Simple mathematical problem (?)

[Skeptoid]

I had a dream about Monty Hall last night! I thought HIS problem was the most controversial.

I was thinking about starting a 0/0 = infinity thread. Or maybe a 0! <> 1 thread.

 

[ceptimus]

...or we could try the, 'Does the moon rotate as it goes around the Earth?' thread.

 

[roger]

or maybe 'does 2+2=5 for sufficiently large values of 2'?

 

[plindboe]

Haven't read the entire thread so not sure if someone else have posted the proof yet.

x = 0.999... <=> 10x = 9.999... of these two equations we get:

10x - x = 9.999... - 0.999... <=>

9x = 9 <=>

x = 1

I can't see any errors in the proof here, and find it strange that such a simple proof has caused such a long thread. Maybe I'm missing something.

 

[LuxFerum]

Here I am. I have tought about it, and I think that I found what the is the real point of this discussion.
And The real point of this discussion is whether the function "tend" to something in infinty,
or the function "is" something in infinity.

My position is that things tend to something in infinity. And I will try to prove it.


Lets start with f(x)=1/x.
What is the limit of this function when x tends to infinity?

When x tends to infinity, f(x) tends to zero. ( this is my opinion.)
When x "is" infinity, f(x) "is" zero. ( this is the other option.)

That just sound the same but it isn't.

Now lets get the function f(x)=1/x and lets multiply it by e^x.

Now, We got the function F(x)=(e^x)/x, What is the limit of this function when x tends to infinity?
The limit of e^x is infinity, the limit of 1/x is zero. But the limit of (e^x)/x is infinity!

Why is that? The function e^x tends to infinity faster than 1/x tends to zero.
And that is why we can't simply say that 1/x is zero in infinity, if we do, we have a paradox.



Now lets get the infinity series.

First serie is Sum 9.10^(-n) , n=1 to infinity. The limit of this serie is 1.
Second serie is Sum 2^(-n) , n=1 to infinity. The limit of this serie is 1 too.

Now lets subtract the first serie from the second.


Then we got this : Sum [ 2^(-n) - 9*10^(-n) ] , n=1 to infinity. (R)

This serie tends to zero when n tends to infinity. (My opnion)
This serie "is" zero when n "is" infinity. ( other opnion).


Look what happens whe we multiply (R) by 10^n.


Then we got this: Sum [ (2^(-n) - 9*10^(-n))*10^n ] , n=1 to infinity. (S)

The amazing thing is, that (S) does not tend to zero when n tends to infinity anymore, instead (s) tends to infinity, when n tends to infinity.



What happens? Sum 9.10^(-n) and Sum 2^(-n) "are" not "1" in infinity anymore. But they still tend to 1 when n tends to infinity.
Before the multiplication by 10^n, (R), we have a small difference between the 2 series. That small difference tends to zero when n tends to infinity.
What whe did when we multiply it by 10^n is increase that difference faster than the diference tends to zero.
That leads to the conclusion that whe simply can't accept that a function "is" in infinity, it simply tends to something in infinity.


No matter how you define 0,9999.... I can always show that anoter sum that tends to one faster than the way you did it.

If you say 0,999 = Sum 9.10^(-n) , n=1 to infinity.
I show you Sum 99.10^(-2*n) , n=1 to infinity.

And the difference between they, show that we can not say they are both 1.
Both tends to 1, with not the same speed.

Therefore 0,9999.... is not 1!
0,9999.... tend to 1.

 

[LuxFerum]

Haven't read the entire thread so not sure if someone else have posted the proof yet.

x = 0.999... <=> 10x = 9.999... of these two equations we get:

10x - x = 9.999... - 0.999... <=>

9x = 9 <=>

x = 1

I can't see any errors in the proof here, and find it strange that such a simple proof has caused such a long thread. Maybe I'm missing something.

10x - x = 9.999... - 0.999... <=>
In this step you just assume that 0,9999...=1
That is what you are trying to prove, you can not assume as true before you prove it.

 

[ceptimus]

10x - x = 9.999... - 0.999... <=>
In this step you just assume that 0,9999...=1
That is what you are trying to prove, you can not assume as true before you prove it.

No. You take x (0.999...) away from 10x (9.999...) and that leaves 9.000...

 

[LuxFerum]

No. You take x (0.999...) away from 10x (9.999...) and that leaves 9.000...

9 or 9.000...something?
I can make that something grow up faster to every way you can made it smaller.

 

[plindboe]

10x - x = 9.999... - 0.999... <=>
In this step you just assume that 0,9999...=1
That is what you are trying to prove, you can not assume as true before you prove it.

No I don't. Here, I'll explain step by step.

First we have;

Equation 1: x = 0.999...

Then we multiply with 10 on each side and we get;

Equation 2: 10x = 9.999...

Then we subtract equation 1 from equation 2. It's an entirely legitimate mathemathical move.

10x - x = 9.999... - 0.999...

9x = 9

x = 1

Simple as that.

 

[ceptimus]

9 or 9.000...something?
I can make that something grow up faster to every way you can made it smaller.

Well, the ... thing means repeating for ever, so there isn't any 'something'.

You take your infinitely long string of 9s, shift it one place to the left, and then take the unshifted version away from the shifted one. That just leaves 9 exactly doesn't it?

 

[LuxFerum]

Well, the ... thing means repeating for ever, so there isn't any 'something'.

Yes there is something.
That something tends to zero, But I can multiply it by something that tends to infinity faster than that something is going to zero.


Did you read the post where I show that

0,9+0,09+0,009...... -(0,5+025+0125...)

is not zero, but tends to zero.?

You take your infinitely long string of 9s, shift it one place to the left, and then take the unshifted version away from the shifted one. That just leaves 9 exactly doesn't it?

You cant work with infinity that way.

Like Walter Wayne post

And a fun one for playing with infinities:
0=(1-1)+(1-1)+(1-1)+ ...
=1 + (-1+1) + (-1+1) + ...
=1

You cant use infinity to make a nine apears and then take the inifity off.

this is what you are doig

infinity - infinity = 0 (S)


but whe know that
Inifinity +9 = infinity (R)

So whe substitute the first infinity in (S) for (R)

then we got

9+infinity-infinity=0
Doing the subtraction of infinites

9=0.


And that is wrong.

 

[plindboe]

Here is another way of writing the proof, without the x's. Maybe that will make it easier to understand.

1 = 9 / 9 = (9.999... - 0.999...) / 9 = (10 x 0.999... - 1 x 0.999...) / 9 = ((10 - 1) x 0.999...) / 9 = (9 x 0.999) / 9 = 0.999

 

[ceptimus]

Lux, You're confusing two different infinities. Here we just have an infinite number of decimal places. That's not the same as the number infinity.

 

[LuxFerum]

Im not. That what I meant I my post where you don't see the diference between tend to something and being something in infinity.


Lets use another number
999....
That is infinity described as an endless string of nines

then whe got that number multiplied by 10
9999..... and still infinity.

substract the 2 infinitys and you got 9=0

That is the same math that you use to prove that 1=0,999....
only withouth the comma.

 

[LuxFerum]

Here is another way of writing the proof, without the x's. Maybe that will make it easier to understand.

1 = 9 / 9 = (9.999... - 0.999...) / 9 = (10 x 0.999... - 1 x 0.999...) / 9 = ((10 - 1) x 0.999...) / 9 = (9 x 0.999) / 9 = 0.999

Your are doing the same mistake,

You use infinite to get another number and then you take the infinite off.
that does not work that way.

 

[plindboe]

Lux, ceptimus is right. The only difference between 9.999... and 0.999 is that 9.999... is 9 larger.

 

[LuxFerum]

Look I will try it again.

            999999999999999999999999...  This is infinity
            999999999999999999999999...  This is infinity too


When we multipliy something by ten, the numbers move one row to the left.
And when we multiply infinity by ten, the result is infinity.


           9999999999999999999999999...  This is infinity multiplyed by 10, and still infinite
            999999999999999999999999...  This is infinity 

When we subtry infinite from infinte the result is zero
sooo 


           9999999999999999999999999...  
           -999999999999999999999999...  
           ----------------------------
           9000000000000000000000000...

So by this result is clear that infinity is equal to zero.

If you can find the error in this, you can see the error in you proof

 

[plindboe]

Lets use another number
999....
That is infinity described as an endless string of nines

then whe got that number multiplied by 10
9999..... and still infinity.

substract the 2 infinitys and you got 9=0

I don't get what you mean here. If you subtract the smaller infinity from the larger infinity, you will still get infinity. It won't be 0. It will be 900000000.......(Infinite 0's).

 

[ceptimus]

Did you try googling for 0.999=1 ? There are lots of websites about this. Also, the normal way of showing Cantor's famous diagonal proof has a section explaining that many numbers can be written in two distinct ways, and that therefore any number written with an infinite string of 9s at the end must be replaced by the alternative form ending in a string of 0s.

 

[LuxFerum]

I don't get what you mean here. If you subtract the larger infinity from the smaller infinity, you will still get infinity. It won't be 0. It will be 900000000.......(Infinite 0's).

yes sorry I saw that latter