Roast me Veeky Forums

Here is a very very simple proof I came up with:

[eqn] S = 0.9999… [/eqn]
Add 0.111… to both sides
[eqn] S + 0.111… = 0.9999… + 0.111… [/eqn]
Now let’s calculate the right side: 0.9999… + 0.111… is equal to:
[eqn] \sum_{n=1}^{∞} (\frac{9}{10^n}) + \sum_{n=1}^{∞} (\frac{1}{10^n}) [/eqn]
Which is equal to:
[eqn] \sum_{n=1}^{∞} \frac{10}{10^n}[/eqn]
Which equals:
[eqn] 1 + .1 + .01 + .001 + … = 1.111… [/eqn]
So now we have this equation:
[eqn] S + 0.111… = 1.111… [/eqn]
Subtract 0.1111... from both sides:
[eqn] S = 1 [/eqn]
Sub for S:
[eqn] 0.999… = 1 [/eqn]

0.999=0.999
Class dismissed

If:

[math]0.999... = 1[/math]

Does that mean:

[math]\frac{1000... \ - \ 1}{1000...} = 1[/math]

And does that mean:

[math]\infty - 1 = \infty[/math]

.99999.... isn't even a real number.

There I said it. Repeating numbers are not real. They are just abstractions. In real life you cannot use a repeating number for any physical purpose.

You can have one rock. You can cut the rock into smaller parts. But you cannot have .33...... of a rock.

Sure you can cut the rock into thirds, but that's just what it is, a third of the rock. It's not .33......, it's just a third.

>does that mean that
[eqn] \frac{1000... \ - \ 1}{1000...} = 1 [/eqn]

Yes

If you take:

[eqn] \lim_{x \to ∞ } \frac{10^x -1}{10^x} [/eqn]

And use l'hopital's rule

[eqn] \lim_{x \to ∞ } \frac{ln(10)10^x}{ln(10)10^x} = \lim_{x \to ∞ } 1 = 1 [/eqn]

So it's 1.

>And does that mean: ∞-1=∞

That is indeterminate form, so no, a statement like is not "true" and cannot tell you anything.

.9999... is arbitrarily close to 1 (for all epislon > 0 in fact). So by definition .999... = 1

>implying abstractions aren't real
>hasn't read Ives' work on abstraction theory

Let's have a thought experiment. Let's say I gave you an object that behaved like a pen. But it was not a pen. When you looked at it, your brain received the image of a pen. When you felt it, it felt as if you were holding the exact shape of a pen. When you used it on paper, you could see what looked exactly like your writing was on the paper.

Now here is a question: how is it not a pen? How? Imagine if every single test you could possibly possibly do on this object has the same result as it would if it were a pen. You could mix it with chemicals, break it, shoot electrons at it, x ray it, and every single test that can possibly happen with always yield the same exact result as doing that same test on a pen. How then, could you possibly distinguish this object from a real pen? Well by definition I just told you that you can't, because every test wouldn't reveal anything different about the mystery object. So there is actually no way to argue that this object is not a pen.

So if 0.9999... and 0.3333... always behave exactly like 1 and 1/3 in every single test that there is, then there is nothing you can do to show that 0.333... is not equal to 1/3.

If you want to really think about abstractions, a pen is an abstraction already, so making it more abstract doesn't change anything. It's like running a certain program on a computer, vs running that program on an emulator on a computer. Are they different? Abstractly, they are the exact same.

For all and intensive prospects yes

>impying [math]\infty -1\neq\infty[/math]

...

cant balive I just missed the fucking get, I was here on 8888800, went to the bathroom and its over

Infinity can't be treated as a number, it's indeterminate. This is what happens when you try:

[eqn]∞ - 1 = ∞ [/eqn]

Subtract infinity from both sides:

[eqn]- 1 = 0 [/eqn]

>[math]- 1 = 0 [/math]

>Infinity can't be treated as a number
Sure you can, a special kind of number that acts a bit different.

Following the logic:

[math]\infty - 1 = \infty[/math]

Then:

[math]\infty - \infty = \infty[/math]

Which makes sense, if you have infinite amount of money, and you keep taking $1 out of your bank out every day for an infinite number of days, you would still have an infinite amount of money.

Infinity IS treated as a number by considering the set of real numbers together with -infinity and +infinity which act and endpoints to the real number line. This set is called the extended real numbers. Your issue is that the rules of algebra don't necessarily work the same way in this set, you might remember from highschool calculus class that infinity minus infinity is indeterminate, which is what you did there.

So, if you subtract infinity from both sides with that logic you get:

[math]-1 = \infty[/math]

That doesn't make sense.

You don't, you just get the same thing again if you subtract infinity from both sides:

[math]\infty - 1 - \infty = \infty - 1 = \infty[/math]

[math]\infty - \infty = \infty[/math]

Perfectly logical senpai. Infinity is as real as any other number.

No, there are many ways to represent infinity. Having a set containing all real numbers is one way. You have x approaching infinity in 5^x.

You can't treat all infinities as the same number because some infinities are different from other infinities. Now you know why it's indeterminate.

You're terribly confused. Try picking up a textbook on elementary analysis when you get the time to do so.

0.999... < x < 1 does not exist.

Oh look some 12 year old is on moms computer again.

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

this
hyperreals dont real

>infinity is a real number

brainlet detected