# Irrational numbers in infinite list of integers

1. Mar 13, 2010

### GunnaSix

Is it safe to assume that the absolute value of sin x is greater than zero for all positive integer values of x? I have no real experience in number theory, and I don't know if you can say that there are no irrational numbers in an infinite list of integers.

2. Mar 13, 2010

### Hurkyl

Staff Emeritus
Can you convince yourself that this conjecture is equivalent to
Every integer is rational​

3. Mar 13, 2010

### Tinyboss

If sin(x)=0 for positive x, then x=k*pi for some positive integer k. If x is an integer, then x/k=pi should raise a red flag.

4. Mar 14, 2010

### ramsey2879

are you talking of integer degrees or radians?

5. Mar 14, 2010

### GunnaSix

Integer radians. Sorry if this was a super obvious question. I guess I was just thinking too much into it.

6. Mar 23, 2010

### Eynstone

If you mean to ask whether the intervals [2npi,(2n+1)pi) (where sinx >0) contain integers, the answer is yes ( as pi>1).

7. Mar 24, 2010

### ramsey2879

SineX = 0 only where X = n*pi radians. But n*pi must be an irrational number or 0. Thus SineX is either greater than or less than 0 for any integer angle n in radians where n>0. I think this is what the poser was asking.

Last edited: Mar 24, 2010
8. Mar 25, 2010

### HallsofIvy

Yes, since any integer, n, can be written as the fraction n/1, all integers are rational numbers. There are no irrational integers. It is true that sin(n) is never 0 for any integer n- but that does NOT mean "greater than 0". sin(4) is negative.

9. Mar 25, 2010

### ramsey2879

While I used the langauge "Sine x" is "greater than or less than 0", the poser asked whether "the absolute value of Sine x is greater than 0 for all integer values of x.". I think you had my language in mind when you overlook the "absolute value" part of the poser's question.