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I think quiet appreciation is enough.

**Dross**- Replies: 12

I am putting a formal request (if such a request is necessary/required/recognised) that someone from site admin put a stop to the on-going joke that has become Anthony R Brown's crusade to convince all of us that 0.999... is not equal to 1.

Let me also say that I am the last person who would want to stop open-minded discussion, alienate or ostracise another person, or make them feel in any way "bullied" or chastised. I may not like someone, but I think I can still have a fair amount of respect for their feelings and points of view.

However, the truth of the matter is that this situation has gotten utterly ridiculous. So far, no fewer than nine topics have been started which concern (either as their main or only subject matter) the 0.999... = 1 issue (see below for the list I got from looking at the first few pages in the "This is Cool" forum), with a total of 1,302 posts between them. Most of the topics were started by Anthony, and most of the posts in them have been either from or too Anthony. On a more subjective note, most of the later posts say the same things - there are simply no new things to say, as the subject has been flogged to death. Furthermore, around half of the topics have already been closed to discussion by a moderator anyway.

This is quite simply a waste of everybody's time.

Recently, Wikipedia deleted his posts, with the following comments being made:

jpgordon (on Wikipedia) wrote:

Anthony's ArgumentsRemoved. There is no reason for Wikipedia to be providing a platform for Anthony; it's time for him to go elsewhere. I'll be removing all his nonsense as it appears. --jpgordon 15:40, 23 October 2007 (UTC)

jpgordon (on Wikipedia) wrote:

I've semi-protected the page for a few hours; he seems to be enjoying himself too much. Go find yourself another playground, Anthony; you're wasting everyone's time here. --jpgordon 20:38, 23 October 2007 (UTC)

(source: http://en.wikipedia.org/wiki/Talk:0.999.../Arguments)

I believe that we are the "playground" referred to in the latter of these comments. These useless discussions make the site and the forum a much less respectable and reputable environment for both existing users and visitors to the site, indeed giving the impression of a playground even though the rest of the forums are places for reasonable, rational and well-debated discussion.

I would like to suggest that Anthony either be banned from the site, and/or have his posts deleted as soon as possible. It might sound harsh to some - to others, it would be a welcome relief.

~~~

List of topics concerning 0.999...:

1) *0.9999....(recurring) = 1?* by i am evil, always will be (856 posts)

2) *Cut A Hole! In A Sheet Of Paper Problem!...* by Anthony.R.Brown (20 posts)

3) *Domestos Kills 99.999...Percent of all Germs!* by Anthony.R.Brown (101 posts)

4) *Infinite 0.9 And Recurring 0.9 : Posts Only! : Unbiased!* by Anthony.R.Brown (8 posts)

5) *INFINITE 0.9 <> 1 PROOF : COMPLETE : By,Anthony.R.Brown,15/02/07* by Anthony.R.Brown (28 posts)

6) *INFINITE 0.9 <> 1 PROOF By,Anthony.R.Brown,15/01/07.* by Anthony.R.Brown (129 posts)

7) *IF TWO NUMBERS START WITH A DIFFERENCE IN THEIR VALUES! By A.R.B* by Anthony.R.Brown (6 posts)

8) *Genius Answer! 1/3 =* by Anthony.R.Brown (67 posts)

9) *I disagree with 0.999... = 1* by Traesk (87 posts)

Identity wrote:

Could you explain that a bit more?

Anthony.R.Brown wrote:

Quote:" Could you explain that a bit more? "

A.R.B

Infinite/Recurring 0.9 = 0.999... " There Are Never Ending .9´s in this Calculation! "

Integer/Whole Number 1 = 1 " There Are No .9´s in this Calculation! "

So... I guess not, then.

And we were doing such a good collective job of not replying to ARB. Now it's just too tempting!

Thus the antiderivative of

is , because when you differentiate you get .Thus the antiderivative of

is , since it is its own derivative.**Dross**- Replies: 4

Two functions, *x* and *e[sup]x[/sup]*, are walking down the street one day when they notice a differential operator walking towards them.

"Oh no," says *x*, "that's a big mean differential operator! If he catches me he'll differentiate me and then I'll just be a constant function!"

"Don't worry," says *e[sup]x[/sup]*, "I'm my own derivative, so he won't change me if he tries to differentiate me - you go hide in that alley and I'll sort him out!"

After hiding in the alley, *x* peeks out to see what's happening, and watches as *e[sup]x[/sup]* approaches the differential operator.

"Hi there, I'm *e[sup]x[/sup]*" he says.

"Hi there," comes the reply, "I'm *d/dy*."

Anthony.R.Brown wrote:

To Dross

It´s Good to see the Fools are Still on this site!....

Now let´s get back to the serous people that come on this site! that are prepared to talk Math! and argue their case! Rather than coming on for five minutes! making ill informed and derogatory remarks! then run away with their tail between their legs! because they have no substance for what they believe in!...

A.R.B

HAHAHAHAHAHAHAHAHAHAHA!!!!!!!!

Oh, dear me! As I wipe away a small tear from my cheek, I remind you all that all we need to do to get rid of Anthony is to ignore him. Stop feeding the local wildlife and it'll stop scratching on your door.

Landof+, we've already tried using logic with ARB. The simple fact of the matter is that he'll abandon all reason and just shout a lot, then claim to have gotten his point across. There's really no use in... well, anything at all with him since he won't listen to reason or anything else.

Anthony.R.Brown wrote:

Quote:" Whyy so many exclamation marks? "

A.R.B

It´s a Bit like the .9´s ! they Never End!!..."

HAHAHAHAHAHAHAHAHAHAHA!!!!!!!!

Domain: Basically you want to see what values of x can go into the function and produce some sort of valid output. Assess things like whether there are any values of x where the denominator is zero, or where the denominator doesn't exist (I assume you're working with the real numbers here, right?)

Range: Find out if there are any maximum or minimum values that will prevent the range from being all the real numbers.

Remember that just because it says you have to find them algebraicly, that doesn't mean you can't draw a graph to give you some direction.

This need not be the case - you could select a green apple first, second or third. There are then two (basically equivalent) ways to look at how to get the correct answer:

1) Since there are three choices as to when the green apple is removed, multiply your answer by 3:

2) Since there are three choices as to when the green apple is removed, find the answer by adding the three relevant probabilities:

Look at statement (2). Remember that y is either greater than, or equal to, zero (by the question).

If y were greater than zero, the inequality wouldn't make sense: it'd be trying to tell you that a positive number (the modulus expression, LHS) is less than a negative number.

The only option, therefore, is that y = 0, and therefore | x - 3 | = 0, thus x = 3.

JaneFairfax wrote:

Its excellent apart from the misspelling of its.

"Bah" to me.

Thank you for the correction - I keep forgetting I shouldn't use "it's" when it's being used as the possessive of "it". I just have this habbit of **if it possesses something, it NEEDS an apostrophe somewhere!** I shall quit it at some point.

**Dross**- Replies: 10

"I did it, sir - but then I measured it's momentum *so* precisely that it could have ended up anywhere!"

Identity wrote:

This might be of interest to those computer programmers out there, or you could just try to solve the problems algebraically

Well, that's me occupied for at least the next few days

Great site - thanks for the link.

I get a magnitude of 8.4N at an angle of 94.97 degrees.

Ricky wrote:

I recommend that no one responds to ARB's posts unless it has some form of coherent thought or content behind it. Perhaps if no one replies, he'll get bored and give it up. But as long as you all keep feeding replies to him, he will stay.

Again, this is just a personal recommendation.

Hear, hear!

I agree. It may be hard - especially since it's inevitable that this sort of boycott will have him posting that it means that we concede that he was right all along (though whether he actually *believes* that he's right, as opposed to just *saying* he's right is something I'm not entirely sure of!). But it might be a bit easier once we realise that he's going to distort whatever we do (posting various proofs *or* the non-posting of anything) into some sort of admission of wrongness on our part, so it doesn't really matter too much what we do.

And it'll hopefully either get rid of him, or make him start making some sense!

I won't be replying to any of his posts hereafter, then.

Anthony.R.Brown wrote:

Quote:" You are not going to drag me into trying to prove you wrong. Here are some things I'd like to point out: "

A.R.B

Now let´s get back to the serous people that come on this site! that are prepared to talk Math! and argue their case! Rather than coming on for five minutes! making ill informed and derogatory remarks! then run away with their tail between their legs! because they have no substance for what they believe in!...

Does anybody else think it's mildly amusing that he seems to implicity include himself in the "serious people" category!

roger wrote:

write an equation in standard form that:

1) passes though (7,2) and parallel to x+y=6

and

2) passes through (1,9) and is perpendicular to 2x-5y=8

Have you made any attempt at these questions? How far have you got?

Both of these need to be in the form y = mx + c (or whatever symbols you're using for "gradient" and "y-intersection" - m and c are what I'm using, respectively).

So here are two further sub-questions that will get you started on the first question:

i) What is the gradient (m in the equation above) of the first line?

If you can answer this, you'll have the m in the equation. Then:

ii) What does c have to be so that y = 2 when x = 7?

Get back with any thoughts/answers/guesses (:P) you have for these.

Anthony.R.Brown wrote:

Quote:

(A) ¹/3 = 0.333... (recurring)

(B) (¹/3*3 = ³/3) = (0.333...*3 = 0.999...)

(C) ³/3 = 0.999...

(D) 0.999... = 1A.R.B

(A) 1/3 = 0.333... (recurring)

The above Calculation Will always Continue as .333...

(B) (1/3 x 3 = 3/3)

The above is the First mistake! (1/3 Will always Continue as .333... ) Now Times 3 must Equal .999...( 3 / 3 Equals 1 so these are not Equal Values!(0.333...x 3 = 0.999...)

The above is a True Calculation! as I have shown above...

(C) 3/3 = 0.999...

The above is the second mistake! 3/3 = 1

(D) 0.999... = 1

The above is the Third mistake! 0.999... Will always Continue as 0.999...

A.R.B

P.S. In case you didn't know what I was referring to by "presupposing your conclusion", it is highlighted above. You are attempting to *show* that 0.999... is not equal to 1, thus you cannot use that fact as you have done in the bold section above (esp. underlined).

I don't care if you think that 0.999... is not equal to 1. You're not making decisions that affect my life because of this erroneous belief. But I *do* hope the penny drops for you at some point.

Anthony.R.Brown wrote:

Quote:" Hello, one and all. I have been gone for a while and apparently you guys are still letting Anthony bait you because he doesn't believe the 0.999... = 1 thing.Some people's minds are just impossible to change about some things; I'm sure it'll be no great loss to humanity if we just allow him to carry on believing. "

A.R.B

You are just as welcome as every other person that has tried to prove my counter proof arguments are wrong?...

So have a go! show Infinte/Recurring 0.9 using one Calculation Only!!...

Erm... actually, I don't feel welcome to do so at all. A proof was given shortly after this quoted post, and you attempted to refute it. What you didn't seem to notice whilst offering this insight is that you presupposed your conclusion (i.e. you used your belief that 0.999... doesn't equal 1 to try and show us that 0.999... doesn't equal 1).

You are not going to drag me into trying to prove you wrong. Here are some things I'd like to point out:

1) You'll probably say that I've given up because I'm admitting that you're right. This is not the case: you are wrong.

2) The reason I'm not going to be dragged into such a discourse is that you simply won't listen to what other people are saying. You argue in ways that are quite simply utterly illogical. You assume an argument isn't true because you don't like it's conclusion. You presuppose your own conclusion, and the falsehood of everyone else's conclusions. You *ask* everyone for their proofs that 0.999... = 1, but you won't take any of them seriously. Likewise, you clearly do not take your *own* "proofs" seriously, as shown by the total and complete absence of any sort of mathematical or logical rigour.

Anthony: may I ask how old you are?

Anthony.R.Brown wrote:

Not clean enough! 100 Percent cant be reached! same as Infinite/Recurring Problems!...

Hello, one and all. I have been gone for a while and apparently you guys are *still* letting Anthony bait you because he doesn't believe the 0.999... = 1 thing. Has this been going for the last six months or so? A year even? Some people's minds are just impossible to change about some things; I'm sure it'll be no great loss to humanity if we just allow him to carry on believing.

For one thing, he's arguing mathematical principles by a technique I'm going to call "appeal to advertising". I mean, come on. Really. I mean... *come ON!!*

Neela2 wrote:

However, the result in the back of the book wants me to get 8/3 squareroot(x-4) + 3

There's certainly no reason to have that result - 8/3 is correct.

It might be a typo, but have you checked everything else? (i.e. read the question correctly, made sure you're looking at the right answer - all the trivial stuff you might not think to check)

Theorem: All positive integers are equal.

Proof:

It is sufficient to show that *a* = *b* for any two positive integers *a* and *b*.

Furthermore, it is sufficient to show that for every positive integer *n*, if *a* and *b* are such that *max(a, b)* = *n*, then *a* = *b*.

Proceed by induction.

For *n* = 1, the only choice for positive integers such that *max(a, b)* = *n* is *a* = *b* = 1, so the proposition holds for *n* = 1.

Now assume the proposition is true for some positive integer *k*.

Then consider the positive integers *a* and *b* such that *max(a, b)* = *k+1*.

Then *max(a - 1, b - 1)* = *k*.

By our assumption, *a* - 1 = *b* - 1, so obviously *a* = *b*.

QED

Heh - should really have given you the benefit of the doubt on that one I guess!

Anyway - now that's been cleared up and all, the error is that because your conclusion takes I to have some numerical value, you can only use your conclusion when you evaluate this integral between limits (i.e. you cannot use an indefinite integral). So you have to say:

Of course, as soon as you evaluate this, the

term becomes zero and doesn't bother us any more.