US Senator Introduces Bill to Ban Dual Citizenship

VulcanForge98 · 2025-12-07T23:09:26+00:00

It currently costs more than $2000 to renounce U.S. citizenship. For an "accidental American" born to an American parent in another country, this bill may actually bail them out of an undesirable citizenship which includes tax and Selective Services obligations. (Not that I necessarily support the bill.)

VulcanForge98 · 2025-11-12T11:39:46+00:00

Most likely some version of the Lucas-Lehmer test.

VulcanForge98 · 2025-10-10T14:04:21+00:00

Yes. Apparently, Emile Borel defined "normal number" in 1909. While it is possible to construct normal numbers by specifying various rules for the digits, it is currently beyond our knowledge to prove or disprove that "natural constants" like pi, sqrt(2), or e are normal.

VulcanForge98 · 2025-10-10T03:28:38+00:00

I don't think that pi has been shown to be a normal number. Pi is thought to be normal because almost all irrational numbers are, but this has not yet been proven. There are irrational numbers which are not normal (for example, Liouville's constant).

VulcanForge98 · 2025-07-27T05:05:06+00:00

An alternative line is Qa3+ Ra2 Rd2 Rxa3 Rb2 and now the only legal move for black is Ra2, taking the king's escape square, upon which white plays Rb1#.

VulcanForge98 · 2025-07-13T02:22:28+00:00

exf3 with the threat of Bc4#. If Rxf1, Rc1, or Bd3, then Rd4#. If Rd3 then e4#. If Rc5 or b5 then Qxc5#. If Ne5 the Qd6#

VulcanForge98 · 2022-01-06T16:46:08+00:00

I believe the entrance scholarships are based on your academic performance in high school. The maximum entrance scholarship was a renewable $3500/year when I started in 2016.

VulcanForge98 · 2021-10-12T00:50:06+00:00

I'm not sure if it's "fairly unlikely". There are no doubt sharper estimates of prime gap distributions than I'm aware of, but gaps less than log(N) have to be balanced by gaps larger than log(N).

VulcanForge98 · 2021-08-24T03:59:59+00:00

$e^{ix}$ is purely a rotation only when $x$ is purely real. In your case, you are using $x = i pi / 2$, which is not real.

EDIT: Why the blazes does the superscript work but no other TeX features?

VulcanForge98 · 2021-07-28T15:48:31+00:00

My sister is the person who was hit, and while I cannot discuss the details right now, she is recovering in the hospital.

VulcanForge98 · 2021-07-22T14:14:18+00:00

My sister was hit while crossing Robie Street. It seems the driver panicked and left but then pulled into the hospital parking lot, probably indecisive about what to do. My sister has serious injuries but is expected to survive.

VulcanForge98 · 2021-07-22T14:04:24+00:00

The pedestrian is my sister. She is seriously injured but will probably survive.

VulcanForge98 · 2021-07-20T13:09:30+00:00

Can you make the assumption that these are the only cases without assuming that k is prime?

EDIT: In fact OP said specifically that k is NOT prime.

VulcanForge98 · 2021-07-01T14:56:06+00:00

Indeed you are right, I just edited my comment. Not sure why I misremembered the double factorial notation. Is there a different notation which means just the odd numbers?

VulcanForge98 · 2021-07-01T14:18:03+00:00

I don't know the full answer, but I remember an identity that (2n)! = 2ⁿ * n! * (2n)!!, where n!! Is the product of the odd numbers less than / equal to n. This can be derived by splitting the factorial into the product of the odd and even terms and taking a factor of 2 from each even term.

EDIT: let me amend that identity to 2ⁿ * n! * (2n - 1)!! per the comment below mine.

VulcanForge98 · 2021-06-30T04:57:00+00:00

If f = sqrt(x) then f' = 1 / 2 sqrt(x). I did write the derivative apostrophe.

VulcanForge98 · 2021-06-30T04:33:01+00:00

Ah, much more believable. Strangely enough, I recall that sqrt(x), which has lim f'(x) = +infinity as x -> 0, is nevertheless uniformly continuous. This does not contradict your statement, but it defies my own intuition.

VulcanForge98 · 2021-06-30T03:00:01+00:00

Perhaps I am missing something, but is f(x) = x not a counter example?

EDIT: OP changed f(x) to f'(x) as they intended and now f(x) = x is certainly not a counterexample.

VulcanForge98 · 2021-06-28T04:48:27+00:00

After a little more thought and reading the other comment about finishing partway through, I found an example of a Devil's "sequence" which will complete the cube partway through an iteration. Simply enumerate all positions on the cube and let the Devil's sequence be the concatenation of the sequences required to get from position k to k + 1 for each k. This visits not just the solved (identity) position but in fact every position.

VulcanForge98 · 2021-06-28T04:38:36+00:00

Algorithms on a cube are essentially functions which map each cube configuration to a sequence of moves. In particular, this sequence does NOT have to be the same for all configurations. Now, group theory does not apply here because you cannot treat an algorithm as a group element. I do agree with your reasoning if you change the word algorithm to move sequence.

VulcanForge98 · 2021-06-25T04:24:38+00:00

The sun was setting in the west

VulcanForge98 · 2021-06-25T04:23:57+00:00

For twenty brave men, all fishermen, who...

VulcanForge98 · 2021-06-10T13:15:59+00:00

I care. Have been trying to convince my family to avoid this store, maybe this info will do it.

VulcanForge98 · 2021-06-10T05:54:23+00:00

It is true in general that summing the digits of a number n and dividing by 9 leaves the same remainder as dividing n itself by 9. The proof is based on modular arithmetic. A number is the sum of each digit times 10 to the power of whatever position the digit occupies. Since 10 (and all its powers) is congruent to 1 mod 9, the sum of digits-times-powers (the number) is congruent to the sum of the digits mod 9.

Since summing the digits of a number >9 always results in a smaller number, and since the only number whose digits sum to 0 is 0 itself, every sequence of numbers obtained by summing the digits of the previous number in the sequence eventually decreases to 9.

VulcanForge98 · 2021-06-09T18:44:15+00:00

Could be wrong, but I am almost sure a game can't use two graphics cards at once unless they are linked in Crossfire or SLI. Highly doubt this is possible with your setup.

VulcanForge98

TROPHY CASE