Date: Nov 18, 2012 10:04 AM
Author: gus gassmann
Subject: Re: Matheology § 152
On 17/11/2012 11:34 PM, Virgil wrote:

> In article

> <126c3310-d023-4f33-9b13-6cac84751832@o8g2000yqh.googlegroups.com>,

> WM <mueckenh@rz.fh-augsburg.de> wrote:

>

>> On 17 Nov., 18:57, Uirgil <uir...@uirgil.ur> wrote:

>>

>>>>>> Consider the following sequence of decimal numbers, consisting of

>>>>>> digits 0 and 1

>>>

>>>>>> 01.

>>>>>> 0.1

>>>>>> 010.1

>>>>>> 01.01

>>>>>> 0101.01

>>>>>> 010.101

>>>>>> 01010.101

>>>>>> 0101.0101

>>>>>> ...

>>>

>>>>>> which, when indexed by natural numbers, yilooks like this:

>>>

>>>>>> 0_2 1_1 .

>>>>>> 0_2 . 1_1

>>>>>> 0_4 1_3 0_2 . 1_1

>>>>>> 0_4 1_3 . 0_2 1_1

>>>>>> 0_6 1_5 0_4 1_3 . 0_2 1_1

>>>>>> 0_6 1_5 0_4 . 1_3 0_2 1_1

>>>>>> 0_8 1_7 0_6 1_5 0_4 . 1_3 0_2 1_1

>>>>>> 0_8 1_7 0_6 1_5 . 0_4 1_3 0_2 1_1

>>>>>> ...

>>

>>> While every real mathematician knows

>>

>> This sequence grows without limit.

>>>

>>>> This can be proved by taking any number n and showing

>>>> that there is a number k such that all for terms a(j) of the sequence

>>>> with k > j we have a(j) > n. Proof: For given n take k = n + 10.

>>>

>>> ow does that work for the sequence a(j) = 0 for all j?

>>

>> Is 0 larger than any number n?

>

> There is no number which is larger than any number.

Ahem. Virgil, that is as nonsensical as anything WM spews forth.

First, you left out the 'n', which in the context of this exchange

clearly is meant to be integer. And yes, there are numbers that are

larger than every integer.

Consider also this: Let x be any number.

Then x+1 is larger than x.

You complete the syllogism...