h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

There is a slight problem with the version of Cantor's Diagonal thingummy which you've used... because '0.1000000000...' and '0.09999999999...' are the same number, as are all similar pairs of numbers, letting your new (diagonally-produced) number contain zeros or nines complicates matters...

The simple solution is to use, for instance, the digit '5' whenever the digit you come across isn't a five, and the digit '6' when it is. No nines or zeros, so no problem...

It might be worth placing a little more emphasis on the fact that the list can be *infinitely* long, and the argument still works... that's the beauty of it, after all...

Good stuff

26199

Specifically, in the second to last paragraph you say:

"So no matter how long the list, we know that it is not possible for it to contain all the real numbers."

However, this would merely mean there are an infinite number, which we already knew. What Cantor proved was that there is an *uncountably* infinite number. i.e. you cannot come up with an enumeration that will eventually include any number you care to mention.

Oops...seems there will be some redundancy in this post. It's just that I wrote it before I saw the older posts...and now I'm too lazy to change it. *grin*
The alternative method that I propose is, at least if I recall the words of my professor correctly, the one actually used by Cantor.

The version of Cantor's diagonal argument used in this article has a subtle, but important, flaw. The problem is due to the fact that we are instructed to add 1 to every digit taken from the list. This may result in the formation of a number already in the list, but written in a different way, thus causing the proof to fail.
What is meant by "a number already in the list, but written in a different way". An example will probably be the best explanation. It can easily be proved that 0.09999... (ad infinitum) and 0.1 are in fact one and the same number. The actual proof could constitute an article of its own, as this is a rather curious fact.
What it all leads to, however, is that we may reach a situation where we, somewhere in a number, form a recurring series of 9s by using the method stated in the article, thus in actuality forming a 1 followed by 0s ad infinitum - a number which may exist in the list!

What, then, is the solution? Rephrasing a few paragraphs will suffice:

"Now create a new number n ... we are going to select highlighted"

Replace with:

Now create a new number n. What digits shall n have? To choose the first decimal of n, look at the first decimal of entry 1. If it is a 2, let 1 be the first decimal of n. If it is not a 2, let 2 be the first decimal of n. The first decimal of entry 1 is 1, which is not 2. We will therefore let the first decimal of n be 2.
Next, look at the second decimal of entry 2. If it is a 2, let 1 be the second decimal of n. If it is not a 2, let 2 be the second decimal of n. As the second decimal of entry 2 is a 9, which is not a 2 either, the second decimal of n will also be 2.
Repeat this procedure for every decimal. This table shows our entries, with the digits we will use to create n in bold:

"After we've added one to ... 7, 9, 3."

Replace with:

After processing these digits, we can see that our new number n will start like this: 0.221212222.


These minor modifications will correct the most important mistake in the article. There are other, minor, ones which have nothing to do with the actual proof. I will address these by writing a new article from scratch (based on this one, however), as handling each issue on its own would just be tiresome. As soon as I'm done, I'll post a reference here. Then readers can look at both, and editors will be free to use material from either. =)

There we go. Now I've learnt GuideML too. =)

http://www.h2g2.com/A479180 - an article on the same subject as this one, but with the errors taken out (hopefully) and some other changes. It's a bit longer, too, as I tend to get carried away when I write about interesting things. =)