23 Jul 06
Originally posted by AThousandYoungBut the limit of (x^2/x) as x -> infinity is 2 (although it is still infinity divided by infinity).
How about the limit of f(x) = x/x as x => infinity? That gives you 1 person per room.
How did you decide infinity/infinity = 0, Jim?
The limit of (X/x^2) as x -> infinity is 0 (although again it is still infinity divided by infinity).
Both by l'Hospital's rule.
23 Jul 06
Originally posted by GastelThe L'Hospital's Rule that I know goes something like this:
But the limit of (x^2/x) as x -> infinity is 2 (although it is still infinity divided by infinity).
The limit of (X/x^2) as x -> infinity is 0 (although again it is still infinity divided by infinity).
Both by l'Hospital's rule.
lim f(x)/g(x) = lim f'(x)/g'(x) where f'(x) is the first derivative wrt x
So for your first example f(x) = x^2 and g(x) = x
Therefore f'(x) = 2x and g'(x) = 1
And the limit would be 2x/1 = 2x (!= 2 unless x = 1)
23 Jul 06
Originally posted by howardbradleyWhoa! My bad. You are absolutely correct, I don't know where my brain is.
The L'Hospital's Rule that I know goes something like this:
lim f(x)/g(x) = lim f'(x)/g'(x) where f'(x) is the first derivative wrt x
So for your first example f(x) = x^2 and g(x) = x
Therefore f'(x) = 2x and g'(x) = 1
And the limit would be 2x/1 = 2x (!= 2 unless x = 1)
23 Jul 06
6 edits
Originally posted by howardbradleyWell, you don't exactly need l'Hopital's rule to see
The L'Hospital's Rule that I know goes something like this:
lim f(x)/g(x) = lim f'(x)/g'(x) where f'(x) is the first derivative wrt x
So for your first example f(x) = x^2 and g(x) = x
Therefore f'(x) = 2x and g'(x) = 1
And the limit would be 2x/1 = 2x (!= 2 unless x = 1)
x^2/x = x
tends to infinity as x tends to infinity!
Besides all this talk of
infinity/infinity
and
something/0
is nonsense. "Infinity" is not a real number so you cannot do arbitrary operations on it. Similarly the set of real numbers form something called a field, and in a field division by zero cannot be done.
So the operations above don't exist, let alone have an answer.
If you really want to start talking about "infinity" and doing operations on "infinity" then you need to learn some set theory and start doing ordinal arithmetic or something. There you really can talk about "infinity + 1".
Search "set theory" or "ordinal arithmetic" in Wikipedia.
The approach above, like in
f(x) / g(x)
where f(x) and g(x) tend to infinity as x tends to infinity, is perfectly fine. But it has nothing to do with the spurious supposed quantity
infinity / infinity.
Instead it is merely some (separate) question in mathematical analysis to work out the answer to the limit of f(x)/g(x) as x tends to infinity.
I should finally point out that a phrase such as "x tends to infinity" does not mean "some quantity x gets closer to some other quantity infinity". Instead the *whole* phrase has a precise mathematical meaning that is quite different. (So don't analyse it using normal English).
23 Jul 06
Originally posted by SPMarsI know you don't :-) Nonetheless it had been introduced, and seemingly used to prove a result that was contrary to the obvious. Since L'Hosptital is not in a position to defend himself, I thought I'd have a go.
Well, you don't exactly need l'Hopital's rule to see
x^2/x = x
tends to infinity as x tends to infinity!
24 Jul 06
Originally posted by GastelBut the limit of (x^2/x) as x -> infinity is 2 (although it is still infinity divided by infinity).
But the limit of (x^2/x) as x -> infinity is 2 (although it is still infinity divided by infinity).
The limit of (X/x^2) as x -> infinity is 0 (although again it is still infinity divided by infinity).
Both by l'Hospital's rule.
No, it's infinity, not 2.
24 Jul 06
2 edits
Originally posted by aginisIf X is a set we let P(X) denote the power set of X, ie. the set of all subsets of X. The set R of real numbers is uncountable (and has cardinality aleph_1 = 2^(aleph_0) = |P(N)| = |R| ).
for all of you enjoying these infinity puzzles.
for any set A let T(A)= {B s.t. |B|=|A\B😏
prove that |T([b]R)| = 2^aleph
notation:
{ } is a set (sets are capitals)
|A| = cardinality of A
R is the set of real numbers (bold represents the basic sets, N,Z,Q,R)[/b]
So if B is a set the only way we can have |B| = |R/B| is if B is an uncountable subset of R for which R/B is uncountable too. So we want to know the cardinality of
T(R) := { B subset of R | B and R/B are uncountable }
a subset of P(R). Well P(R) is a disjoint union of 4 subsets:
T(R) and
U := { B subset of R | B is countable and R/B is uncountable } and
V := { B subset of R | B is uncountable and R/B is countable } and
W := { B subset of R | B is countable and R/B is countable }.
Now clearly W is empty, and U, V are in bijective correspondence (via complementation) and so have the same cardinality, that of the set
X = { B subset of R | B is countable}
of countable subsets of R. Now it's easy to show |X| = |R| since
|R|
24 Jul 06
8 edits
Sorry I typed this into notepad and it seems to have got truncated. Moreover I can't seem to complete the original details here...
Well anyhow you show the set of countable subsets of R has the same cardinality as R itself, and then it follows easily that |T(R)|=|2^R| or 2^aleph_1 if you like.
25 Jul 06
1 edit
Originally posted by SPMarsI'm not sure that you are correct in saying |X| = |R|
Sorry I typed this into notepad and it seems to have got truncated. Moreover I can't seem to complete the original details here...
Well anyhow you show the set of countable subsets of R has the same cardinality as R itself, and then it follows easily that |T(R)|=|2^R| or 2^aleph_1 if you like.
There is a better way to answer the question, although i get your idea.
the problem is that in our discrete math class we weren't allowed to use the idea that if
|A| + |B| = 2^aleph
|A| < 2^aleph
then |B| = 2^aleph
this is because we don't know whether there exists a set with cardinality x s.t. aleph < x < 2^aleph.
HINT: the easiest way to prove that a set A has cardinality x is to show
that.
x =< |A| =< x
i.e. find a set B subset of A s.t. |B| = x and
set C s.t. |C| = x and A is a subset of C.
25 Jul 06
7 edits
Originally posted by aginisWhat do you mean by aleph?
hmmmmm, there is a better way although i get your idea.
the problem is that in our discrete math class we weren't allowed to use the idea that if
|A| + |B| = 2^aleph
|A| < 2^aleph
then |B| = 2^aleph
this is because we don't know whether there exists a set with cardinality x s.t. aleph < x < 2^aleph.
HINT: the easiest way to prove that a set A has ...[text shortened]...
i.e. find a set B subset of A s.t. |B| = x and
set C s.t. |C| = x and A is a subset of C.
aleph_1 := 2^aleph_0 := |P(N)|
I assume?
To be honest I don't see the problem with what I've done:
We have a disjoint union P(R) = T(R) U Y where Y is a subset of P(R) that has cardinality |R|. The only possibility is that |T(R)| = |P(R)|, or am I missing something?
Even if there was a set x with
| R | < |x| < |P(R)|
then we can't have |T(R)|=|x| since then |P(R)|=|T(R) U Y| < |P(R)| which is a contradiction.
25 Jul 06
Originally posted by SPMarsyes it could be that |T(R)| = x and x + aleph = 2^aleph
What do you mean by aleph?
aleph_1 := 2^aleph_0 := |P(N)|
I assume?
To be honest I don't see the problem with what I've done:
We have P(R) = T(R) U Y where Y is a subset of P(R) that has cardinality |R|. The only possibility is that |T(R)| = |P(R)|, or am I missing something? Even if there was a set x with | R | < x
plus as i just edited the above post its not clear that |X|=aleph
(yes aleph is what you call aleph_1)
26 Jul 06
6 edits
Originally posted by aginisR is the set of real numbers.
I'm not sure that you are correct in saying |X| = |R|
N is the set of natural numbers.
X is the set of countable subsets of R (including the finite ones).
If C and D are sets let C^D denote the set of functions from D to C.
If C and D are sets then C < D means "there is an injection from C to D" and C=D means "there is a bijection from C to D".
We have
R < X < R^N = (2^N)^N = 2^(NxN) = 2^N = R.
The first injection is got from mapping the singleton sets into X and the second injection is got from mapping a countable subset to a function that lists its elements in increasing order (in the finite case we repeat the maximum element infinitely often).
Since we have R < X and X < R it follows from the Schroeder-Bernstein Theorem that X = R.