.999~ = 1

Just think of a function that always adds a 9. (0.9, 0.99, 0.999, ...). Running it infinitife times, it's limit will be 1. Working with infinity is always a little bit weird. Just remember that infinity can't be used as some incredible huge number, you'll have to follow other rules.

Other example:

(1+(1/n))^n will be e for n->infinity, but not for any huge number you type.

And yeah, another reason is that no pc or whatever will be able to use the number 0.9999...to infinity, it's limited.

Vex it depends on which binary mode e.g. Hexidecimal: 0001=1, 0010=2, 0011=3, 0100=4, 0101=5, 0110=6, 0111=7, 1000=8, 1001=9, 1010=A(10), 1011=B(11), 1100=C(12), 1101=D(13), 1110=E(14), 1111=F(15). thats what i use in hexi with binary= 256, 128, 64, 32, 16, 8, 4, 2, 1 and when converting TO hexi you would use [8, 4, 2, 1]

DD_Draco, you didn't notice the decimal point. another example, in base 5: 0.44444~ = 1. (binary is actually base 2)

if we take for instance:
1 - 0.9 = 0.1 = 1/10
1 - 0.99 = 0.01 = 1/100
1 - 0.999 = 0.001 = 1/1000

if we follow this pattern we get:
1/infinity, now just try to imagine what that number is. its small isn't? in fact it is so small it may as well be zero. it makes more sense that way.

put that into the pattern we get:
1 - .999~ = 0
1 = .999~

-----------------------------------------------------

and while we are on the topic of maths problems, check tis out:

WARNING! not for those without senior maths education or those with heart conditions, pregnancy, spinal injuries or been under operation in the last 48 hours.

a = b

Multiply by a
a^2 = a * b

Subtract b^2
a^2 - b^2 = a * b - b^2

Distribute and Difference of two squares
(a + b)*(a - b) = b*(a - b)

Divide by (a - b)
a + b = b

Substitue a for b since a = b
a + a = a

Divide by a
1 + 1 = 1

Simplfy
2 = 1

It's a classic. Though I knew it with a*a=a*a

Anyway, the problem was that you cannot divide with (a-b) if (a-b)=0 and same with a (not sure about the second one though). Anyway, some conditions I'm too lazy to check right now.

~Daelin

i noticed the decimal point, even with it being infinate it will never be a whole numeral it could go from 0.444~ to 0.999~ but never reach 1 unless you added the required number (e.g. again 0.999~+0.0~1=1).

Re: .999~ = 1

Afronight_76.

10(.999~) - .999~ = ?

10 * .999~ = 9.999~

9.999~ - .999~ = 9

Those are the wrong step.

This is the fact:
10(x) - x = 9(x)

so,

10(.999~) - .999~ != 9 //not equal

but

10(.999~) - .999~ = 9(.999~)

9(.999~) = 8,999~ != 9

Then,

1 != .999~

10(.999~) - .999~ = 9(.999~)

9(.999~) = 8,999~ != 9

Then,

1 != .999~

aright i understod all you postet, leopard. exepting the ! thing. can you explein?

Mastermindaright i understod all you postet, leopard. exepting the ! thing. can you explein?

!= means not equal to.

heres some more evidence that .999~ = 1.

1/9 = .111~
2/9 = .222~
3/9 = .333~
...
...
7/9 = .777~
8/9 = .888~
therefore, 9/9 = .999~. but we know that 9/9 simplifies to 1.

therefore .999~ = 1. case sovled.

Xeridanus
1/9 = .111~
2/9 = .222~
3/9 = .333~
...
...
7/9 = .777~
8/9 = .888~
therefore, 9/9 = .999~. but we know that 9/9 simplifies to 1.

therefore .999~ = 1. case sovled.

There are missing a very small value on that statements, so .999~ = 1. How about this:

1/9 = .11111111111~~~~~~1
8/9 = .88888888888~~~~~~9
-------------------------- +
9/9 = 1.00000000000~~~~~~0

However, counting in floating number is a lil bit inaccurate. That's why we sometimes ignore a very small value.

yes, i completely agree with you. .999~ = 1