my teacher give me a problem

PRoof that 0.999999........ equals 1

The proof :

0.9999.........=p

9.9999.........=10p

9+0.999.......=10p

9+p =10p

9 =9p

p =1

But I still don't believe that 0.99999........................ = 1:mad: ????!!!!!!

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- Jan 6th 2007, 06:45 PMSingular0.99999999999........=1
my teacher give me a problem

PRoof that 0.999999........ equals 1

The proof :

0.9999.........=p

9.9999.........=10p

9+0.999.......=10p

9+p =10p

9 =9p

p =1

But I still don't believe that 0.99999........................ = 1:mad: ????!!!!!! - Jan 6th 2007, 07:12 PMThePerfectHacker
Nothing wrong.

The definition of a decimal is the convergent value. Meaning the value it approaches. Since .99999... converges (approaches) to 1 the value is 1.

Note,

1.00000... also converges to 1.

Thus,

.9999.... = 1.000...

There is nothing wrong. You are assuming that decimal representations are unique. They are not. - Jan 7th 2007, 01:44 AMCaptainBlack
- Jan 7th 2007, 01:58 AManthmoo
I find it simple enough like this

$\displaystyle \frac{1}{9}=0.1111111...$

$\displaystyle 0.1111111...*9=0.999999....$

but in fact...

$\displaystyle \frac{1}{9} * 9 = 1$ as $\displaystyle \frac{9}{9}=1$

Hope that helps you believe! - Jan 7th 2007, 03:27 AMBubbleBrain_103
All of the previous are totally correct proofs. The first time one sees it, however, they can seem like deceiving tricks.

Maybe this will help to convince you. Two real numbers are different only if they have a non-zero difference. So try to answer yourself this question:

What is 1 - .99999.... ?

Convince yourself that, no matter what positive number you pick, the difference is always smaller than that positive number. Thus the difference must be zero. - Jan 7th 2007, 04:41 AMCaptainBlack
The problem with all the symbol manipulations like this that can be used

to demonstrate the 0.999..=1, is that the operations being performed are

not defined (as far as the reader knows at any rate) for these "infinite"

decimals.

These are extrapolations of what happens with rationals and finite decimals

and have to be given a proper meaning before you have a proof.

RonL - Jan 7th 2007, 04:47 AMgalactus
Here's a response I seen on Danica McKellar's math site. I thought it was a good

answer:

**"First, we start by labeling:**

x= .999999999999..... (repeating infinitely)

Then if you multiply both sides by 10, you get:

10x = 9.99999999999..... (repeating infinitely)

Well, what if we subtracted the top equation from the bottom one? We'll get another true statement. Before I do that, let me review why this will work. Say that a = b , and c = d. Then do you agree that "a-c = b-d" is also a true statement? After all, a = b , and c = d. So once you agree with that, let's continue and go ahead and subtract the top statement from the bottom one. We get, for the left hand side of the equations:

10x -x = 9x

And for the right hand side:

9.99999..... - .99999999...... = 9

And we conclude that "9x = 9" is also a "true statement." Divide both sides by 9, and you get x = 1. Since we started with x = .9999999 (repeating infinitely), we've now shown that the "infinite decimal of .9999 repeating" equals the whole number 1. Yes, I did it much faster on the TV show. :)

What I didn't talk about on the show is the somewhat philosophical issue this proof brings to light. Our mathematical system was invented by the human mind, which assumes that space (and the number line) is infinitely divisible and that there *is* such thing as a theoretical "point" that takes up no space, and has no volume or mass. So, this little proof shows that - according to a math system that assumes such things - indeed, .999999.... (repeating infinitely) equals 1. There are those who would say this isn't true, but then they are forgetting that we are dealing with a "language" that was invented by the human mind, and also that infinity is a slippery concept. :)" - Jan 7th 2007, 04:47 AMCaptainBlack
That a number is smaller than any finite number does not proove that it

is zero. If we are in the standard reals it does, but not in the non-statandard

reals. The problem here is that this is all handwaving, define what we mean

by the recuring decimal and there is no problem.

RonL - Jan 7th 2007, 06:31 AMearboth
Hello,

your problem is a few thousand years old.

Have a look here: Zeno of Elea - Wikipedia, the free encyclopedia

EB - Jan 7th 2007, 06:59 AManthmoo
^For an essay :p

0.999... - Wikipedia, the free encyclopedia

The real thing...but seriously stay at Math Help Forum to understand what wikipedia means! - Jan 7th 2007, 07:32 AMBubbleBrain_103
I'm familiar with the so called hyper-reals. However, the user I'm assuming was inquiring about real numbers. And even if we were operating in the hyperreal number system, the difference between two reals, 1 and .9999..., is again a real number, and the argument still applies.

Showing that a real number is zero by showing that it is both non-negative and smaller than any positive is quite a standard argument in analysis. I really wouldn't call it handwaving, often it's the only reasonable approach. - Jan 7th 2007, 12:18 PMearboth
Hello,

thanks for the link.

To introduce infinite periodic decimals I used the paradoxon of "Achille and the tortoise" which was described by Zenon. There are other paradoxa described by this philosopher which will give very nice introductions to problems of analysis.

Therefore - and only therefore - I refered to this site of wikipedia.

EB - Jan 7th 2007, 01:29 PMCaptainBlack
There is no diffrerence even in the non-standard reals because of

what the notation 0.999.. is definded to mean, that is it is the sum

of the appropriate series.

Quote:

Showing that a real number is zero by showing that it is both non-negative and smaller than any positive is quite a standard argument in analysis. I really wouldn't call it handwaving, often it's the only reasonable approach.

reals for which it is true. I doubt the OP has anything but the hazyest

idea of what the reals are so an argument should hold whatever variant

of the reals they have in mind (in fact anyone who maintains that 0.999..

does not equal 1 has some non-standard interpretation of the reals in

mind, but that is irrelevent for the purposes of this argument)

RonL - Jan 11th 2007, 06:03 PMRanger SVO
I just wanna add my 10-cents worth

- Jan 11th 2007, 06:34 PMCaptainBlack