.999~ = 1

[Afronight_76]

.999~ = 1

Proof?

Let an equation be 10(x) - x = y

10(1) - 1 = 9

Agree with this statement? Let's substitute 2 for that statement and see what we get.

10(2) - 2 = 18

So obviously every input will get a different output.

Let's substitute in .999~

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

10 * .999~ = 9.999~

9.999~ - .999~ = 9

The exact same answer for 1 and .999~ so we know that .999~ = 1

im still wondering how this is still possible

 

[VGsatomi]

Thats why rounding is G-d's gift to mathmaticians....
-VGsatomi

 

[DeadDogNoGoWoof]

.999~ = 1

Proof?

Let an equation be 10(x) - x = y

10(1) - 1 = 9

Agree with this statement? Let's substitute 2 for that statement and see what we get.

10(2) - 2 = 18

So obviously every input will get a different output.

Let's substitute in .999~

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

10 * .999~ = 9.999~

9.999~ - .999~ = 9

The exact same answer for 1 and .999~ so we know that .999~ = 1

im still wondering how this is still possible

Damn you!

 

[Ki]

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

10 * .999~ = 9.999~

Explain how you did that.

 

[VGsatomi]

ah, thats right....

.999 x 10= 9.990

 

[Ki]

Didn't Blizzard try to pull something similar to this on April Fools Day?

 

[Sansui]

let ... = the last integer repeating/


1/3 = .333...

2/3 = .666...

1/3 + 2/3 = .999...
1/3 + 2/3 = 3/3

and 3/3 = 1 and .999...

 

[Ki]

That is a good question.

I think it has to do with the fact that 100 can't be divided by 3. When you say 1/3 you are dealing with a perfect "scale" (figuratively speaking). I guess technically 1/3 wouldn't be .333 extended; it's just as close as it will get.

If you ever put together a puzzle and find a piece that doesn't fit exactly but relatively close, I think you can squeeze it in.

 

[Daelin]

That's because 10/3=3.(3). That's equal to 3.33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333

And so, we lose a 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 or whatever, the 1 being somewhere close to infinite. I think something like this should be possible: 0.(0)1

~Daelin

 

[Jacek]

It is well known that every mathematical equation can't be solved for sure (i mean that 1/3 in computers is 0.3333333333333333333333......4 and that is wrong, but it is only way to calculate something. You can't use 0.3333... just 0.3333...4.

 

[Goliath the God]

Jeez, since when did we get interested in mathmatics?

0.333 , is not the 1/3 of 1.. so just forget it.

0.999 is not = 1.. It's Close to, but not Quite!


That's simply why you cant divide 1 by anything else than Pair Numbers, or itself. :roll:
While .333, .666 and .999 can be divided by only 1, 3, the double of itself or itself.
:roll:

 

[Capcy]

hmmm MATHS my most popular subject... Okay you want to find if 0.999~ = 1 well lets see:

Your formula is:
10(0.999~)-0.999~=1
hmmm i find this to be wrong.
Reason being:
0.999~ being infinate is still 0.0~1 off from reaching the whole numeral being 1 all it needs is 0.0~1 to make itself a whole number.

heres a way to test it:
0.999+0.001=1

Hope this is helpful.

Regards,
Damned-Dragon

 

[olofmoleman]

*gets stuck on "dude wtf?!? mode"*

 

[Daminon]

I have always found math very hard but somehow I sometimes finds it enterresting as long as it's not about solving some equation.

This whole thing of the 1/3 is possible depends on how you want the awnser to be. In float it has infinity decimals but sometimes you want it to be in fractional number and in that case (1/3)*3=1. Then we can talk about roots and pow... 0.5root(10)=100

And for those who didn't know: Sqrt(-1)=i and sqrt(i)=Sin(45)+iCos(45). I like imaginary vaules. :mrgreen:

 

[Vexorian]

0.9... to infinity is just a theory based on the fact that 0.3... to infinity existed, in practice there is no such a number like 0.9... that's why when you try to get that number you get 1 as result.


And this is merely caused by the base we use, 10. in binary 0.1... to infinity equals 1

 

[DarkShadow]

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.

 

[Capcy]

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]

 

[Xeridanus]

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

 

[Daelin]

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

 

[Capcy]

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).

 

[Leopard]

.

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~

 

[Mastermind]

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? :?

 

[Blade.dk2]

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

!= means not equal to.

 

[Xeridanus]

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.

 

[Leopard]

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.

 

[Xeridanus]

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