Question

Digit manipulation proof, elucidate me.

\[0.999... = 1\]

I understand how the infinite series and sequences proof works however, why is the digit manipulation proof valid?

\[x = 0.999... \\ 10x = 9.999... \\ 10x = 9 + 0.999... \\ 10x = 9 + x \\ 9x = 9 \\ x = 1 \]

I will expand in the thread below.

Answer 1

We already know that \(x = 0.999...\).
Step 2 where \(10x = 9.999...\) should not come after:
\(10 + 0.999... = 9.999...\)

Not what ever they did?

Answer 2

As in we know what \(x\) is, what are we doing by manipulating it?

Answer 3

10*0.99999.....= 9.9999999.....

Answer 4

10+.99999= 10.999999

Answer 5

Oops silly my mistake.

Answer 6

:)

Answer 7

However why are we solving for \(x\) since we already know what the value is?

Answer 8

where did you read this proof?

Answer 9

@zzr0ck3r here:
https://en.wikipedia.org/wiki/0.999...#Digit_manipulation

Answer 10

tell them to prove that 10*.9999.... = 9.99999....

Answer 11

thats a standard trick for converting repeated decimals to fractions

Answer 12

@ganeshie8 care to share this standard trick?

Answer 13

if you want more practice, try converting \(0.123123123123123\ldots\) to fraction

follow the exact same steps :)

Answer 14

i think this video explains it better than wiki
https://www.khanacademy.org/math/algebra/solving-linear-equations-and-inequalities/conv_rep_decimals/v/coverting-repeating-decimals-to-fractions-1

Answer 15

http://www.mathsisfun.com/9recurring.html

Answer 16

However this step: \(9x = 9\)
We know that \(x = 0.999...\)
\[9x = 9 \\ 9 * 0.999... \ne 9\]
It should return a value less than 9 right?

For example: \[9 * 0.99 = 8.91\]

Answer 17

9x = 8.999... = 9

Answer 18

Round up dat figure ;)

Answer 19

\(0.99999999999999999999 \ne 1\)

however

\(0.999\ldots = 1\)

Answer 20

I know that to be true via the infinite series and sequences proof. However how is the current proof true?

Answer 21

Orange = Black

Answer 22

after reaching 9x = 9
you move ahead, dont look back. divide 9 both sides

Answer 23

if you look back, you will get a repeating decimal again and you will be in circles

Answer 24

\[x = 0.999... \\
10x = 9.999... \\
10x = 9 + 0.999... \leftarrow here\\
10(0.999) \ne 9 + 0.999... \\
10x = 9 + x \\
9x = 9 \\
x = 1\]

Never mind that, so the proof can only be read forward not backward? Fascinating.

Answer 25

you read it back wards if you want to prove that 1 = 0.999999....

Answer 26

Prove that an untrue statement is true o-o

Answer 27

there is a mistake here :

\[
x = 0.999... \\
10x = 9.999... \\
10x = 9 + 0.999... \leftarrow here\\
\color{red}{10(0.999) \ne 9 + 0.999... \\}
10x = 9 + x \\
9x = 9 \\
x = 1
\]

Answer 28

both sides will be equal if you replace "0.999" by "0.999..." on left hand side

Answer 29

\[1:x = 0.999... \\
2:10x = 9.999... \\
3:10x = 9 + 0.999... \leftarrow here\\
4:10(0.999) \ne 9 + 0.999... \\
\text{Step 4 makes no sense since we know from the beginning that x = 0.999...} \\
5:10x = 9 + x \\
6:9x = 9 \\
7:x = 1\]

Oh wait I see my mistake, I jumped, I took from step 1 when I was working backwards instead of taking from step 7.

Thank you @ganeshie8 it makes sense now.

Answer 30

looks good!

Answer 31

food for thought :

...999999999999999999999
+1
--------------------------
...000000000000000000000

Answer 32

it may not be that exciting to you as you're good with computers and programming

Answer 33

Is repeating numbers any different than the concept of infinity?

Answer 34

i think it can be explained easily using overflow/carry...

Answer 35

perform below operation on a 4 bit computer and see the contents of accumulator :

```
ADD F 1
```