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October 6

Using "curve" to refer to a straight line

In a log-log plot, plotting any monomial will give a straight line. However, it seems a little odd to refer to the plot of e.g. $x^{2.5}$ as a line, but equally strange (if correct) to refer to that plotted straight line as a curve. What would you do? Oliphaunt (talk) 08:03, 6 October 2011 (UTC)[reply]

In what context? --Trovatore (talk) 08:21, 6 October 2011 (UTC)[reply]

A paper we're writing. We've fitted some data sets with curves of the form

ax^{b}

and plotted the result on a log-log scale. I'm going back and forth on this; perhaps "line" is the way to go after all. Just interested in others' opinions. Oliphaunt (talk) 08:42, 6 October 2011 (UTC)[reply]

Arguably it is a curve, but is rendered as a line because it is a log log plot. For that reason, I'd go with curve. --Tagishsimon (talk) 12:31, 6 October 2011 (UTC)[reply]

My (possibly imperfect) understanding is that, mathematically, a line (even a y=mx+b Euclidean/Cartesian one) *is* a curve, in the same sense that a square is a rectangle. It's not incorrect to call a line a curve, even though in most situations you'd likely use the most specific descriptor which applies. -- 174.24.217.108 (talk) 16:55, 6 October 2011 (UTC)[reply]

Yes, the article on curve mentions that, and I alluded to it by interjecting "if correct" :-) —Oliphaunt (talk) 06:43, 7 October 2011 (UTC)[reply]

It is perfectly fine to call a straight line a curve. Just like every Irishman is a man and every odd number is a number. — Fly by Night (talk) 19:57, 7 October 2011 (UTC)[reply]

I realise it's fine; I was just wondering what you'd do. It's also fine to refer to such a representation

of a monomial as a line. So this is choosing between two 'fine' options. The consensus seems to be that people would pick "curve". —Oliphaunt (talk) 12:33, 9 October 2011 (UTC)[reply]

I would also recommend "curve". CRGreathouse (t | c) 17:46, 10 October 2011 (UTC)[reply]

.9 repeating and 1+1=3

I read that on the .9 repeating page that it's possible to create a new number system where .999~ does not equal 1. But such a number system would be fairly useless. With that logic, is it also possible to create a new number system where 1+1=3? 198.151.130.149 (talk) 15:35, 6 October 2011 (UTC)[reply]

Technically yes, although that starts to call into the question what you mean by "3". (Usually 1+1=2 is take to be the definition of "2", roughly speaking, so if you also have 1+1=3, you either would have 2=3, or you start to mess with the basic properties of equality.) Note that redefining addition from its "normal" definitions is not always as silly as you seem to think it is. For example, doing things like defining 1+1=0 is the key step which gets you to modular arithmetic. - By the way, -- 174.24.217.108 (talk) 16:49, 6 October 2011 (UTC)[reply]

Number Game

There was a game that I've heard of, that I was hoping you cats could help me find the name of:

You have a large group of people, everyone kicks into the pot (something small, a dollar or whatever) and then everybody writes a positive integer on a piece of paper. Whoever wrote the smallest UNIQUE integer wins the pot. I'm looking for the name because I was hoping that there'd been some scholarship on the optimal strategies or something.

Thanks!209.6.40.167 (talk) 15:53, 6 October 2011 (UTC)[reply]

Well, you know no one will write down 1, out of fear that someone else will have chosen to pick that number as the smallest positive integer, causing them to lose. And, knowing that no one will pick any number less than or equal to n, it follows that no one will write down the number n+1 either, for the same reason. Hence, by induction^†, no one will write down anything.

(No, I don't know what the game is called.)

^†Actually, strong induction, which is completely different from weak induction in that it has the word "strong" before it.

--COVIZAPIBETEFOKY (talk) 17:24, 6 October 2011 (UTC)[reply]

See Unique bid auction. —Bkell (talk) 18:09, 6 October 2011 (UTC)[reply]

Let N be the number of people in the group. Choose a random number k≥1 from the geometric distribution Pr(k)=(N−1)⁻¹(1−N⁻¹)^k. Bo Jacoby (talk) 21:01, 6 October 2011 (UTC).[reply]

However, that approach, even if it was to be optimal (and it is by no means clear that it would be in real-world auctions against humans) only works in you know N. An interesting related problem would be how to estimate N by looking at the winning prices in a succession of auctions, as online auction operators do not typically reveal N, for obvious reasons. -- The Anome (talk) 07:33, 7 October 2011 (UTC)[reply]

The more difficult game is of course when the second lowest unique number wins. – b_jonas 08:44, 7 October 2011 (UTC)[reply]

The game is easily rigged. If a syndicate of floor(N/2)+1 players agree to choose the integers from 1 to floor(N/2)+1 in some predetermined manner, then one of then is bound to win the $N pot. Splitting the winnings equally between syndicate members gives each member a certain gain of a little under $1. As the game is so easily rigged, there is no point in playing. Gandalf61 (talk) 10:01, 7 October 2011 (UTC)[reply]

Applying the logic a bit above why should anybody share anything in the last round of the game if there isn't going to be another round? And thus why should anyone share in the second last round if they know the person wont share in the last round... ;-) Dmcq (talk) 10:13, 7 October 2011 (UTC)[reply]