Infinity is a concept. It is NOT a number. It can be quite mindboggling.

For example take the Infinite Hotel theorem. Assume that a special hotel has an infinite number of rooms, each having one guest in it.

What happens if we want to fit one more guest in? We move the guest in room 1 to room 2, the guest in room 2 to room 3 and so on, then we place the new guest into room 1.

Then what happens if an infinite number of guests arrive? We move the guest in room n to room 2n, so 1 goes to 2, 2 goes to 4 and 3 goes to 6 and so on. We then place the new guests in the new rooms.

Now you can see why infinity is so complex. (But it isn't a complex number. )

 

The OP is once again reminded that an actual "infinity" can and does not exist. As has been pointed out several times, infinity is a concept, useful in math.

Trying to somehow use examples to change infinity from a concept to a reality can't possibly work. For one thing, it is trivial to demonstrate that there are infinitely many different sized infinities. One size does not fit all.

The discussion is starting to become quite unproductive.

Dude, as you can see, I'm actually trying to prove that a number that has an infinite number of digits (i.e. 0.9999...) does exist and not equal to a number that has a finite number of digits (i.e. 1)

Also, please go to the other thread for infinity discussion.

 

I could have sworn this thread was locked yesterday.

or Am I seeing things.....

nope !!! I remember checking the thank you buttons to see if edit is there.
no edit...just thanks in the last post.

 

Dude, as you can see, I'm actually trying to prove that a number that has an infinite number of digits (i.e. 0.9999...) does exist and not equal to a number that has a finite number of digits (i.e. 1)

Been done before, dude. Try concept of "limits".

What's your angle? That wheel has long been invented.

 

Oops wrong, wait...

 

What ever u do, what ever u show

0.99 is never considered as 1

 

I'm not talking about 0.99, I'm talking about 0.9999...

\(\sqrt{0.999...}\)=\(\(3/\sqrt{10})+(3/\sqrt{100})+(3/\sqrt{1,000})+...+(3/\sqrt{10^infinity})\)

Since the limit of 3/10^infinity is zero, then \(\sqrt{0.999...}\)=0
but \(\sqrt{1}\)=1 and not 0

 

I never did understand these posts about things like this.

SUPPOSE: someone tells you, that you are 100% correct in all your assumptions. What happens then?


....



That's what I thought....




mental masturbation feels good but accomplishes nothing.

Congrats

 

I never did understand these posts about things like this.

SUPPOSE: someone tells you, that you are 100% correct in all your assumptions. What happens then?


....

That's what I thought....


mental masturbation feels good but accomplishes nothing.

Congrats

Then it means that 0.999... is not equal to 1; It also means that Midy's theorem is wrong, that Cantor set is just mental masturbation etc.

Also, since you don't understand the above proof let me explain in simpler terms:
If 1 is equal to 0.999... then the square root of 1 should also equal the square root of 0.999... but as you can see, the square root of 0.999... is 0, and the square root of 1 is 1. And because we know that 0 is not 1, then we just shown that 0.9999... is not equal to 1.

 

(I'm glad) I didn't enter the previous thread about a pie which was fruitless.



If you genuinely want to consider whether 0.9 recurring is mathematically the same as 1 then you need to study a bit of number theory.

There are, in fact, more real numbers than can be represented by the decimal (or any other base) system. The proof of this, and the construction of such a number, can be found towards the beginning of many first year university algebra / number theory texts.

In fact you will find that the whole point of the real number system is that between any two real numbers we can always insert an infinite number of additional real numbers. That is one meaning of the statement, in mathematics, that the real line is complete.

 

(I'm glad) I didn't enter the previous thread about a pie which was fruitless.

If you genuinely want to consider whether 0.9 recurring is mathematically the same as 1 then you need to study a bit of number theory.

There are, in fact, more real numbers than can be represented by the decimal (or any other base) system. The proof of this, and the construction of such a number, can be found towards the beginning of many first year university algebra / number theory texts.

In fact you will find that the whole point of the real number system is that between any two real numbers we can always insert an infinite number of additional real numbers. That is one meaning of the statement, in mathematics, that the real line is complete.

Yes, thanks for the advice but I'm already a graduate. I'm just revisiting the "whys"

Anyway, 0.999... can also be expressed as: (9/10)+(9/10^2)+(9/10^3)...+(9/10^infinity) - with this, we can use limits for its square root.

 

I'm not talking about 0.99, I'm talking about 0.9999...

\(\sqrt{0.999...}\)=\(\(3/\sqrt{10})+(3/\sqrt{100})+(3/\sqrt{1,000})+...+(3/\sqrt{10^infinity})\)

Since the limit of 3/10^infinity is zero, then \(\sqrt{0.999...}\)=0
but \(\sqrt{1}\)=1 and not 0

I had decided to sit myself out of this moquery of a "mathematical" thread, but this is just too much!

I mean, come on! Are you even trying? At least get your highschool algebra right if you are to talk about abstract meanings like infinite and infinite series...

\(0.\bar{9}=\frac{3}{10}+\frac{3}{100}+ \frac{3}{1000}+ \cdots\\
\text{but its root is}\\
\sqrt{0.\bar{9}}=\sqrt{\frac{3}{10}+\frac{3}{100}+ \frac{3}{1000}+ \cdots}\\
\text{and by no means this attrocity:}\\
\sqrt{0.999...}=\frac{3}{sqrt{10}}+\frac{3}{sqrt{100}}+\frac{3}{sqrt{1,000}}+...!!!\\
\)

Let alone the fact that this sums up somewhere at 1.373 not 0!

All the people who post in your threads could produce actual work elsewhere in the forum! Don't deprive AAC from its manpower!

 

How did you get 0.999..^0.5 = 0?
It has to be higher than 0.9^0.5 which is about 0.95

Einstein was wrong about a lot of things, just quoting him doesen't make it true.

 

Oh yeah, just want to say that adding infinity [...] in the middle and having a definite end is something which I didn't just made up.

There can be no such thing as having an infinite number of repeating digits followed by another digit as you have stated in the following quote.

Thus does this also make 0.9999...9 and 0.9999...8 the same number? (considering that ellipses [...] represents an infinite and unending process of 9s.

The fact that 0.9999...8 ends with an 8 means that the number is not repeating; it stops at some point. Therefore, the 9's cannot be repeating forever (infinitely).

All "0.9999...8" means is that the 9's continue for an unknown number of iterations before being truncated by the 8. Without more information, we only know that this number consists of many 9's after the decimal point, followed by an 8. It could be 0.99998, 0.999998, 0.9999998, 0.99999998, or 0.99999999999999999999998.
Because the number does finally end, it can only represent one of the infinite possibilities that can be represented by "0.9999...8".

Only when the ellipses appears at the end of a series of digits, does it indicate that the digits repeat forever.

If one needed to express a number like this having, say 257,649 "9's" followed by an "8" we could write "0.9999...8" and specify that there are 257,649 "9's"

That's not very "mathematical," so we could instead write:

\(9 \times 10^{-1} + 9 \times 10^{-2} + 9 \times 10^{-3} + ... + 9 \times 10^{n-1} + 9 \times 10^n + 8 \times 10^{n+1}\), n = 257,649

Note that in this case, the ellipses does NOT indicate infinity (or an infinite number of repetitions), but rather that the sequence continues until the following term is reached.


But this is clumsy, so we would probably use summation notation instead:

\(\frac{8}{10^{n+1}} + \sum_{x=1}^n \frac{9}{10^x}\) , n=257,649

 

Yar yeh nai ho sakta jab 0.1111111*9 karin to answer 1 aata ha na ka 0.9999999

 

Yar yeh nai ho sakta jab 0.1111111*9 karin to answer 1 aata ha na ka 0.9999999

Did you fall asleep on your keyboard?

 

Did you fall asleep on your keyboard?

LMAOF
Good one.

Actually it's hindi.
Somebody ought warn him about language rules

 

Space can be divided infinitely.
1/9 = .111111111..... infinite.
I don't see why you want to prove that 1/9 = 1. There's no point in it.
If you wanna see why there's no point in it, I'll give you a hint: Euclidean geometry.

 

is similar to this argument
(1/4) = (0.2500...) = (0.2...)
4(1/4) = 4(0.2...)
1 = 0.8...

Which is erroneous.

The method isn't in error, your numbers are. the "..." means the number is repeating (bar)

1/4 = 0.250.... , not 0.2.... Since 2 is a different number than 5.

Therefore, 4* 1/4 = 1.000....

 

Space can be divided infinitely.

This is by no measure known for certain.



There is a lot of confusion in this thread. Mostly instigated by the OP who seems to take great joy in breaking the rules of mathematics.

@OP, you claim to be a graduate. Of what? If you need a refresher in math, undergrad texts in number theory, algebraic theory and mathematical analysis should provide more insight on how to overcome your dilemma.

1=0.(9) or 1=0.999... are not correct ways to represent the equation in the attachment (sorry, don't know how to include equations in-line).