Why does .99999999... = 1?

LaVoS

Senior member
Aug 1, 2001
224
0
0
i could be wrong but look at this... seems right to me...Assign 0.99999... as the variable x.
x = 0.99999...

Multiply x by 10.

10x = 9.99999...

Subtract x from 10x and solve for x.

10x = 9.99999...
-x = -0.99999...
-----------------
9x = 9
x = 9/9
x = 1

Didn't we assign x as 0.99999...? Yes, we did. Then how did the
answer magically change to 1? It's not magic. It's just that
0.99999... was equal to 1 all along.

k? theres that one AND this one...

go to Calculator on your computer (or a normal calculator its up to you)

type in
1/3 (use fractions to decimles o get .33333333...)
then multiply that by 3
and you will get 1.... i thought .33333333333... * 3 = .99999999999...
why? i think that math is screwed up.... cos .9999999999 clearly does not = 1...

 

Kelvrick

Lifer
Feb 14, 2001
18,438
5
81
What grade are you in? In calculator, its an approx because then the repeating .3333 represents 1/3, and 1/3 *3 = 1.
 

Mucman

Diamond Member
Oct 10, 1999
7,246
1
0
You have to remember it is 0.9999... not just 0.9999

There is a difference :)

I remember seeing a good proof of it in the highly technical forum :)
 

Shantanu

Banned
Feb 6, 2001
2,197
1
0
It has to do with computer's general inability to handle vulgar (common) fractions, unless a specific instruction set is written telling it how to handle such fractions. Best not to worry about it too much. Better yet, know the math yourself so you won't be dependent on your calculator so much.
 

Martin

Lifer
Jan 15, 2000
29,178
1
81
10x-x = 9x.

x = 0.999999

therefore

9x = 9*0.99999 = 8.9999991

8.9999991 /9 = 0.999999

your method is wrong.
 

geno

Lifer
Dec 26, 1999
25,074
4
0


<< type in
1/3 (use fractions to decimles o get .33333333...)
then multiply that by 3
and you will get 1.... i thought .33333333333... * 3 = .99999999999...
why? i think that math is screwed up.... cos .9999999999 clearly does not = 1...
>>


That's all in the code of the Calculator. It retains that you typed in 1/3, and squares that. Type in .333333333333333333333333 (as far as it goes) and square that, watch what it looks like.
 

Martin

Lifer
Jan 15, 2000
29,178
1
81
for the second one, most calculators can't handle fractions. Mine can and answer comes to 1. as long as you dont use finite decimals to represent fractions, you wont get anything screwed up.
 

wildwildwes

Senior member
Jul 18, 2001
320
0
0


<< i could be wrong but look at this... seems right to me...Assign 0.99999... as the variable x.
x = 0.99999...

Multiply x by 10.

10x = 9.99999...

Subtract x from 10x and solve for x.

10x = 9.99999...
-x = -0.99999...
-----------------
9x = 9 <--------- 9x is not 9, it is 8.99991
x = 9/9
x = 1

Didn't we assign x as 0.99999...? Yes, we did. Then how did the
answer magically change to 1? It's not magic. It's just that
0.99999... was equal to 1 all along.

k? theres that one AND this one...

go to Calculator on your computer (or a normal calculator its up to you)

type in
1/3 (use fractions to decimles o get .33333333...)
then multiply that by 3
and you will get 1.... i thought .33333333333... * 3 = .99999999999...
why? i think that math is screwed up.... cos .9999999999 clearly does not = 1...
>>

 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
From the view point of abstract mathmatics, and the world of Real Analysis.
.99... =1 Where .9... represents an infintly repeating string of 9s.

There are many ways to prove this. Above 3* 1/3= 1 and the algebraic addition is also valid, both point to the truth.

This link shows what equality means to a mathematician.

Did you understand that?

Here is a proof using the fact that a geometric series has a sum.

Here is a proof that uses the concept of equality as stated above.

Now these proof are good and fine if you have sufficient math to understand them. IF you are a high shcool student who has yet to be exposed to high mathematics, Please look at these and attempt to understand them. When you do calculus you will see much more of this type of proof, get used to it.

IF you are of mindset which simply cannot accept the concept then please read every post in the link provided by Viper GTS above. Then feel free to post your opinion.
 

MustPost

Golden Member
May 30, 2001
1,923
0
0
LINK To HT forum thread.
Im not exactly sure why it needed 10 pages. I stopped reading it sometime whenthere were two pages of the thread there.
 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
BTW No calculator or computer can deal with an infinte string of digits. They all have a fixed word length which limits the accuracy of the result. To see this simply create a loop with a single precision variable and increment it by .1 and display the total each time though the loop. It is surprising how quickly the round off errors show up.
You may ask why is the sum of x = x + .1 being rounded! Good question. The sad fact is that when .1 is converted to binary fraction by your binary computer it become an infinitly repeating base 2 number.

.1 (base 10) = .000110011001100... (base 2) it must be truncated at the computers word length.
 

yoda291

Diamond Member
Aug 11, 2001
5,079
0
0
<goes catatonic mumbling "mommy" make the nines stop">



<< 10x = 9.99999...
-x = -0.99999...
-----------------
9x = 9 <--------- 9x is not 9, it is 8.99991
x = 9/9
x = 1
>>



Wild, you have got to be kiddin me. take 9+x-x and trust me, you'll end up with 9.

I got tired of it around page 8 or 9 on the old thread
 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0


<< what if the way we humans understand mathmatics is fundementaly flawed? >>



We created it, so I would think we understand it. It is not a case of flawed understanding, it is an artifact of the construction of the Real number line.
 

MF1

Senior member
May 29, 2000
298
1
0
Marty and wild,

LaVos set x=0.99999... meaning that 9 is repeated infinitely.

10x = 9.99999... ; notice "..."


10x - x = 9.99999... - 0.99999...
9x = 9 ; notice that "..." is gone
x = 1
 

b0mbrman

Lifer
Jun 1, 2001
29,471
1
81
Proof is too easy.

1/3 = 0.333...
2/3 = 0.666... (mark of the beast :< )
1/3 + 2/3 = 0.333... + 0.666...

0.333... = 0.3 + 0.03 + 0.003 + ...
0.666... = 0.6 + 0.06 + 0.006 + ...
(0.3 + 0.03 + 0.003 + ...) + (0.6 + 0.06 + 0.006 + ...) = 0.999...
0.333... + 0.666... = 0.999...

1/2 + 2/3 = 0.999...

1/2 + 2/3 = 3/3
3/3 = 1

1 = 0.999...

Probably way longer than it had to be but i got there...
 

GoldenBear

Banned
Mar 2, 2000
6,843
2
0
In calculus we almost never find the exact number, but rather it's a strong approximation. Hence the concept of limits.

You can also prove the .9999... stuff through the infinite sum equation:

S = a/(1-r) - since to get .9999... you add .9 to .09 to .009 to .0009 and such.

a = .9 and r = .1

.9/(1-.1)

= .9/.9 = 1

I didn't bother reading those links so if this is posted already, eh..
 

SinNisTeR

Diamond Member
Jan 3, 2001
3,570
0
0
its impossible, the reason you get that in the calculator is because of the flaw in the calculator. this is so simple all you need is common sense.