Hey director.
This seems like a pointless question (no offence), but you could use the fact that [n,n+1] = (n,n+1) + {n} + {n+1} which implies (n,n+1) = [n,n+1] \ {n} or {n+1}.
I have the following question in my textbook, it appears at the end of a section that talks about Mathematical Induction:
Use an induction argument to show that for each natural number n, the interval (n, n+1) fails to contain any natural number.
Not exactly sure how to structure the proof.
Since this is induction, I think I'm going to have to deal with two consecutive intervals. Interval k and interval k+1.
I show that (1, 2) intersected with N is empty then I assume the interval k contains no natural numbers and show that interval k+1 also contains no natural numbers? I don't know if this makes sense to me.
Maybe I'm looking at this from the wrong angle. Are only the end points the inductive ingredients in the proof?
Thanks
Base case: n = 0
(0,1) does not contain any natural number.
Recall that (0,1) = {x in R: 0 < x < 1}
For (0,1) to contain a natural number, we must have A = N∩(0,1) ≠ Ø. If we proceed by way of contradiction, we suppose that A is non-empty. By the well-ordering theorem of natural numbers, A must contain a least element, k.
We then have 0 < k < 1. Since 0 < k, we must have k = S(k'), for some natural number k', that is k = k'+1. Hence k = k'+1 < 1, implies k' < 0, a contradiction, as 0 is the smallest natural number by definition.
Inductive step:
Suppose that for n = m ≥ 0 , we have: (m,m+1) contains no natural numbers, and again suppose by way of contradiction that (m+1,m+2) DOES. Then there is some natural number r with m+1 < r < m+2.
Hence m < r-1 < m+1, so r - 1 cannot be a natural number, which means we MUST have r = 0 (for if r = S(k) for some natural number k, then r = k+1, and thus r -1 = (k+1) - 1 = k, a natural number). But r > m+1 > m ≥ 0, and r cannot both be greater than AND equal to 0. Thus there is no such r, and we are finished.