# Proving an inequality - |sin n|>c

1. Sep 27, 2007

### khari

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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?

2. Sep 27, 2007

### 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.

Last edited: Sep 27, 2007
3. Sep 27, 2007

### 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.

4. Sep 27, 2007

### 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.

5. Sep 27, 2007

### Dick

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.

6. Sep 27, 2007

### 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.

7. Sep 27, 2007

### Dick

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

8. Sep 27, 2007

### 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.

9. Sep 27, 2007

### 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.

10. Sep 27, 2007

### 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.

11. Sep 27, 2007

### Dick

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

Last edited: Sep 27, 2007
12. Sep 27, 2007

### 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.

Last edited: Sep 27, 2007
13. Sep 27, 2007

### 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?

14. Sep 27, 2007

### 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.

Last edited: Sep 27, 2007
15. Sep 27, 2007

### 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.

16. Sep 27, 2007

### 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.

17. Sep 27, 2007

### Dick

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

18. Sep 27, 2007

### 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.

19. Sep 27, 2007

### 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.

20. Sep 27, 2007

### bel

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