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May 25

Math Problem

My friend and I have some trouble while arguing who is right and who is wrong. The math problem, a very simple problem, is (+17)-(-3). What is the answer? 74.113.188.65

If that's arithmetic, 17-(-3) = 17+3 = 20. Black Carrot 00:42, 26 May 2007 (UTC)[reply]

Negative_numbers#Addition_and_subtraction Black Carrot 00:43, 26 May 2007 (UTC)[reply]

You and your friend have encountered the famous "double negative equals positive" conundrum. Knowing the correct answer to this specific example is less important than understanding the reasoning. What are the proposed answers, and what is the thinking behind them, for each of you?

Meanwhile consider the following line of thought. Suppose the correct answer is A, so that

(+17)-(-3)=A.\,\!

If we add equal quantities to both sides, we will preserve the equality. In particular, we can add (−3) to both sides:

(+17)-(-3)+(-3)=A+(-3).\,\!

switch the places of minus -3 and plus +3

(+17)+(-3)-(-3)=A+(-3).\,\!

On the left, we have both added and subtracted (−3), so the two should cancel. On the right, we have added (−3), which is the same as subtracting (+3). Thus

17=A-3.\,\!

In other words, if we subtract 3 from the correct answer we should get 17. Finally, we can add (+3) to both sides. Thus we arrive at a value for A without ever explicitly resolving a double negative. Whatever the resolution, it must be consistent with this answer if it is to be consistent with the other ordinary laws of arithmetic. --KSmrq^T 08:43, 26 May 2007 (UTC)[reply]

Fixed point theorem

Gah, this unexpectedly popped up in my Analysis exam today, and I had no idea what to do. Anyway, I tracked down the proof in Fixed point property, and I understand what's going on - but I'm not sure why the multiple counter-examples I've thought of don't prove it wrong. The [0,1] x [0,1] "box" surely has multiple functions that do not intersect with f(x) = x, and so there are multiple functions that do not have fixed points, surely? Icthyos 22:41, 25 May 2007 (UTC)[reply]

Well, I myself am rather confident in the Brouwer fixed point theorem. Perhaps if you gave one of the counter examples we could figure out the problem you have with the theorem. nadav (talk) 22:59, 25 May 2007 (UTC)[reply]

Perhaps I'm misunderstanding, since I've not done much deep Analysis, but what about, say, f(x) = x + (1/2)? It can be restricted to a function from [0,1] -> [0,1] (though obviously not bijective, but that's neither here nor there as far as I can tell), and it has no fixed point...that I can see. Icthyos 23:14, 25 May 2007 (UTC)[reply]

How is it a function from

[0,1]\!\,

to

[0,1]\!\,

? Is

f(0.75)\in [0,1]

? -- Meni Rosenfeld (talk) 23:22, 25 May 2007 (UTC)[reply]

In case you meant

f(x)={\begin{cases}x+{\tfrac {1}{2}}&0\leq x\leq {\tfrac {1}{2}}\\1&{\tfrac {1}{2}}\leq x\leq 1\end{cases}}

Then

f(1)=1

. -- Meni Rosenfeld (talk) 23:37, 25 May 2007 (UTC)[reply]

I think you are confronting the conditions of the theorem, which is good. In the geometry of the Euclidean plane, a translation is an example of a continuous map of the plane to itself that leaves no point fixed; so is a glide reflection. If we then convert point w to w/(1+||w||), which is a bijection, we can create a continuous map of the open unit disk to itself without fixed point. Restricting attention to the closed unit disk, consider the map taking every point to the origin except the origin itself, which maps elsewhere; clearly this example has no fixed point, but neither is it continuous.

These examples tell us that the theorem demands a continuous function of the closed disk to itself, or the equivalent. Too often when we learn a mathematical fact we ignore the fine print, the preconditions. We over-generalize, like children who pluralize every noun by adding "s", including "cats" and "dogs", but also "*foots" and "*childs". This is presumably a natural attempt to simplify; but like the children, we eventually must learn better. --KSmrq^T 09:38, 26 May 2007 (UTC)[reply]

Ahh, thank you all. I was confusing myself with my counter-example, by only treating a sub-interval of the function's domain, and ignoring what happened when x > (1/2) - obviously at these points, f(x) > 1. I was having trouble imagining it before, but I see now - if you've got a function that doesn't have a fixed point at 0, then it has to cross the line f(x) = x at some point, or have a fixed point at 1. Oh well, there go six exam marks, heh. I scribbled something down about the intermediate value theorem, though not with any degree of relevance - but at least I see it was used in the fixed point property proof.

(I swear, I'll learn TeX at Honours next year...) Icthyos 10:51, 26 May 2007 (UTC)[reply]

TeX isn't hard, at least not the part you need in order to write formulae in Wikipedia - Help:Formula is a good reference, which you can use together with copying other people's markup until you get accustomed with it. -- Meni Rosenfeld (talk) 12:48, 26 May 2007 (UTC)[reply]

Learn superscripts ("c^2" → "c²"), subscripts ("z_n" → "z_n"), and the names of a few special characters ("\pi" → "π"); that's probably 80 percent of what you'll use at first. Later you can pick up fractions ("\frac{p}{q}") and matrices ("\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}") and other more challenging notation (like "cases"). Don't worry too much that you may never fully master TeX; almost nobody does, not even here. --KSmrq^T 09:36, 27 May 2007 (UTC)[reply]

I'd like to think at least Knuth has mastered TeX, considering that he wrote the damn thing! —Preceding unsigned comment added by 149.135.29.71 (talk • contribs) 12:08, 2007 May 27

If he has, that falls under "almost nobody". But I would not be too sure. Yes, Knuth designed and programmed TeX initially, but the TeX most of us use today has (thankfully) evolved. Also, it is a large, complicated piece of software which can be extended with macro packages and the like, and such creations can surprise even their creator. Consider that Knuth designed TeX for his personal use, to produce new editions of his books on computer science; I doubt he ever contemplated using it for, say, commutative diagrams.

What I often find useful is knowing how TeX "thinks". It has a surprisingly primitive view of its input, and its output is built entirely on a one-dimensional boxes-and-glue model. (Michael Plass' clever line-breaking approach using dynamic programming, assisted by effective (English) hyphenation, contributes to producing pretty paragraphs; but these rarely require consideration.) --KSmrq^T 21:48, 27 May 2007 (UTC)[reply]