This exercise was given out to students at my university, and almost no one was able to solve it

This exercise was given out to students at my university, and almost no one was able to solve it.
Even the solution given by the assistant was wrong.
I have a solution by now, but I'm wondering what Veeky Forums can come up with.

If one of the limits (consider q_n) were defined, they would have to be integers:
Just take epsionm you can pull out q_n of the limit. The limit equals
q_m^(-1) \lim_{n\to infty} p_n
The limit \lim_{n\to infty} p_n exists and is thus again some integer.

Maybe I use some stronger theorem here and don't notice it, but I have the feeling my steps are true.

What you are saying about integer sequences is true, but you are forgetting that if a sequence does not diverge towards infinity, it still does not have to converge.

I proceed by contradiction.
Case 1: suppose pn converges to k and qn converges to c

First, as the natural numbers are closed (topological definition of closed), so k and c are natural numbers.

Then the limit as pn/qn goes to infinity would be k/c which is a rational number, which is a contradiction.

Case 2: pn converges to k, qn diverges:
Then pn is bounded from above and thus the limit to infinity of pn/qn = 0 which is a rational number and thus a contradiction.

Case 3: pn diverges and qn converges. Then qn is bounded from above and thus the limit to infinity of pn/qn diverges, which is a contradiction.

Assume that is not the case, then at lest one of the sequences has an finite accumulation point. If p_n has an finite accumulation point, then that accumulation point has to equal to some natural N, so thekre is a subsequance p_{n_k}=N, but then if the limit of p_n/q_n exists it is an rational number, since lim p_n/q_n= lim p_{n_k}/q_{n_k} (if q_nk is constant fo a while the rational number is positive, if it is non constant then the limit is 0). For the other case you argue exactly the same way, but then you know that the limit cannot be zero.

see Looks good, but I find your argument that lim p_{n_k}/q_{n_k} is rational not convincing (it is true though).

if q_{n_k} is not constant after a while, then p_{n_k}/q_{n_k} =0, since we assumed that the limit p_n/q_n exists (oscillation of q_{n_k} would imply that the limit of p_n/q_n does not exist).

I meant "oscillation of p_{n_k}" not "oscillation of q_{n_k}"

Ah that makes a good argument. One can write q_nk = p_nk * (q_nk / p_nk) where the RHS converges towards an irrational number, but the LHS is always natural.

Jesus, the first one is correct, I'm sorry.

If pn is bounded above while qn is not then pn/qn decreases to zero.
If qn is bounded while pn is not, then the sequence pn/qn goes to infinity.
If both are bounded, then the sequence is not cauchy.

Ok, so here is one for you to thin about. pick any three 3 digit numbers. Write down a matrix A such that its rows are the digits of those numbers. Show that det A is divisible by 17.

You can compute a value of epsilon such that every (hypothetical) value of p_n, or q_n below a natural number M is eliminated from consideration. More importantly, you can pick M to be arbitrarily large, and you will always find a sufficiently small epsilon. Label the irrational number by x. What you need to do is consider the M-1 natural numbers below M. Because the set is finite, you can find the natural number A between 1 and M such that |x - B/A|, given the appropriate choice of B from the integers, is minimized. By choosing epsilon to be smaller than the minimized value, then it is impossible for any choice of a, from 1 to M, and b, from the integers, to satisfy the condition |x - b/a|N(epsilon(M)), p_n is larger than M, and q_n is larger than M. M is arbitrarily large, so the limit of p_n is infinity, and q_n is infinity.

Am I missing something? Clearly 100, 110, 111 don't work.

Yes I forgot to add that all three should be divisible by 17.

I mean, because x can't be 0, p_n would have to approach infinity because q_n is approaching infinity, and the ratio of p_n/q_n must approach x, a non-zero number.

>If both are bounded
i meant pn/qn is an eventually constant sequence, as all of its terms will be drawn from a finite set with an order less than or equal to the order of the cartesian product of the set of terms of pn and qn.

I don't see how you get that q_n is larger than M so easily. Seems good otherwise.

What about sequences that are not bounded but go to infinity, e.g. 1,2,1,3,1,4,1,5,.. ?

>sequences that are not bounded but don't go to infinity
is what i meant.

Take x to be the irrational number. The function |x - b/a|, where a is from 1 to M, and b is the variable, is bounded below by 0. Because the function diverges to infinity for large |b|, we know its infimum coincides with its minimum, therefore the minimum b can be chosen from the integers.We can repeat the process for all a between 1 and M-1, so we can create a set of these minimum values, then choose the minimum value from this set.This is possible because given a set of M-1 real numbers, you can always find the minimum real number, and divide that real number by 2 (or whatever). This small number can be our epsilon, and, because the sequence converges, we can find N such that n>N allows |x - p_n/q_n|

The set of minimum values refers to the set of |x - b/a|, where each a between 1 and M is represented (once), not the b terms themselves.

Mutiply the first column by hundred, the second by 10. add the second and third columns to the first one, we get that the determinant of the new matrix is 1000 times the first one, but as the first column in the new matrix is those three three digit numbers, the determinant of the new matrix is also an integer multiple of 17. 1000 is not divisble by 17, hence the determinant of the original matrix has to be divisible by 17.

Yeah forgot about those.
If one them isn't bounded then it's either zero, infinity or indeterminate.

doesnt this just follow from R being a banach space?

Let pn = qn?

I see, I mixed up the roles of a and b. I wasn't sure wether (in this case) p_n getting large followed by your first argument or not.

Nevermind, overlooked the irrational condition.

Scrathc hat.
Just take qn to be bounded and pn to be unbounded, and the limit to be irrational which is a non zero number, giving us a straightforward contradiction. For the other case where qn is unbounded and pn is not just invert the fraction, an operation which makes sense as none of them are zero.

Assume that [math](q_n)[/math] does not go to infinity. Then, we can extract a subsequence [math](q_{n_k})[/math] that is bounded. Since [math](p_{n_k}/q_{n_k})[/math] is convergent (and therefore bounded), it follows that [math](p_{n_k})[/math] is also bounded. Hence, both subsequences assume a finite number of values (as they are bounded and integer-valued), hence we can further extract and find constant subsequences [math](p_{n_k'})[/math] and [math](q_{n_k'})[/math] (by the pigeonhole principle if you will).
But [math](p_{n_k'}/q_{n_k'})[/math] converges to the same limit as [math]p_n/q_n[/math], hence the original limit is rational, contradiction.

If I'm not missing anything, you have now shown that both of them must be unbounded and neither can be bounded.

Good proof. To finish off, this argument used on the reciprocal sequence [math]q_n / p_n[/math] shows that [math]p_n[/math] also goes to infinity.

Simple problem and I'm not even a mathfag.
Just do this but don't be sloppy and cover all the cases. E.g. it can't be true either that Pn or Qn is bounded but not well defined (while the other is) as the limit of the quotient would not be well defined either, going against our assumption.

>not even a mathfag
I can tell because you think a proof by exhaustion is acceptable here.

Woop die doo then also show that it's possible the quotient can exist and be irrational when the two limits diverge. Not difficult either, you can literally pull an example out of your ass.

If math majors have troubles with this like OP says I really don't know what you fucks are doing all day.

Take your choice of proof technique OP. Similarly, straightforward, or check. Pic related is how real math is done.

so this is what peak performance looks like, huh
DAMN