Okay, just to tell you first, this is the difference between A and B Honor Roll!

Its worth 10 extra points.

Its the old debate, on whether 1=0.999 recurring.

We need to present 4 proofs of it.... I have these:


First Proof:
0.333… = 1⁄3
3 × 0.333… = 3 × 1⁄3
0.999… = 1


Second Proof:
let c=0.9 recurring
Therefore:
10c=9.9 recurring
Subtract 1c from both sides
9c=9 since 1c = 0.9 recurring
therefore:
1c=1

1=0.9 recurring



Okay, I need 2 more proofs, I know what the equations look like but I must be able to explain how they work, since they use Calculus and Analyctic Geometry equations to prove it as an Eighth grader it is beyond me on how to explain them

Anybody know how to explain 2 more proofs?

Ill be sure to look. Ill edit this if i find anything.

Nice to see a fellow Minnesotan on crypt. Pm me sometime too

Thank you.

Go Vikes! or not they scored 3 points today lol.

you cant prove something thats not true...

http://en.wikipedia.org/wiki/Recurring_decimals

I can prove to you right now that 4 equals 7.

It does not mean it is true.

4(x+y) = 7(x+y)

Cross out the (x+y) and you have 4 = 7 right?

Wrong.

.9 inifinity will never ever equal 0.

0.9 to one significant figure is one. That is to say it is rounded to one.

A rational number does not equal another rational number.

I am essentially repeating myself over and over.

Honors Algebra 2 has taught me something at least.



Gunny

Jason is right.

.9 repeating is not equal to 1.

0.333… = 1⁄3
3 × 0.333… = 3 × 1⁄3
0.999… =/= 1

If you say c=0.9 then 10c=9 not 9.9; that alone disproves your second proof

You're better off proving how it's not equal simply because it isn't

also, just think about it

1 - 0.999... = 0. (infinate 0s) 1

numbers are only equal when their difference = 0

S_a = \sum_{n=0}^{a} \frac{0.9}{10^n}

S_a = 0.9 \sum_{n=0}^{a} \frac{1}{10^n}

By standard result:

S_a = 0.9 \frac{10^{-a-1} - 1}{10^{-1}-1}

From definition:

\lim_{a \rightarrow \infty} S_a = 0.99999 \ldots

So applying this on the sum of the geometric series:

\lim_{a \rightarrow \infty} 0.9 \frac{10^{-a-1} - 1}{10^{-1}-1} = 0.9 \frac{-1}{-0.9}

0.9 \frac{-1}{-0.9} = 1

Therefore:

.99999 \ldots = 1

Let x = 0.999999 (recurring of course)

x = 0.99999999999
10x = 9.999999999

10x - x = 9x
9.9999999 - 0.99999999 = 9

9x = 9
Simplifies to x = 1

That's the only proof I know of how 0.9 recurring = 1. (Even though it really isn't lol)

0. (infitite 0s) does equal 0. You can't have a 1 after an infinite ammount of 0s, because there is no place to put the 1. Also, kind of like what you said: Try to give me a number that you can add to .99 repeating to make it equal 1, you can't.



And like 1 of the examples in the 1st post,
1/3 = .3 repeating
2/3 = .6 repeating

1/3 + 2/3 = 1
while
.3® + .6® = .9®



he said that c= 0.9 recurring, so like .9999
to get 10c you just move the decimal point to the right once, so
10c=9.9999999

since c=.999999, you can just take of the decimal from the 10c by subtracting 1c, so
9c=9
c=1, when before you had stated that c=.9999



So, I do believe that .99 repeating does equal 1.

As .99 approaches infinity, it approaches or "equals" 1. So, .99999 isn't short of 1, since it continues forever.

Another way of looking at is as follows:

1.00000....
- .99999....
--------------
0.00000....

A number with a 0 in every decimal place is 0.

Dudes, I got different feedback so stop posting lol.

Its not why its correct, its why people may think its correct.

And I already got the extra credit, so stop posting lol.


Slayeroftime- Thank you very much, although I got the same feedback from a different source.


Gunny... lol... thats completely different.

Closed.