Operations over infinite decimals numbers

DarkFalz

Hello, i would like to ask the following question that has been troubling me.

lets say i have 1/3 = 0.33333333333(3)

it may seem clear that 1/3 - 1/3 = 0, but when operating over the decimals this doesnt seem clear. How can i perform an operation over infinite digits and consider it as finite? If i say 1/3 - 1/3 = 0, then i am saying that i was able to subtract all the the infinite digits, or not?

The same for PI, if i say PI - PI = 0, am i not saying that i was able to subtract all the digits of pi?

Maybe i am missing something, but this has been troubling me =( it would also lead to a contradiction to say that PI - PI != 0 since that would mean that they are not the same quantity, something i just asserted when i wrote them, yet i cant perform their subtraction to check that it is 0 in fact... =S

Thanks in advance

tiny-tim

Hello DarkFalz!

Quote by DarkFalz

How can i perform an operation over infinite digits and consider it as finite?

You can't.

You'll get an infinite number of digits in your result.

Of course, if they're all 0, that's pretty easy to interpret!

coolul007

Change bases and it won't be infinite anymore, sometimes..

DarkFalz

Well, but can 1/3 exist in reality? can i truly have 1/3 of a cake? Or its just that i can have 1/3, but i cant measure it to confirm it? It has been confusing me!

Also, if it is true that 0.333333... - 0.3333333 is 0 because 3-3 = 0 in each position, then is it true that 1/3 + 1/3 + 1/3 != 1 because

0.33333...
+
0.33333...
+
0.33333...
=
0.99999... which is not 1 ?

(i applied the same rule that makes 0.3333.. - 0.3333.. be 0)

acabus

Quote by DarkFalz

Well, but can 1/3 exist in reality? can i truly have 1/3 of a cake? Or its just that i can have 1/3, but i cant measure it to confirm it? It has been confusing me!

Also, if it is true that 0.333333... - 0.3333333 is 0 because 3-3 = 0 in each position, then is it true that 1/3 + 1/3 + 1/3 != 1 because

0.33333...
+
0.33333...
+
0.33333...
=
0.99999... which is not 1 ?

(i applied the same rule that makes 0.3333.. - 0.3333.. be 0)

0.999... does equal 1, see www.physicsforums.com/showthread.php?t=507001

Robert1986

Quote by DarkFalz

(i applied the same rule that makes 0.3333.. - 0.3333.. be 0)

Uh-Oh. Prediction: this thread will be closed in 24 hours.

Summary of converstaion:

Person1: .99... isn't 1 because it never actually "gets" to 1.

Person2: What do you mean "gets to". .9 repeating is a number, it isn't getting anywhere.

Person1: Yeah, but how can it be?

Person2: Here's a proof using series <insert proof>

Person1: I don't know about series, you're wrong.

Person3: <Some intuitive argument>

Person4: <Another intuitive argument>

Person5: <Some vague discussion about something that doesn't really relate to anything>

Person1: No, y'all are all wrong because you never actually "get" to 1.

Lather, rinse repeat.

Bacle2

This argument works for the Standard Reals:

For any e>0 , 1-0.999999... <e ; just take enough 9's.

This implies d( 1, 0.9999...)=0 (to rigorize, e.g., take a limit)

In a metric space , d(x,y)=0 iff x=y. Then 1=0.999.....

pwsnafu

Quote by Robert1986

Uh-Oh. Prediction: this thread will be closed in 24 hours.

That's nothing compared to what the Wikipedia guys go through.

But we are getting sidetracked.

DarkFalz, numbers don't exist physically. They are mental constructs which have agreed upon behaviour, and from time to time, allow us to represent quantities from nature.

Secondly, you have a misconception where the operations on the reals are done one digit at a time. This probably originates from long multiplication you learned at primary school. We teach this algorithm because kids don't want memorise their 278-times-tables. With infinite decimal expansions, we don't calculate term by term, the calculation occurs across all the digits at the same time.

DarkFalz

The only thing that is still confusing me, is this step and subsequent rule:

In Proof #2, you say 10x=9.999... But this 9.999... has one fewer nine than 0.999...
Another popular argument. This time, the confusion arises from not grasping infinity. 0.999... has an infinite number of nines. If we somehow remove a nine from this sequence, then we would still have an infinite number of nines. So there are an equal number of nines in 0.999... and 9.999...

The same thing happens here: consider two sets of numbers, A and B, where
A={0,1,2,3,4,...}

and
B={1,2,3,4,...}

Both sets are infinite. And actually, both sets have an equal number of elements. But A doesn't contain 0, so it has one fewer element than B? Yes, but this reasoning only applies to finite sets. For infinite sets, it's quite possible to have one element less and still have an equal number of elements. Indeed, consider the following correspondence:

0↔1, 1↔2, 2↔3, 3↔4,..., n↔n+1,...

i remove a 9 and they remain the same? what the hell? infinity is very strange, is this truly intuitive to everyone else?

pwsnafu

Quote by DarkFalz

i remove a 9 and they remain the same? what the hell? infinity is very strange, is this truly intuitive to everyone else?

It tends to be a shock when you first see it. Imagine what the mathematicians at the time thought when they discovered stuff like this.

Also read the article on Hilbert's Hotel.

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coolul007 Recognitions: Gold Member	Change bases and it won't be infinite anymore, sometimes..

Bacle2 Recognitions: Science Advisor	This argument works for the Standard Reals: For any e>0 , 1-0.999999... <e ; just take enough 9's. This implies d( 1, 0.9999...)=0 (to rigorize, e.g., take a limit) In a metric space , d(x,y)=0 iff x=y. Then 1=0.999.....