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May 20[edit]

Anti-Hilbert problems?[edit]

If you search Google for "anti-Hilbert problems", you find many references saying that John Horton Conway had a set of "anti-Hilbert problems (open questions whose pursuit should emphatically not drive the future of mathematical research)" and that one of them is the question of whether Beggar-My-Neighbour always terminates.

My question is, what are the others? SartorialBenefit (talk) 01:58, 20 May 2012 (UTC)[reply]

Nice idea, this list! I suspect it could be considerably longer than Hilbert's; maybe natural numbers are not enough ;) In any case, to avoid confusion, I would number them by negative numbers. Also, here's a possible item: how many Smith_numbers are there? --pm a 10:07, 20 May 2012 (UTC)[reply]

The article says "W.L. McDaniel in 1987 proved that there are infinitely many Smith numbers". But the number of Smith brothers is unknown. -- ♬ Jack of Oz ♬ ^{[your turn]} 22:54, 20 May 2012 (UTC)[reply]

Hey, I discovered the smallest Smith 7-tuple at 164736913905. But I agree that my weird number crunching hobby shouldn't drive the future of mathematics so I will just be amused that somebody mentioned Smith numbers at all. PrimeHunter (talk) 00:22, 21 May 2012 (UTC)[reply]

And what does this have to do with "what are the others" "anti-Hilbert problems"? Are these "Smith numbers" an instance of an "anti-Hilbert problem"?

I'm curious about the other problems too; here and here are the two possibly distinct references I found, but they only mention Beggar-My-Neighbor. With a bit of luck, I may be able to ask Conway this firsthand within the next year. Cyrapas (talk) 23:21, 3 May 2014 (UTC)[reply]

Silly syntax has me stumped[edit]

Our page on Hilbert systems lists the following as a standard axiom schema:

$(x=y)\to (\phi [z:=x]\to \phi [z:=y])$

From this schema, how can we derive the following?

$(x=y)\to ((u=v)\to (\phi [z:=x,w:=u]\to \phi [z:=y,w:=v]))$

If we apply the axiom schema to one substitution before the other, the axiom is no longer applicable, because we no longer have a single $\phi$ that is being applied to both variables. If we cannot prove the second schema, then how do we prove obvious sentences like this:

$\forall x,y,u,v\ (x=y\to u=v\to x\in u\to y\in v)$

« Aaron Rotenberg « Talk « 19:47, 20 May 2012 (UTC)[reply]

Apply the schema to get:

(x=y)\to (x\in v\to y\in v)

(u=v)\to (x\in u\to x\in v)

The rest is propositional logic.

Taemyr (talk) 10:49, 25 May 2012 (UTC)[reply]

How far does light travel in a light year?[edit]

'Light year' The opening sentence indicates just under 10 trillion kilometers or just under 6 trillion miles. That seemed clear enough for me! Then there is a chart with more exact numbers that would indicate is was just under 10 million kilometers or just under 6 million miles. What am I missing and which would be more accurate for my classroom project? — Preceding unsigned comment added by Zydeco38 (talk • contribs) 21:31, 20 May 2012 (UTC)[reply]

Wolfram Alpha says trillion, as does Google, verifying the article. The article cites this IAU page, which gives a value of 9,460,730,472,580.8 km. « Aaron Rotenberg « Talk « 22:33, 20 May 2012 (UTC)[reply]

A light year is a unit of distance, not a unit of time. The question is like asking "How far does a car travel in a mile?" - the answer to which is, of course, a mile. So the answer to your questioh is: a light year. What you really want to know is: How many miles/kilometers are there in a light year? -- ♬ Jack of Oz ♬ ^{[your turn]} 22:43, 20 May 2012 (UTC)[reply]

Yes, the exact answer is 9.460 730 472 580 8 million million kilometres which is about 5.8786 million million miles (avoiding usage of the still-slightly-ambiguous "trillion", though there's no ambiguity in your country). You might be reading the comma as a decimal marker in the chart. It is used as a thousands separator there. Dbfirs 23:03, 20 May 2012 (UTC)[reply]

By definition the light year is 299792458(m/s)·365.25(days/julian year)·24(hours /day)·60(minutes/hour)·60 (seconds/minute). One year is (according to Wolfram Alpha) 365 days, while one julian year is 365.25 days. So in one year the light travels 365/365.25=0.999316 light years. Bo Jacoby (talk) 08:41, 21 May 2012 (UTC).[reply]

You could use more precise values: 365.2422 days in a year, and 86400.002 seconds in a day. Plasmic Physics (talk) 09:08, 21 May 2012 (UTC)[reply]

You could, but astronomers don't. The light year is a conventional unit defined according to a conventional year, not a real year (sidereal, tropical, anomalistic, draconic, lunar, Sothic, ...). —Tamfang (talk) 08:19, 22 May 2012 (UTC)[reply]

According to these values, the answer with the correct sigfigs, is: 9.460529 petametres. Plasmic Physics (talk) 09:26, 21 May 2012 (UTC)[reply]

In international miles, it is 5.878500 × 10¹² miles. Plasmic Physics (talk) 09:48, 21 May 2012 (UTC)[reply]

How many AUs are there in 1 light-day? Plasmic Physics (talk) 09:08, 21 May 2012 (UTC)[reply]

1 AU is almost exactly 499 light seconds, so there are 86400 / 499 ≈ 173 AU in 1 light day. Gandalf61 (talk) 09:20, 21 May 2012 (UTC)[reply]

The figure I quoted above (from the light year article) is based, of course, on the Julian Year of 365.25 years which is apparently the standard year used for this astronomical purpose (and leap-seconds are ignored). The quoted accuracy appears excessive, but it is based on multiplying exact values, so I suppose we can justify it. The figure of 365.24219 quoted (rounded) by PP above is the mean solar year (averaged over equinoxes and solstices), but our system of leap years is based on the mean time between northern vernal equinoxes which is currently about 365.2424 days. These values are slowly changing, hence the use of the Julian approximation for a fixed value for the standard light year. Of course, this year, light will travel about 1.002053388 standard light years. Dbfirs 17:15, 21 May 2012 (UTC)[reply]