# 0.999... = 1

#### Andrej64

Registered Member
Take a look at this series of equations:

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1
0.999... = 1

"999..." means infinity of digit "9".

Is there any mistake or does that mean that 0.999... really equals 1? If so, is there any point in using 0.999...?

Is there any mistake or does that mean that 0.999... really equals 1?
Yes it does! Take a look!

1/9 = .1111111111......
2/9 = .2222222222.......
3/9 = .3333333333......
....
....
7/9 = .77777777777.....
8/9 = .88888888888....
9/9 = .99999999999.... = 1

interesting

Every infinitely repeating decimal has a value that can be expressed more compactly in fractional notation.

.142857142857142857..... can be more compactly expressed as 1/7.

.2323232323..... can be more compactly expressed as 23/99.

.999999999.... can be more compactly expressed as 9/9, as Zeno said. It can also be expressed as 1/1.

You could think of it this way, too:

It seems reasonable that 1-0.999... = 0.000...0001 where there are an infinite number of zeros.

So this number is essentially 1/10<sup>∞</sup> = 0.

So 0 = 0.000...001

0 + 0.999... = 0.000...001 + 0.999..999
0.999...999 = 1

Ok, so I made that up on the spot.

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Let´s take a & B and let it be that t = a + b. Then...
a + b = t
(a + b)(a - b) = t (a - b)
a2 - b2 = t*a - t*b
a2 - t*a = b2 - t*b
a2 - t*a + t2/4 = b2 - t*b + t2/4
(a - t/2)2 = (b - t/2)2
a - t/2 = b - t/2
a = b

Mathematics can prove anything, don´t you just love it!

If f and g are 2 derivable functions, then (fg)'' = (f'g + fg')' = f''g + f'g' + f'g' + fg''. (f' is the derivative of f, f'' is the second derivative, fg is f times g).

Notice that the term f'g' appears twice. We all agree that the two f'g's are the same, right?

Well let's calculate the second derivative of x<sup>7</sup>e<sup>x<sup>2</sup></sup> with respect to x, with f(x) = x<sup>7</sup> and g(x) = </sup>e<sup>x<sup>2</sup></sup> (or e^(x^2)).

(f(x)g(x))'' = (7x<sup>6</sup>e<sup>x<sup>2</sup></sup> + 2x<sup>8</sup>e<sup>x<sup>2</sup></sup>)' = 42x<sup>5</sup>e<sup>x<sup>2</sup></sup> + 14x<sup>7</sup>e<sup>x<sup>2</sup></sup> + 16x<sup>7</sup>e<sup>x<sup>2</sup></sup> + 4x<sup>9</sup>e<sup>x<sup>2</sup></sup>

Since we already agreed that f'g' = f'g', we have that 14x<sup>7</sup>e<sup>x<sup>2</sup></sup> = 16x<sup>7</sup>e<sup>x<sup>2</sup></sup> for all real values of x, and therefore 14 = 16.

I got a whole bunch of "proofs" like this during one very cool maths course at university.

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Naat said:
Let´s take a & B and let it be that t = a + b. Then...
a + b = t
(a + b)(a - b) = t (a - b)
a2 - b2 = t*a - t*b
a2 - t*a = b2 - t*b
a2 - t*a + t2/4 = b2 - t*b + t2/4
(a - t/2)2 = (b - t/2)2
a - t/2 = b - t/2
a = b

a-t/2 = (a-b)/2
b-t/2 = (b-a)/2

Nice try

Yea, and your derivative is just wrong. Lol. Where the hell did you get 16x<sup>7</sup>e<sup>x<sup>2</sup></sup>?

Absane said:
Yea, and your derivative is just wrong. Lol. Where the hell did you get 16x<sup>7</sup>e<sup>x<sup>2</sup></sup>?
What do you think the derivative of 2x<sup>8</sup>e<sup>x<sup>2</sup></sup> should be?

Andrej64 said:
Take a look at this series of equations:

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1
0.999... = 1

"999..." means infinity of digit "9".

Is there any mistake or does that mean that 0.999... really equals 1? If so, is there any point in using 0.999...?

The only thing I can spot with this argument, although fun, easy, and correct, the main assumption I see with this is that 0.999...999 is a number we operate with like we might 1 or Pi. Same probably goes for my argument, too. I remember seeing a technical proof of it at one point but it involves plenty of calculus.

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przyk said:
What do you think the derivative of 2x<sup>8</sup>e<sup>x<sup>2</sup></sup> should be?

(16x^7)*e^x^2 + (4*x^9)*e^x^2

I'm too lazy for "sup" right now :/

Good so far - those are the third and fourth terms of (fg)''...

Well.. I did not actually go and calculate dy/dx or anything. Perhaps if I feel like it I will do something about it. Paper and pencil are too much for me tonight. For now, I will just assume 14=16

As a corollary, -2 = 0 = 2 and, well, all the rational numbers are equal to each other, and consequently, the reals, complexes, quaternions, matrices. Equality for all, demonstrated through mathematics. No one can be richer than anyone else if everyone's income is equal. No one can be more popular than anyone else if everyone's number of friends is equal. The great quest begun by Einstein in search of a unified theory in physics is finally at an end. The universe is zero, and one, and all other digits. It is simultaneously everything, and nothing.

We must start rewriting the textbooks. There is much work to be done.

You would be amazed at what you can do if you just change the assumptions on the reals. Changing them to see what happens is quite interesting. I decided to make 1 the identity element of addition and all was fine. I changed the assumption that 0 != 1 and all that did was collapse the real system, but as far as I can remember all the other assumptions played out well for this change

Naat said:
Let´s take a & B and let it be that t = a + b. Then...
a + b = t
(a + b)(a - b) = t (a - b)
a2 - b2 = t*a - t*b
a2 - t*a = b2 - t*b
a2 - t*a + t2/4 = b2 - t*b + t2/4
(a - t/2)2 = (b - t/2)2
a - t/2 = b - t/2
a = b

(x)^2 = (-x)^2
x = -x

Cool.

Naat said:
Let´s take a & B and let it be that t = a + b. Then...
a + b = t
(a + b)(a - b) = t (a - b)
Naat,

You multiplied the whole equation by (a - b), and this could be done only if (a - b) <> 0.
a2 - b2 = t*a - t*b
a2 - t*a = b2 - t*b
a2 - t*a + t2/4 = b2 - t*b + t2/4
(a - t/2)2 = (b - t/2)2
a - t/2 = b - t/2
a = b
which means (a - b) = 0 and so the second step is invalid.

Andrej64 said:
(x)^2 = (-x)^2
|x| = |-x|

In my earlier quote i made it bold the last 3 steps of 'Naat" and followed with what would happen in that way.