Analysis I Group 9 handout 1. FOR FUN ONLY. NOT EXAMINABLE.
How do we know there exist n ∈ N with sin(n) arbitrarily close
to zero?
Several times we have asserted that there are integers n for which sin(n) is arbitrarily close
to 1 (or -1 or 0 or whatever). How do we know this? In this handout I’ll give an argument that
for any > 0, ∃n ∈ N such that | sin(n)| < .
First of all, convince yourself that | sin(x)| ≤ |x| (formal proof not required, a picture is
sufficient). Now remember that sin(x) = sin(x − 2qπ) for any q ∈ Z. Hence, for any n ∈ N,
q ∈ Z,
| sin(n)| = | sin(n − 2qπ)| ≤ |n − 2qπ|.
So, what we need to do is show that we can find n and q so that |n − 2qπ| < . You already
know that there are positive integers with | nq − 2π| < . However, you cannot just multiply
through by q to get what we need – a more cunning argument is needed. Let’s do the proof in
a more general setting.
Lemma Suppose that x > 0 and > 0. Then, there exist n, q ∈ N such that |n − qx| < .
Before we dive into the proof, recall that the integer part [y] of a positive real number y is
the largest integer that is at most y. For example [2.65] = 2, [e] = 2, [π] = 3, [4] = 4 etc. The
fractional part y − [y] is just the number that follows the decimal point, e.g. 2.65 − [2.65] = 0.65,
e − [e] = 0.71828 . . . and π − [π] = 0.14159 . . ..
Proof. Let N > −1 . Define ak = kx − [kx], 0 ≤ k ≤ N , to be the fractional part of kx. Notice
we have defined N + 1 terms here, each of which is an element of the half open interval [0, 1).
In particular, each one lies in one of the intervals
[0, 1/N ), [1/N, 2/N ), . . . , [(N − 1)/N, 1).
There are only N intervals here, so one of them must contain at least two of the ak . That is,
there are distinct integers k1 , k2 ∈ {0, 1, 2, . . . , N } and s ∈ {0, 1, 2, . . . , N − 1} such that
ak1 , ak2 ∈ [s/N, (s + 1)/N ).
Since both ak1 and ak2 lie in an interval of length 1/N , their difference must be less than
1/N . That is,
|ak1 − ak2 | < 1/N
(try to derive a contradiction for |ak1 − ak2 | ≥ 1/N ).
Inserting the definition of ak1 and ak2 ,
|(k1 x − [k1 x]) − (k2 x − [k2 x])| < 1/N.
Now, we rearrange to get
|x(k1 − k2 ) − ([k1 x] − [k2 x])| < 1/N.
Both of the bracketted terms are integers. We can assume that k1 > k2 . Thus, x(k1 − k2 ) > x.
If [k1 x] − [k2 x] is negative, then x(k1 − k2 ) − ([k1 x] − [k2 x]) ≥ x. Show that this possibility can
be ruled out by taking N > x−1 .
Hence, taking n = [k1 x] − [k2 x], q = k1 − k2 and noting that 1/N < , we see
|qx − n| < as required.
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