Proving an inequality - |sin n|>c

[khari]

Homework Statement

Is it true or false that \exists c>0 s.t. \forall n \in N, |sin n|> c, Justify your answer

Homework Equations

The Attempt at a Solution

I figure it is true, because n != (k-1)\pi \forall n,k \in N, so |sin n|>0 \forall n
It seems fairly obvious to me that if |sin n|>0 then there are an infinite number of possible values I can assign for c and have it be between 0 and |sin n|. Is this self evident enough to simply state? Or is there a way to prove this formally? Or am I simply wrong?

 

[bel]

If the Natural Numbers are defined as the set of positive integers, as opposed to the definition where the Natural Numbers are the set of non-negative integers, since no natural number is a integer multiple of 2\pi, and |sin(n)| only equals to zero when n= 2k\pi, and is between in the range (0, 1] everywhere else, it is true.

 

[khari]

Yeah, that's basically where I'd gotten to. My problem is that I'm not sure if it can be simply stated like that or if there's a more formal way to prove it. Even if it is obvious enough to state, I'd still like to see a formal proof, just because I need to get as familiar with proofs as possible, because I'm terrible at them.

 

[bel]

No, there isn't a particular "formal" way to do proofs, (excluding those types already formalised, such as mathematical induction proofs,) other than transforming most of the words into symbols, which you seem quite adept at doing already and adding "Q.E.D." at the end.

 

[Dick]

If the Natural Numbers are defined as the set of positive integers, as opposed to the definition where the Natural Numbers are the set of non-negative integers, since no natural number is a integer multiple of 2\pi, and sin(n) only equals to zero when n= 2k\pi, and is between in the range (0, 1] everywhere else, it is true.

Just because a sequence of numbers is in the range (0,1] doesn't mean it has a positive lower bound. That's a long ways from a proof.

 

[bel]

Sorry for leaving out the absolute signs. Added, it does, by the axiom of the completeness of the reals, if a set has a lower bound, it has a maximum lower bound which is real.

 

[Dick]

I don't think you left anything out. But why can't the lower bound be zero?

 

[bel]

This is because the the argument of the sine function in question can never be multiples of two pi, which are the only arguments for which the function can be zero, and it can't because all natural numbers are integers and pi is an irrational number, and any integer times two (which is an integer) times an irrational is an irrational, which the argument would never be, hence the set is open on the lower bound.

 

[Dick]

I accept that the argument is never a multiple of 2pi. And so |sin(n)| is never zero. That's easy. What I don't see is why that shows it has a POSITIVE lower bound. I don't think it does. And just because I can't prove that just now doesn't convince me that it is otherwise.

 

[bel]

The maximum lower bound, as defined by the axiom of the completeness of the reals, is zero in this case, hence the lower bound is not negative. The lower bound is zero, but because the set is opened, the function is never zero, given the arguments.

 

[Dick]

Now you are making no sense whatsoever. The OP wants to show or disprove that the lower bound is positive, not zero.

 

[bel]

I don't really see what's wrong with my argument, so you would probably have to point it out to me. If I missed anything in the statement I first gave, it was the citing of the axiom of the completeness of the reals (and the absolute sign, which I later added), once applied, the proof is complete.

 

[Dick]

Ok. The sequence 1/n for n>=1, n integer is in (0,1]. The reals are indeed complete so it has a limit. The limit is zero. Zero is not in (0,1] and is certainly not positive. What makes you think |sin(n)| is different?

 

[bel]

Well, for one thing, the sequence |sin(n)| does not converge, when you start taking limits to infinity, you can't apply the axiom since infinity is not a number. It's the same as asking why some infinite series of rational numbers converge to irrational numbers. We're not taking limits here. If you taking any number n (n an element of the natural numbers) and apply it to 1/, you will never get to zero, but zero is the maximum upper bound, that is guarenteed by the openess of the set on the side of the zero and the axiom aforementioned.

 

[Dick]

Maybe I should concentrate on trying to figure out why the lower bound is in fact zero (which I suspect) rather try unsuccessfully to convince you that you don't have a proof that the lower bound is positive.

 

[bel]

The maximum lower bound is zero, that was guarenteed by the axiom aforementioned, but the values are never zero, hence all values are positive.

 

[Dick]

I know that! Why does that show there is a c>0 such that |sin(n)|>c. We are going in circles.

 

[bel]

I sort of feel we are as well, let's hope someone will come in and say something more constructive then either of us have had to contribute. By the way, since |sin(n)|>0, the openess of the set means that there is always a neighbourhood (i.e., another open set) smaller than |sin(n)| but bigger than zero (we have identified our biggest open set in the given set, and the biggest open set always contain all other open sets). We can choose our "c" from that neighbourhood.

 

[khari]

Well it's been interesting for me so far, and has given me a little doubt about my conclusion or at least that my statement of my conclusion is quite lacking.

 

[bel]

Hahah, I can't say that was all too constructive =), but I'm glad it was interesting for you all the same.

 

[khari]

Well, it did correct a misunderstanding I had about the completeness theorem, which I had added to my explanation after I posted this thread, but before you brought it up. I had it stated that because |sin n| > 0, it must have a positive lower bound, which is apparently incorrect, because it can be 0 as well.

 

[Dick]

I sort of feel we are as well, let's hope someone will come in and say something more constructive then either of us have had to contribute. By the way, since |sin(n)|>0, the openess of the set means that there is always a neighbourhood (i.e., another open set) smaller than |sin(n)| but bigger than zero (we have identified our biggest open set in the given set, and the biggest open set always contain all other open sets). We can choose our "c" from that neighbourhood.

I promised myself to stop responding to these nonsequiturs but the set of values for |sin(n)| for n an integer is NOT OPEN. How can a countable set in the reals be OPEN? khari, keep doubting your conclusion. It's almost certainly untrue.

 

[Dick]

Well, it did correct a misunderstanding I had about the completeness theorem, which I had added to my explanation after I posted this thread, but before you brought it up. I had it stated that because |sin n| > 0, it must have a positive lower bound, which is apparently incorrect, because it can be 0 as well.

I think that's the whole point.

 

[khari]

I think that's the whole point.

It sure does seem to be. I've much to think about, and not much time to think it in. I'm going to check back with this thread in the morning to see if anything conclusive has turned up. Otherwise I guess I'm going to hand it my answer as is, confused or not. Thankfully this assignment is just for bonus marks, and not many at that.

 

[Dick]

It sure does seem to be. I've much to think about, and not much time to think it in. I'm going to check back with this thread in the morning to see if anything conclusive has turned up. Otherwise I guess I'm going to hand it my answer as is, confused or not. Thankfully this assignment is just for bonus marks, and not many at that.

I wouldn't hand in a bad proof if you knew it had a hole in it. When I was teaching stuff like that made me think the student didn't even know wrong from right. Better to just say, I don't know, than to make a grader read through a bunch of gibberish and then hate you. Really.

 

[bel]

That's very interesting. What I was thinking was that you could take any integer, and multiply it by two, then by pi, then subject that result to the floor function, and then find 2k(pi)-floor(2k(pi)) and it could always be lower and it all depends on the digits in pi, it's sort of like asking if there is always a longer string of zeroes in the digits of pi. So I guess I couldn't prove it this way and I did have some implicit assumption I was unaware of, thanks for shewing this to me though, or I would have thought I was right.
Nevertheless, it is true that the lower bound is never zero, since 2k(pi)-floor(2k(pi)) is never zero, and though the set (of the range of the function |sin(n)|) may not be open, it would not be closed at zero, but at some number above zero, would that then guarantee that there is such a "c" value then?
And no, I don't really think this is non-sequitur, though I agree it seems somewhat circular.
I don't understand how the set could ever be closed at or below zero though, it is a set of the range of an absolute function, isn't it?

 

[Dick]

If anyone still cares, the proof is actually pretty easy. Let a be any irrational number. First show that for any e>0, there are integers m and n such that m*a+n<e. To do this divide [0,1] into a finite number of intervals of size <e. Now m*a+n=m'*a+n' iff m=m' and n=n'. Furthermore for any m there is an n such that m*a+n is in [0,1]. That means that there are an infinite number of different numbers of the form m*a+n in [0,1]. Since there are only a finite number of epsilon intervals, one interval must contain two such numbers (pigeonhole). Let M*a+N be the difference of those two numbers. |M*a+N|<e.

This tells you that numbers of the form m*a+n are in fact, dense in R. I'm leaving the details and application of this to the sin(n) problem as an exercise.

 

[Dick]

Nobody does care, right? I'm so hard of working my fingers to the bone trying to prove something and then nobody cares because the homework deadline has expired. Boo, hoo.

 

[khari]

Sorry I haven't checked back since, I got caught up in an Optics lab and forgot. Yes, the proof you gave is essentially the same as was given given in class:) I don't think anyone got it right, or if they did I don't think anyone proved it. Thanks for your help. And yes, I am interested in the proof whether the deadline has passed or not. Thanks again.