don't think it is sufficient to prove the well ordering principle however since the well ordering principle deals with sets of positive integers while #5 deals with only one real number x. --- How can I approach #9 on page 36? i think i start with: a line of length (root 1) an be constructed for the positive integer 1. but then assuming k can be constructed, how do i prove the assertion holds for (k+1)? let's see... i believe root n < n for all n>1 is this on the right track? is the answer to prove that the inequality holds for all positive integers? or maybe i'm totally off Please Register or Log in to view the hidden image!
Fair enough. With 10e you can show that if you can never simplify to an odd number, there will be a reduction to absurdity because your integers a/b will necessarily get smaller and smaller without end, when we know that 1 is the smallest possible positive integer. As far as proving 11, combining 10e with 10d allows you to arrive at an immediate contradiction. I was actually having some trouble showing that all integers must either be even or odd, I was thinking contradiction but nothing came right away so I skipped to the others. I'll start on 12 soon.
For 10 b i began by stating that one is odd. then i used something similar to induction to show that if n is even then n+1 is odd and vice versa. then state that 0 is even. then treat the set of negative integers by showing that if n is even then -n is even and if n is odd then -n is odd. i need help with #9 on p36 though (see previous post)
Here's my attempt at 12. When he spoke of the archimedean and LUB properties of the rational number system, I assumed he meant to use the definitions for the reals, with "rational" replacing every occurence of "real". For every rational a/b there exists a positive integer n such that n > a/b. This is justified because the positive integers cannot be bounded above. If a/b > 0 and if p/q is an arbitrary rational number, there exists a positive integer n such that na/b > p/q. In this case, choose rational pb/qa, so that there must exist n > pb/qa, and it follows that na/b > p/q. Then consider the statement: "Every nonempty set S of rational numbers which is bounded above has a supremum; that is, there is a rational number B such that B = sup S" Then consider the set S of rationals r such that S = {r| r < sqrt 2} Then suppose it has some rational B = sup S, which by definition must be greater than or equal to all r in S and must be less than all other upper bounds for S. B = sqrt 2 is not possible because sqrt 2 is not rational. If B > sqrt 2, then by a previous theorem there must exist a rational z such that sqrt 2 < z < B. But then z is a rational upper bound less than B, which can't be. If B < sqrt 2, by the same theorem there must exist rational z that B < z < sqrt 2. But then z is a rational in S larger than B, which can't be. B, the LUB of S, then, cannot exist.
that's more thorough than mine (i think). Please Register or Log in to view the hidden image! i was content to say sqrt 2 was the supremum of S but that there would always be a rational greater than any r in S and therefore S had no maximum
Sorry for neglecting the thread for so long. I've been adapting to dorm life this past week. As for number 8 on page 36, I've shown how to derive An <= A1C^(n-1) from the initial information, and I've shown that its true for n=1. However, has the "induction" already been performed by the assertions stated in the question, or am I missing something here?
As for number 9, I'm not sure what he means with the straightedge and compass bit; reminds me of something I read in the intro to a modern algebra book.......
About number 10, preliminary thoughts; again the role of induction seems a little unclear. By solving for r in the first equation and substituting in the inequality, one can arrive at: q <= n/b < q+1 N and B were already chosen and their existence is not being debated. By an earlier theorem, the integer q exists and is unique. It seems that now because we know that N, Q, and B exist, r can simply be defined as r = n - qb. This seems to work for all numbers because this explanation was for arbitrary N and B. Where does induction come in?
