.9999999 = 1
 

I just learned this today in math class and I am finding it fascinating. We talked about how .999999 with a repeating decimal equals 1. And did a proof for it. It was awesome.

It was crazy are there any other things like that in math?

The current US federal budget or lack of ^^/


No but really:
Squaring a 2 digit number ending in the number 5.


65*65
1. Multiply the first digit one more than itself. (in this example one more than 6 is 7)
6*7 = 42
2. Put 25 after the answer. The answer will always end in 25.
65*65 = 4225

How is this done?
Squaring a binomial. It only works however when the number you are squaring ends in 5.

1 = ............................................1= 1*1
1+2+1 =.....................................4 =2*2
1+2+3+2+1=..............................9 =3*3
1+2+3+4+3+2+1=..................... 16=4*4
1+2+3+4+5+4+3+2+1=..............25=5*5
and so on

I like math parlor tricks.

That is pretty cool and I could care less about gdp.

I think that is the consensus from the government too.

Exactly. I feel that it really does not matter because china will not ask for their money anytime soon since it would just end up in us calling in debts which no one really has or wants to pay back.

But damn I can not believe .9999999 = 1 can be proven with a geometric sequence.

You really should be writing that with "..." to show that you mean an infinitely repeating decimal, i.e. .999... = 1. When I first saw this thread I thought you were going to mention some example of really bad roundoff error. :D

But if you think about it a bit, it does seem intuitively reasonable that 0.999... = 1. Since you can approximate 0.999... as close to 1 as you like simply by listing enough 9s after the decimal, there's no reason to suspect that the infinitely repeating decimal is not the same as 1.

And here's another way to think about it, if you remember that 1/3 has the infinitely repeating decimal 0.333.... Well, obviously 3 * 1/3 must equal 1. But if you take the repeating decimal representation of 1/3 and multiply that by 3, you would get 0.999.... Therefore 0.999.... must be equal to 1.

I'm sure the geometric sequence is a much more rigorous way to demonstrate the equality, though.

Good point with the ... I could not find a way to do a repeating decimal.

Also If you think about it 1 minus .99999.... people may try and say .000000.. with a 1 somewhere on the end but the act of placeing the 1 ends it and then it is not repeating forever.

but if you look at t as an additive like .9.../10^1 +.9.../10^2 + .9..?10^3.... then you see a pattern of .9.../10^k where the pattern begins at k=1 and continues indefinitely then you can represent it as .9.../10^2/.9.../10^1 which equals 1/10 and if you plug that r value into the formula .9.../1-r you get .9.../.9... or 1.

I really headache with mathematics

Some difficult math concepts come easily to me and then I will have trouble with basic things like factoring and forgetting to use order of operations. I just need to slow down and think more with it.

You can easily define a math where the infinitesimal exists, and consequently show that .999... doesn't equal 1. Proof of nonexistence is only valid within the relevant universe of discourse, and it is unfortunate that you went through a proof without defining its limitations.
My own opinion on the matter is that if your math contains pi, 0.999... doesn't equal 1.

That is actually a unanimously accepted proof, it is just a fact. I think you might not believe at first sight because you are under the notion that any given number has only one decimal expansion, which is not the case. It is as easy to prove in base-3 as it is in base-10. Defining pi would have nothing to do with this, as it only shows that pi is defined and you can represent it with decimal notation. I could define pi as a continued fraction if I like or from bessel functions like Ramajan did: these represent the same number, but will likely have differing based off of convergence.

As a challenge, if .999... is not equal to 1, what is the difference of the two of them (i.e., what is 1-.999...). Recognizing what

lim 1/(10^N) is as N approaches infinity should help you in this endeavor.

If I went with nanicoar then I presume you would tell me 1 - .999... = what? .oooooooooo...1? It would not be a valid answer ever to throw a 1 at the end of the repeating 0's because this would end the whole sequence of endlessly repeating 0's.

At the same time it can be so close to 1 by infinity that the human mind can't distinguish it from 1.

I never used pi.

A pretty old one but quite a good if you consider this uses only basic algebra.
a=b
a² = ab
a²+ a² = a² + ab
2a² =a² + ab
2a² - 2ab = a² + ab - 2ab
can also be written as:
2(a² - ab) = 1(a² - ab)
cancelling (a² - ab):
2 = 1

Now that is some cosmic shit right there.

Wait. You divided by 0. Invalid answer.

that's right^^ but for such a simple task not too bad huh?
there are some other, way longer versions of it so in basic check every step if it's right but don't keep in mind that a=b in the beginning.

I often forget that removing the same thing from both sides or cancelling out is division. I just see it as cancelling out like it is its own thing. But it would be perfect otherwise.
I do not think it can be allowed because people's minds would be literally blowing.