0.99r = 1

[SideWinder]

 

[SlyReaper]

They would just say it equals 0.00r1

To which the obvious answer is "you can't have a 1 in the infinityeth place, you dolt. Infinity does not work that way". The "you dolt" part is the most important part of that.

 

[lukeharvest]

Her first example

.999r = x
9.999r = 10x

Then she does -.999r on both sides, but no matter how far you repeat it, at the end it will be 1, they just round it down to suit their argument.

10 - .999r doesn't equal 9. it'll be an infinitely small number, 9.000r1 Not 9.

It isn't 10-0.999..., it's 9.999r - 0.999r. Besides, that's by far the weakest proof of 0.999... = 1 because you can question manipulating infinite decimals. A much better proof is expressing 0.999... as a sum

9(1/10) + 9(1/10)^2 + 9(1/10)^3 + ...

Which is clearly 0.999...

You can then use the sum of an infinite geometric series to get the answer as 1.

 

[ilcp]

Here we go again

Not unless a don happens to close this thread

 

[Arnie]

To which the obvious answer is "you can't have a 1 in the infinityeth place, you dolt. Infinity does not work that way". The "you dolt" part is the most important part of that.

So how would you define a number that is infinitely small but not 0?

 

[SlyReaper]

So how would you define a number that is infinitely small but not 0?

It is 0.

 

[Moothead2]

 

[RomanNose]

Or you just do
0.33333333r=1/3
0.99999999r=3/3
0.99999999r=1

 

[Randomface74]

1 - 0.99r

/thread

 

[SlyReaper]

Or you just do
0.33333333r=1/3
0.99999999r=3/3
0.99999999r=1

Well yes, but first you have to prove that 0.33r is equal to 1/3. It is, but if we're doing rigorous maths here, you still have to prove it. And it's easier to prove 0.99r = 1 than 0.33r = 1/3 in the first place.

 

[Flubble]

0.999r + 0.111r = 1

 

[lukeharvest]

Or you just do
0.33333333r=1/3
0.99999999r=3/3
0.99999999r=1

The problem with that proof, and the x = 0.999... and doing 10x-x proof is that it isn't anywhere near as rigorous as the infinite geometric sum proof, because you're multiplying an infinite decimal.

 

[SlyReaper]

0.999r + 0.111r = 1

Nope. It's 1.111r.

 

[Flubble]

Nope. It's 1.111r.

i said 0.111r

add one to start 1.xxxx all the rest shift to 0?

 

[dowie]

Her first example

.999r = x
9.999r = 10x

Then she does -.999r on both sides, but no matter how far you repeat it, at the end it will be 1, they just round it down to suit their argument.

10 - .999r doesn't equal 9. it'll be an infinitely small number, 9.000r1 Not 9.

she took .999r away from 9.999r leaving 9

.999r = x

so taking x from 10 x leaves 9x

she was left with

9 = 9x

1 = 1x = 0.999r

its not that hard to understand if you give it a bit more thought its just a different decimal representation of the same real number. You just seem thrown by the idea that not all numbers will have a unique representation within a given number system.

 

[RomanNose]

A more interesting subject, does infinity exist?

 

[SlyReaper]

i said 0.111r

add one to start 1.xxxx all the rest shift to 0?

No, you're failing at basic arithmetic here. Check it again.

 

[SlyReaper]

A more interesting subject, does infinity exist?

It's not interesting at all. Mathematically, of course it does. In the real world, not so much.

 

[lukeharvest]

A more interesting subject, does infinity exist?

Is it a number - no. Does it exist, yes.

 

[RomanNose]

There was this documentary on it

It says that some mathematicians actually don't think it exists and some think it loops back around to zero.