fmam3 said:
Suppose x_n = p_n / q_n \to x where (x_n) is a sequence of rational numbers and x > 0 is an irrational number. For contradiction, suppose that \lim_{n \to \infty} q_n \ne +\infty; that is \exists M > 0 such that \forall N_0 \in \mathbb{N}, \exists n > N_0 such that q_n \leq M.
No, as pointed out by Elucidus, if a sequence does not tend to +infinity, it does not mean that it's upper bounded. But instead, there's some (at least 1) sub-sequence in it that is bounded above. And you'll work on that sub-sequence, constructing from the negation of:
\forall N > 0, \exists k \in \mathbb{N} : n > k \Rightarrow a_n > N,
which should be:
\exists N > 0, \forall k \in \mathbb{N}, \exists n > k : a_n \leq N
Consider a subsequence (x_{n_k}) = (p_{n_k} / q_{n_k}) constructed as follows. Set n_k = N_0 + k and clearly, by this choice, we have n_1 < n_2 < \ldots. And hence, by induction, we can have a subsequence (x_{n_k}) = (p_{n_k} / q_{n_k}). But since q_n \leq M, it follows that q_{n_k} \leq M for \forall k \in \mathbb{N}.
Ok, fine. But remember that it should be a sub-sequence of a sub-sequence (where q
n is upper-bounded) instead.
Furthermore, since (x_n) is convergent, the subsequence (x_{n_k}) is convergent and so it is bounded and thus we can pick some k_1, k_2 \in \mathbb{N} such that x_{n_k} = p_{n_k} / q_{n_k} \in (k_1, k_2). Now, see that k_1 < \frac{p_{n_k}}{q_{n_k}} < k_2, rearrange, k_1 \cdot q_{n_k} < p_{n_k} < k_2 \cdot q_{n_k} \leq k_2 \cdot M. And since p_{n_k}'s are integers, it implies that although there are infinitely many p_{n_k}'s, they can only take on finite number of values (i.e. some of the values repeat themselves).
This part is fine. Although in this inequality, you may omit the k
1q
nk < p
nk part, since it does not help much, more over q
nk varies, and so does k
1q
nk. Putting it there is okay, but no use, so we may just drop it. :)
k_1 \cdot q_{n_k} < p_{n_k} < k_2 \cdot q_{n_k} \leq k_2 \cdot M
Now consider \inf\{|x_{n_k} - x| : k \in \mathbb{N}\} = \inf \{|p_{n_k} / q_{n_k} - x| : k \in \mathbb{N}\}.
No, we don't find the inf of it, instead, we find the min. Because that set is finite, so there exists the minimum.
The main different between the inf, and the min of a set is that. The inf does
not contain in the set, while the min does.
And if you are using inf in this case, you cannot prove that its value is not 0.
I gave you an example in some earlier post. Consider the set:
A = \left\{ \frac{1}{n}, n \in \mathbb{N} ^ {*} \right\}
0 is nowhere to be in A. But the infimum of A is indeed 0. :)
And since 0 is not in A, so this set does not have minimum value.
Since q_{n_k}'s are integers and are bounded, so it can only take on finite number of values; similarly, we had shown that p_{n_k}'s are integers and can only take a finite number of values. Thus, there are only finite number of values for x_{n_k} for \forall k \in \mathbb{N} and all of these values are rationals but x is irrational. Thus it follows that \inf\{|x_{n_k} - x| : k \in \mathbb{N}\} \ne 0. It thus follows that 0 \ne \inf\{|x_{n_k} - x| : k \in \mathbb{N}\} \leq \lim_{k \to \infty} |x_{n_k} - x| or that \lim x_{n_k} \ne x, but since we assumed \lim x_n = x --- contradiction, if a sequence converges, every subsequence converges to the same limit.
Fine. :D
----------------------
Now, there's only one little problem left is to show that:
If a sequence does not tend to +infinity, then there's some sub-sequence of it, which is bounded above. Let's see how far you can get.
fmam3 said:
Sorry for the late reply --- I was busy yesterday so couldn't get around to go on this forum. But thank you for your great responses and help!
----------------------
Comments:
(2) In the proof below, I'd modified your (k, k+1) for some k \in \mathbb{N} argument. The reason is the following. We have that p_{n_k} / q_{n_k} are rationals for every k but it is very possible that it takes on a value, say, p_{n_k} / q_{n_k} = 1000; after all, \frac{1000}{1} is a rational. Then in that case, if we still write k < p_{n_k} / q_{n_k} = 1000 < k + 1, with strict inequality, I don't think such a k \in \mathbb{N} exists. That is, we can take k = 999 but k + 1 = 1000 which means the last inequality must be a "\leq".
Yes, you can, but remember that x is irrational, so x cannot be either k, or k + 1.
And moreover, since your sequence tends to x, i.e, it'll be "close" enough to x when n is large enough. So if it happens that at some
n0, a
n0 = k, or a
n0 = k + 1, it just means that
n0 is not "large" enough.
I think it does no harm to generalize it to some k_1, k_2 \in \mathbb{N} and write the interval as (k_1, k_2). What do you think?
Yes, of course, you can. It's all up to you, as long as you can prove that the number of choices for p
n is finite.
