0.9 repeater = 1

SquishyDave · Jul 9, 2003

Ok 0.9 repeater equals one, thats standard stuff, so who can show me a proof for it?

I already know the simplelest proof, but there are others, and I am interested in seeing what they are.

And remember 0.9 repeater equals EXACTLY 1. Don't think I mean it's close enough to make no difference, coz that's wrong.

rwald · Jul 9, 2003

Well, there's the handy fact that if you have 0.abc...n with the whole thing repeating, is has the same value as (abc...n)/(999...9). So 0.9 repeating = 9/9 = 1.

That's all that I know of.

SquishyDave · Jul 9, 2003

rwald said:
Well, there's the handy fact that if you have 0.abc...n with the whole thing repeating, is has the same value as (abc...n)/(999...9). So 0.9 repeating = 9/9 = 1.

That's all that I know of.

Great thanks, but I didn't follow you the whole way through that.

Ummm I get the first part of what you're saying, I think, but can you give an example using another number apart from 0.9 repeater? Or fill it out a bit more for us slow types?

Yahweh · Jul 9, 2003

0.9999999999999999 = 1

Rounding to 1 significant figure is sufficient enough for me.

SquishyDave · Jul 9, 2003

Yahweh said:
0.9999999999999999 = 1

Rounding to 1 significant figure is sufficient enough for me.

agreed, sufficient enough for me too, but there are at least 2 and possibly more mathematical proofs that show 0.9 repeater (thats infinitely repeated) does exactly equal 1.

The one I know is that

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

1/3 = 0.3 repeater

so

0.3 repeater + 0.3 repeater + 0.3 repeater = 0.9 repeater and 1

Brown · Jul 9, 2003

As has been mentioned, everyone agrees that 1/3 + 2/3 = 1, but if you add the decimal equivalents of 1/3 and 2/3, you get
0.33333... + 0.66666... = 0.99999...
so 0.99999... must be equal to 1.

Here's another proof.

Let x= 0.99999...
Then 10x = 9.99999...
and 10x - x = 9.99999... - 0.99999...
or 9x = 9
So x = 1. But, as defined above, x is also equal to 0.99999...
So 1 = 0.99999...

rwald · Jul 9, 2003

Here's another example of what I was saying: 0.45 repeating (0.45454545...) = 45/99. That's what my rule means.

Of course, you could ask where I got that rule from, and I honestly don't remember. I once knew a way to turn any repeating decimal (including ones where the first digit doesn't repeat) into a fraction, but I've forgotten. Someone else remember?

[edit]I just read Brown's proof, above. I think we should just go with that...[/edit]

SquishyDave · Jul 9, 2003

Brown said:
Let x= 0.99999...
Then 10x = 9.99999...
and 10x - x = 9.99999... - 0.99999...
or 9x = 9
So x = 1. But, as defined above, x is also equal to 0.99999...
So 1 = 0.99999...

Beautiful

thanks Brown. That's nicely elegant.

rwald said:
0.abc...n with the whole thing repeating, is has the same value as (abc...n)/(999...9). So 0.9 repeating = 9/9 = 1

0.45 repeating (0.45454545...) = 45/99

Also wonderful rwald, I tried it out a few times and understand it now, that's three different proofs.

Does anyone know if there are anymore?

[edit]Sure rwald, his looks more elgant, but don't be hard on yourself, yours holds true aswell, people often disbelieve when I tell them this, so the more proofs I can burry them with the merrier[/edit]

rwald · Jul 9, 2003

Here's another one to try. I'm going to

Code:

 it, so that the columns will line up clearly.

[code]Let's let 1 = x and 0.999... = y
Take the difference x - y, one term at at time:
  1.000...
- 0.999...
----------

Do the first term:
  1.0
- 0.9
-----
  0.1

Do the next term:
  0.10
- 0.09
------
  0.01

And the next term:
  0.010
- 0.009
-------
  0.001

Notice a pattern? Carry it to infinity, and you get:
  1.000...
- 0.999...
----------
  0.000...

So, if x - y = 0, then x = y and 1 = 0.999...

SquishyDave · Jul 9, 2003

oohhh good one.

That's four, they won't know what hit em.

Nucular · Jul 9, 2003

Right, I'm not a mathematician. Far from it. So, not being a mathematician, I had kind of heard rumours of these sorts of things (x=y stuff), but not really paid them too much attention, assuming they were simple tricks relying maybe on different bases or, you know, weird ways of counting or something.

But to see these proofs of a very simple x=y statement, even though it's just replaced my previous 'x pretty much=y' assumption, is amazing.

What is the significance of this? In terms of logic and the universe? I've got no time to think at the moment, but it feels like the significance is huge.

Edited to remove an errant extrapolation

BillyJoe · Jul 9, 2003

Nucular said:
What is the significance of this? In terms of logic and the universe? I've got no time to think at the moment, but it feels like the significance is huge.

Surely its just a trick?

Or perhaps incomplete logic......

The statement,

1 = 0.99999999999999999999999999999.....

is only approximately true or true only when you reach infinity which is.....never!

regards,
BillyJoe

whitefork · Jul 9, 2003

Sum of a geometric series

from http://www.jimloy.com/algebra/9999.htm

1/9 = .1111.....

because the series 1/n + 1/n^2 + 1/n^3 ..... = 1/(n-1)
and if n = 10, the 1/10 + 1/100 + 1/1000 .... = .11111.....
so .111.... = 1/(10-1) = 1/9

and

9 * 1/9 = 1

9 * .1111..... = .999999

therefore .9999..... = 1

Nucular · Jul 9, 2003

Isn't that last one the same as SquishyDave's first one, only with ninths instead of thirds? Presumably it could work for any fraction?

And I don't really see how it can be 'just a trick', as the numbers are clearly there, unambiguously contradicting each other.

Can you use a more complicated version of the same proofs to do it with other numbers? Could you eventually prove, say, that 1=2?

Edited to add: and surely it's not only true when you reach infinity, which is never - the proofs posted above don't require infinite calculations (well, the one based on the separate terms might). They just demonstrate that you do the calculation one valid way and it equals 0.999..., and you do it another valid way and it equals 1.

ceptimus · Jul 9, 2003

I don't think it's a trick. It's just that the decimal notation we have invented, and use for representing numbers, has more than one way of representing the same value.

Nobody gets exited about 1 = 1.0 = 1.00 = 1.000 but that's four different ways already of writing the same value.

ceptimus.

ceptimus · Jul 9, 2003

Here's the classic

let a = b

[Multiply both sides by a]
a² = ab

[Add (a² - 2ab) to both sides]
a² + a² - 2ab = ab + a² - 2ab

[Factor the left, and collect like terms on the right]
2(a² - ab) = a² - ab

[Divide both sides by (a² - ab)]
2 = 1

ceptimus.

Paul C. Anagnostopoulos · Jul 9, 2003

Isn't the correct statement that the limit of 0.999... is 1?

~~ Paul

BillyJoe · Jul 9, 2003

Nucular said:
....and surely it's not only true when you reach infinity, which is never - the proofs posted above don't require infinite calculations (well, the one based on the separate terms might). They just demonstrate that you do the calculation one valid way and it equals 0.999..., and you do it another valid way and it equals 1.

(abc...n(GETS THERE AT INFINITY = NEVER)/(999...9). So 0.9 repeating(GETS THERE AT INFINITY = NEVER) = 9/9 = 1.

0.3 repeater(GETS THERE AT INFINITY = NEVER) + 0.3 repeater(GETS THERE AT INFINITY = NEVER) + 0.3 repeater(ETS THERE AT INFINITY = NEVER) = 0.9 repeater(GETS THERE AT INFINITY = NEVER) and 1

0.33333...(ETS THERE AT INFINITY = NEVER) + 0.66666...(GETS THERE AT INFINITY = NEVER) = 0.99999...(GETS THERE AT INFINITY = NEVER)
so 0.99999...(GETS THERE AT INFINITY = NEVER) must be equal to 1.

Let x= 0.99999...(GETS THERE AT INFINITY = NEVER)
Then 10x = 9.99999...(GETS THERE AT INFINITY = NEVER)
and 10x - x = 9.99999...(GETS THERE AT INFINITY = NEVER) - 0.99999...(GETS THERE AT INFINITY = NEVER)
or 9x = 9
So x = 1. But, as defined above, x is also equal to 0.99999...)ETS THERE AT INFINITY = NEVER)
So 1 = 0.99999...(GETS THERE AT INFINITY = NEVER)

In other words. 0.999999999999999...... NEVER equals 1

regards,
BillyJoe

BillyJoe · Jul 9, 2003

ceptimus

ceptimus said:
Here's the classic

let a = b

[Multiply both sides by a]
aÂ² = ab

[Add (aÂ² - 2ab) to both sides]
aÂ² + aÂ² - 2ab = ab + aÂ² - 2ab

[Factor the left, and collect like terms on the right]
2(aÂ² - ab) = aÂ² - ab

[Divide both sides by (aÂ² - ab)]
2 = 1

ceptimus.

[Divide both sides by (aÂ² - ab)] ???

aÂ² - ab = 0

Cannot devide by zero.

Also....DON'T CHANGE THE SUBJECT

regards,
BillyJoe

BillyJoe · Jul 9, 2003

Paul C. Anagnostopoulos said:
Isn't the correct statement that the limit of 0.999... is 1?

Yes, and the limit is NEVER reached so 0.999... NEVER equals 1.

Okay, I have one friend.

ANYONE ELSE?

0.9 repeater = 1

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