.999~ = or != 1

[SirNitram]

verilion... You are attempting to argue a subset of Math, Calculus, is wrong. This is crazy talk.

[Durandal]

0.999~ = 1, due to a technicallity.

It's not a "technicality." Stop making it sound like we're letting a murderer go because a cop forgot to read him his Miranda rights. 0.999~ = 1 because they're different representations of the same number.

[Colonel Olrik]

Durandal and Kuroneko, both your sigs give me a headache. Part of my brain immediatly starts solving the damned expressions everytime I read one of your posts, and I just can't stop the process. It's very sad

[Durandal]

Durandal and Kuroneko, both your sigs give me a headache. Part of my brain immediatly starts solving the damned expressions everytime I read one of your posts, and I just can't stop the process. It's very sad

It works out, trust me.

[Kuroneko]

Durandal and Kuroneko, both your sigs give me a headache. Part of my brain immediatly starts solving the damned expressions everytime I read one of your posts, and I just can't stop the process. It's very sad :(

But see, he keeps on posting statements that are essentially 0 = 0 or 1 = 1, etc. This is all obviously nonsense; I can't just take such an affront silently. I have to correct his erroneous ways and show that in fact 0 = 1.

[Spanky The Dolphin]

0.999~ = 1, due to a technicallity.

It's not a "technicality." Stop making it sound like we're letting a murderer go because a cop forgot to read him his Miranda rights. 0.999~ = 1 because they're different representations of the same number.

Bleh, I meant that more as "technically, 0.999~ = 1," but my word order got messed up. I didn't mean to make it sound like a compromise or such.

[phongn]

Durandal and Kuroneko, both your sigs give me a headache. Part of my brain immediatly starts solving the damned expressions everytime I read one of your posts, and I just can't stop the process. It's very sad

But see, he keeps on posting statements that are essentially 0 = 0 or 1 = 1, etc. This is all obviously nonsense; I can't just take such an affront silently. I have to correct his erroneous ways and show that in fact 0 = 1.

We've got to get Durandal to clean up your sig, it looks ugly

[Zoink]

I see it like this:

A have a line, which I consider as an infinity of real number points.

I draw a point and define it as "0.999~".

I can draw another point and label it "3.5" (for example). The distance between them isn't important, I can note that between the two points, I can place an infinity of points, just like I should.

Now I consider where to put the point "1". I note that there can be no number between "0.999~" and "1". I must in fact place "1" in the same location as "0.999~" (ie. they are the same point) or else it is suggested that there is in fact points between the two. I can see no other conclusion other than 0.999~ and 1 are the same point, just different names for that point.

[Durandal]

Durandal and Kuroneko, both your sigs give me a headache. Part of my brain immediatly starts solving the damned expressions everytime I read one of your posts, and I just can't stop the process. It's very sad

But see, he keeps on posting statements that are essentially 0 = 0 or 1 = 1, etc. This is all obviously nonsense; I can't just take such an affront silently. I have to correct his erroneous ways and show that in fact 0 = 1.

You do some interesting gymnastics with calculus, but (dy/dx)(dx/y) = dy/y != 1.

EDIT: And yes, if you want to give me the LaTeX for your sig (or just a PDF, assuming you use pdflatex), I'll make it look nice 'n purty for you.

[Admiral Valdemar]

Maybe I should put some equations in my sig too, y'know, just to feel popular. I know some interesting Michaelis-Menten ones.

[Durandal]

Maybe I should put some equations in my sig too, y'know, just to feel popular. I know some interesting Michaelis-Menten ones.

Two people out of hundreds doing it is hardly what I'd call "popular."

[Kuroneko]

But see, he keeps on posting statements that are essentially 0 = 0 or 1 = 1, etc. This is all obviously nonsense; I can't just take such an affront silently. I have to correct his erroneous ways and show that in fact 0 = 1.

You do some interesting gymnastics with calculus, but (dy/dx)(dx/y) = dy/y != 1.

What are you talking about? (I'll use ' for differentiation for brevity.)

(fg)' = f'g + fg', thererefore Int[f'g] = fg - Int[fg'] (basic integration by parts). Observe that Int[ cot x dx ] = Int[ (sin x)'(1/sin x) dx ] = sin(x)(1/sin(x)) - Int[ (sin x) (1/sin x)' dx ] = 1 - Int[ (sin x)(-cos x/sin^2 x) dx ] = 1 + Int[ cot x dx ].

Theforefore, Int[ cot x dx ] = 1 + Int[ cot x dx ]. The raw calculus part is completely and utterly correct. Any fool can see that by subtracting Int[ cot x dx ] from both sides, we get 0 = 1. (Those who cannot see this are insufficiently foolish.)

EDIT: And yes, if you want to give me the LaTeX for your sig (or just a PDF, assuming you use pdflatex), I'll make it look nice 'n purty for you. :)

Really? I'll take you up on that in a little while, though can I put a couple in? I'm thinking of putting the ones that have been my sigs on automatic rotation.

[Admiral Valdemar]

Maybe I should put some equations in my sig too, y'know, just to feel popular. I know some interesting Michaelis-Menten ones.

Two people out of hundreds doing it is hardly what I'd call "popular."

It was humour, I don't think there's an equation for that.

I checked your sig in IE last night too, hot damn, does IE fuck up the rendering of .png files.

[Durandal]

What are you talking about? (I'll use ' for differentiation for brevity.)

When you expand int(cot x dx) into int(cos x/sin x dx) and then into int[d(sin x)/dx * dx/sin x], you reduce it to sin x/sin x = 1. I don't think that's allowed. It should be reduced to int[d(sin x)/sin x] ... and I don't think that's a valid integral.

EDIT: Well, if we let u = sin x, we can see that the integral becomes int(du/u), and the anti-derivative is ln(u) + C = ln(sin x) + C, which does not equal 1 except under the condition {x|x = 0, pi, 2(pi), ... n(pi)}.

In my original criticism, I forgot that you were integrating.

Really? I'll take you up on that in a little while, though can I put a couple in? I'm thinking of putting the ones that have been my sigs on automatic rotation.

Sure. I'm not at my home machine right now anyway.

[Death from the Sea]

.111~ = 1/9
.222~ = 2/9
.333~ = 3/9
...
.888~ = 8/9

All of the above are exact and can easily be proven by doing the division 1/9, 2/9, etc.

Therefore: .999~ = 9/9 = 1

but he asked for .9 and .9 = 9/10

[Durandal]

.111~ = 1/9
.222~ = 2/9
.333~ = 3/9
...
.888~ = 8/9

All of the above are exact and can easily be proven by doing the division 1/9, 2/9, etc.

Therefore: .999~ = 9/9 = 1

but he asked for .9 and .9 = 9/10

No, he asked if 0.999~ = 1, and that is true.

[Death from the Sea]

.111~ = 1/9
.222~ = 2/9
.333~ = 3/9
...
.888~ = 8/9

All of the above are exact and can easily be proven by doing the division 1/9, 2/9, etc.

Therefore: .999~ = 9/9 = 1

but he asked for .9 and .9 = 9/10

No, he asked if 0.999~ = 1, and that is true.

shit my short term memory is FUBU because of term papers , your right it is .999

[Kuroneko]

EDIT: Well, if we let u = sin x, we can see that the integral becomes int(du/u), and the anti-derivative is ln(u) + C = ln(sin x) + C, which does not equal 1 except under the condition {x|x = 0, pi, 2(pi), ... n(pi)}.

Right, or rather logarithm of |sin x|. But why would I want to solve the integral when I can prove 0=1 instead? (Heh.)

Besides, it is not the integral that I'm equating 1 with, but merely one term out of integration by parts. To summarize,

Int[ f'(x)g(x) dx ] = f(x)g(x) - Int[ f(x)g'(x) dx ], and therefore

Int[ cos x / sin x dx ] = sin x/sin x + Int[ cos x / sin x ]
by the assignment f(x) = sin x and g(x) = 1/sin x.

Edit: typo.

[Spanky The Dolphin]

shit my short term memory is FUBU because of term papers , your right it is .999

No, it's 0.999~. 0.999 and 0.999~ are different numbers.

[CmdrWilkens]

shit my short term memory is FUBU because of term papers , your right it is .999

No, it's 0.999~. 0.999 and 0.999~ are different numbers.

But .999~ and 1 aren't...which has been the point of the whole thread

[IRG CommandoJoe]

So then if .999~ = 1, then would 3.999~ = 4 and 5.999~ = 6 and so on?

[CmdrWilkens]

yes

[Durandal]

Right, or rather logarithm of |sin x|. But why would I want to solve the integral when I can prove 0=1 instead? (Heh.)

Indeed.

Besides, it is not the integral that I'm equating 1 with, but merely one term out of integration by parts. To summarize,

Int[ f'(x)g(x) dx ] = f(x)g(x) - Int[ f(x)g'(x) dx ], and therefore

Int[ cos x / sin x dx ] = sin x/sin x + Int[ cos x / sin x ]
by the assignment f(x) = sin x and g(x) = 1/sin x.

Edit: typo.

Oh so that's what you're doing. I'd long since forgotten about the Hell that was integration by parts.

[Kuroneko]

Oh so that's what you're doing. I'd long since forgotten about the Hell that was integration by parts.

Personally, I dislike that u,v,du,dv formulation. Yes, it is prima facie more general, but only at first glance--but if I think of integration by parts as simply the integral of the product rule (fg)' = f'g + fg', then everything comes out easier.

[SirNitram]

If this keeps up, I'm posting the reason why the population of the universe is zero.