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IMO, .999... is not 1:

So, what is .9999...? for fractions over 1 it is (0+.9999...)/1 for fractions over 10 it is (9+.99999...)/10 etc for other fractions. So, you can't compute .9999... without using .9999... and adding it to another number. So .9999... will always be 1-.(infinite0's)1 which is not 1. As for computations a computer can only hold a finite set of .9's so .999... is 1-.(maximum 0's - 1)1 which is even less of 1 than .9999...

mathematician wrote:Speaking as a mathematician, you can take it from me that 0.999999 recurring does equal one. The differential and integral calculus, for a start, would come crashing down around our ears if it didn't.

And it is certainly not an imaginary number.

Bravo

Brackets wrote:IMO, .999... is not 1:

So, what is .9999...? for fractions over 1 it is (0+.9999...)/1 for fractions over 10 it is (9+.99999...)/10 etc for other fractions. So, you can't compute .9999... without using .9999... and adding it to another number. So .9999... will always be 1-.(infinite0's)1 which is not 1. As for computations a computer can only hold a finite set of .9's so .999... is 1-.(maximum 0's - 1)1 which is even less of 1 than .9999...

Would you care to make precise "1- (infinite0's)1"? If you can do it you will have opened up a whole new field of mathematics.

.99999 == 1

.99999 / 3 = .33333
1/3 = .333333

.33333 * 3 = 1

therefore (.99999 / 3) * 3 = 1

t0xic wrote:.99999 == 1

.99999 / 3 = .33333
1/3 = .333333

.33333 * 3 = 1

therefore (.99999 / 3) * 3 = 1

That is just wrong.
1/3 == .333333...
1/3 != .333333
1/3 * 3 == 1
.333333 * 3 != 1
.333333... * 3 == 1

and so on
.

What do they teach you in school nowadays? Certainly not how to convert a vulgar fraction into a decimal it would seem. 1/3 = 0.3recurring, and that is just a fact. I knew that by the time I was eleven.

Sorry i left out the "..." lol

Oh. I understand now..

1/3 + 1/3 + 1/3 = 3/3
That makes sense. The only way to represent it is with a fraction. It has to be correct since 1/3 is .3333333333~.

A computer can do it too.
double r = (1.0/3.0) + (1.0/3.0) + (1.0/3.0);
printf("%f\n", r);

It will print 1.0 which is correct.

only problem being that 1/3 * 3 != .999999...

By the way, how can you write 1/3 in binary? just curious. Something like .01010101... maybe? or are there other ways of representing it? (except 1/11)

No. It could not equal .9~..

Math really would collapse right on top of our heads if (1/3)*3 != 1.

The only way to represent .3~ with accuracy is with 1/3.e

We have to deal with a finite number.. I do not think there is math that can handle infinity? I am not math expert. So if .3~ is able to be represented correctly then we use 1/3.

There is actually no expression like:
1. .3333333333333....
2. .3~

So if using any of those you would not be able to insert them into a equation so they are not valid which I think is what most people have been saying already.

You can not write 1/3 in binary as a single number. I think the only way to do it would be by using a fraction. I think what might be happening is the x87(compatible) floating point circuitry uses 80-bit precision internally so that it hold something like:

.3333333333333333333 +
.3333333333333333333 +
.3333333333333333333

Then to store it there needs to be a truncation of the number and the x87 must be rounding up. So that if it stopped at something like:

.333..3 +
.333..3 +
.333..3 +
.999..9

The last digit is rounded up:

...1
.990
.1
.900
1.000


I think the x87 cheats here, and it does not actually prove why .99999~ == 1. Instead it just happens to be a result of rounding, right?

Seriously, Isn't this topic getting considerably long... And it clearly isn't going anywhere...

Brynet-Inc wrote:Seriously, Isn't this topic getting considerably long... And it clearly isn't going anywhere...

Go check out the post in the:
Announcements, Test Requests, & Job Openings Forum.
http://www.osdev.org/phpBB2/viewtopic.php?t=14048

I think you missed that one, Brynet-Inc. You are slacking up on your job. We are going to have to fire you and hire someone else. That guy needs to good talking too about using his brain when he decides where to post a question at.

1/3 as a bicimal is 0.01010101 recurring. Of course I went to school in the sixties, so it took me about ten seconds to work that one out.

0.3 recurring is an infinite geometric progression:

3/10 + 3/100 + 3/1000 + 3/10000

If it converges the formula for the sum of a general geometric progression in the form of:

S = ar + ar^2 + ar^3 + ar^4........ is S= ar/(1-r)

In this case a=3 and r=1/10

Kevin McGuire wrote:Go check out the post in the:
Announcements, Test Requests, & Job Openings Forum.
http://www.osdev.org/phpBB2/viewtopic.php?t=14048

I think you missed that one, Brynet-Inc. You are slacking up on your job. We are going to have to fire you and hire someone else. That guy needs to good talking too about using his brain when he decides where to post a question at.

I sense an insult in there somewhere.. but I can't break your encryption scheme.. You'll slip up one day!

mathematician:
I want to eat what is in your brain. I will leave a little so hopefully what I ate will grow back and then I can eat some more.