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I am a software engineer at Beckman Couleter, Inc. I enjoy reading, playing piano, and dancing.


11h
awarded  Caucus
Dec
16
comment Is Cantor's diagonal argument dependent on the base used?
@IttayWeiss Hence why I said "imagine that it's somehow possible"--which of course it isn't. The point was to give a general, non-technical idea of what "almost all" means, not to provide a superior technical definition.
Dec
16
comment Is Cantor's diagonal argument dependent on the base used?
@ZachEffman Ah, true. And in any case I should have said "probability that you will select a number from the list" rather than "select a rational."
Dec
16
comment Is Cantor's diagonal argument dependent on the base used?
@AkashC The above explanation is correct but pretty technical. "Almost all" means, essentially, that if you imagine that it's somehow possible to randomly select a real number within a certain range such that every real in that range has an equal probability of being selected, then the total probability that you will select a rational number is 0. (See the Wiki page on "Almost surely," which is a similar concept in probability; the first example given, regarding throwing a dart a a unit square, is somewhat similar to my example of picking a random number within a range on the continuum.)
Dec
15
comment Don't see the point of the Fundamental Theorem of Calculus.
@PaulDraper I agree with "not obviously related," but the better word there is "unexpected." For this reason, I prefer heropup's answer. Anecdotally, having explained calculus to a couple of people who "hate math" and never took calculus, heropup's approach of simply not treating integration and differentiation as related, then explaining that the FTC demonstrates a relationship, tends to produce exactly the kind of "aha!" moment that it ought to (and which I didn't experience when I first learned calculus).
Dec
15
comment Don't see the point of the Fundamental Theorem of Calculus.
@PaulDraper I understand Kundor's point; I just don't think that it's helpful to assert that related operations are "unrelated," despite knowing that the relationship must be proven. There is no "bizarre reason" for the symbol sharing; the symbols weren't universally settled upon until well after the FTC was established. I'm just saying that the claims here are both wrong and confusing, which is the worst of both worlds.
Dec
12
comment Don't see the point of the Fundamental Theorem of Calculus.
I'm surprised there's so little sympathy here for your predicament. I had exactly the same confusion, and I had quite a good calculus teacher. (@BrunoJoyal: Yes, really.) I can't imagine this is an uncommon question. (I suppose your original phrasings are a bit rantish, but you've improved the post dramatically since then.)
Dec
12
comment Don't see the point of the Fundamental Theorem of Calculus.
@IlmariKaronen ...and yet the point remains that they're not unrelated. Note that without bounds, the antiderivative sign also represents the "indefinite integral"--so of course it's related to the definite integral!
Dec
6
awarded  Yearling
Dec
1
comment Is there a shape with infinite area but finite perimeter?
Excellent observation! I suppose, then, that the "real" question would be whether there exists a closed curve of finite length such that both the interior and the exterior of the curve have infinite area--which would appear to be prohibited by the isoperimetric inequality, as stated in Travis's answer.
Nov
7
revised Is irrational times rational always irrational?
Minor clarifications/capitalization/grammar fixes, and removed unnecessary material
Nov
7
suggested approved edit on Is irrational times rational always irrational?
Oct
3
comment If $A$ is dense in $S$ and $S$ is dense in $T$ , then $A$ is dense in $T$
Thanks @tohecz.
Oct
3
comment If $A$ is dense in $S$ and $S$ is dense in $T$ , then $A$ is dense in $T$
...any idea why the subset statement isn't rendered in that last comment?
Oct
3
comment If $A$ is dense in $S$ and $S$ is dense in $T$ , then $A$ is dense in $T$
@tohecz $\epsilon - d(t,s)$ may look complicated, but the very simple intuition behind the proof is that points in $T$ are "close to" points in $A$ because they're also "close to" points in $S$, and points in $S$ are "close to" points in $A$. I.e., in metric spaces, "closeness" is transitive. This is easy to visualize and reason about. I do like the simplicity and generality of the $S \bar \subseteq A \bar$ statement, and I appreciate that the proof works in non-metrizable topological spaces, but I also like being able to visualize statements about metric spaces.
Oct
2
comment If $A$ is dense in $S$ and $S$ is dense in $T$ , then $A$ is dense in $T$
@tohecz What do you mean by "doesn't catch the points at all"?
Oct
2
comment If $A$ is dense in $S$ and $S$ is dense in $T$ , then $A$ is dense in $T$
I like this because it seems (to me) more direct than the proof based on the subset/closure definition of density.
Sep
24
awarded  Autobiographer
Sep
19
revised Associative law is not self evident
Two guesses were made; clarified that the *first* guess is the correct one
Sep
19
comment Associative law is not self evident
@Joe Ah yes, the only-one-answer "shortcoming." That is a fight you will not win. (To give the policy its due, I do think it's helped keep SE sites remain Q&A focused rather than devolving into discussion forums.)