#theephyran

[maths-help]

Hello, clever math people.

theephyran:

Hello, I hope this question isn't too silly, as I don't have any mathematical training beyond high school and my own curiosity, but I have a personal worry about Cantor's theorem, and while I'm sure this cannot have been overlooked by mathematicians, I think it is overlooked in informal presentations of the proof, so I don't know if an informal answer is possible, but I'll have a go anyways: How can we be sure that the number denoted by the decimal representation generated in the diagonalization argument which is not identical to any other number on the list of real number presupposed, does not indeed have a decimal representation which is already on the list - in the same way the number 1 can have a decimal representation both as 0.999... and 1.000...?

I don't know if I've been clear enough to have my question answered, or if it is possible to explain without presupposing massive amounts of college-level mathematical knowledge, but if you could maybe point me towards a Wiki or a textbook of some sort, I would already be grateful. I suppose this isn't exactly the sort of question you usually receive either, so if it takes longer, or if you don't answer at all, I won't be offended :P.

It's not a silly question at all!

So the answer comes in two parts. First, we know that the only time a number has two decimal representations is when one representation ends in a string of nines and the other one ends in a string of zeroes, like 0.999... and 1.000...

(I think if you wanted to prove this, you'd take the two representations, start truncating them at the tenth place, hundredth place, etc, and take the difference for each of them. The longer the representations get the smaller the difference has to be, and that constrains your possibilities a lot.)

Then second, we just make sure the number we're constructing won't end in a string of nines or zeroes! For example, in this post I used the numbers 4 and 5, so there won't be any problems.

Cheers!

P.S. For those of you who are curious, we did a series for Cantor's diagonalization earlier: 1 2 3 4

[twocubes]

(This is a question about this post I wrote a while ago)

Well, the thing I did is indeed wrong in the real numbers. But! It turns out that the rational numbers don't only live in the real numbers, they also live in the p-adic numbers for any given prime p, which are numbers that emerge from a different notion of convergence on the rational numbers than the one that gives rise to the reals.

Informally, a sequence of rational numbers "converges p-adically" if for any n, there is some point onwards from which pⁿ divides all the differences between the terms. This means, among other things, that we can let numbers go on forever "in the wrong direction".

So, while normally you might write an expression like 1.121212...₃ and it would make sense as a real number in base 3 (the limit of the sequence of rational numbers 1₃, 1.1₃, 1.12₃, 1.121₃...), in the 3-adic numbers you might write ...222121.1₃ and it would be a perfectly valid 3-adic number (the "p-adic limit" of the sequence of rational numbers 0.1₃, 1.1₃, 21.1₃, 121.1₃,... ).

The weird thing then is that things can cancel "in the wrong direction". For example since 2₃ + 1 = 10₃, 223 + 1 = 100₃, 222₃ + 1 = 1000₃ and so on and so forth, we have ...2222₃ + 1 = ...0000₃ = 0, and so ...2222₃ = -1. (This implies that ...1111₃ = -1/2).

The full story, without all the hand-waving, is kind of long and incredibly fascinating, and I'm not remotely qualified to tell it. (In fact, I'm kind of worried that I might have done something wrong already, since I haven't actually officially learned any of this yet. I haven't even finished the book I'm about to recommend!) So I'll just drop a link to the p-adic numbers article on wikipedia and recommend "p-adic Numbers, An Introduction" by Fernando Quadros Gouvea (from which I lifted the trick in that post anyways :¬P) because it is a nice book.

The thing to understand is that strictly speaking, when writing a limit each of whose terms is rational, it is ambiguous which notion of convergence we should be using. It turns out that while the series 1+3+3²+3³+... = ...1111₃ diverges in the reals, it converges to -1/2 in the 3-adic numbers, and so we can use that fact to solve the problem from before. (Although, of course, it's kind of silly to do so this way...)

The usual development goes from ℕ to ℚ to ℝ to ℂ. Here, we have glimpsed a whole bunch of number systems that contain ℚ but aren't in ℝ. So, it turns out, there was a fork in the road there. This is what happens when you go the other way .‿.

[outofthedarkness]

theephyran said: I’m reading The Idiot. Due to the unfortunate fact that I have been expected to interact with the outside world, I am only about a quarter or a third of the way in - AND ONE DAY HAS BARELY PASSED, WHAT IS THIS MADNESS? It is nighttime now, so I must read.

And then all of a sudden you'll jump six months and the you'll jump a couple weeks and you'll be totally confused about the progression of time.

But it's all worth it because it's such an amazing book!

[maths-help]

Well, quite a lot of us use this online equation editor. You can write pretty much anything with it (including both symbols you are searching for) e.g:

However, you will only be able to download those equations as an image which makes it hard to edit them later on. Of course, you can always get the symbol only as an image and put it into the text.

Another way to solve this problem is to search for a symbol you need here and then copy it from the HTML column of the grid and paste it , e.g.

⇒ → ⊃ ∧

I'm not sure whether there are other ways but we're always happy about suggestions!

-- Frauke

EDIT:

squigly-protractor replied to your photo: Well, quite a lot of us use this online equation...

My teacher has recommended MathType, though I haven’t tried it myself. I believe there’s a free trial and a cheap full version.

Thank you! Although I've never used it either, it seems like a very helpful program with a lot of features (at least from what I can tell from pictures and reviews).

This is the website and this is where you can get the free trial version.

[maths-help]

Flattery will get you nowhere, all, but interesting problems will :P

Blindly, we're looking for distance, so it makes a lot of sense to use the distance formula:

We have

so then the distance is

And we want to minimize this expression. Let's complete the square under the radical to get

Now, the (t-4)^2 term is always going to be at least zero, so to minimize we just set t = 4, which means our minimum distance is √18 = 3√2.

- H2

[astrotastic]

你学习中文吗？I ask because that would make you all the more awesome to me, and that's never a bad thing. You will remain just as awesome if you don't though. Btw - you are awesome :)

當然!! I'm learning to read 繁體字 and it's kind of hard, but makes learning 简体字 a lot easier. Plus, by learning traditional characters, I'm able to read so much more Chinese literature. I think I want to study it more in college.