Math Is Fun Forum

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: 2+2 = 3.99 recurring??

Assuming that this trend continues into the realm of infinite 9's after the decimal is a lot safer than assuming 3.999... = 4.

We never assumed it to be true.

1/3 + 1/3 + 1/3 = 1
1/3 = .(3)
.(3) + .(3) + .(3) = 1
.(3) + .(3) + .(3) = .(9)
.(9) = 1
.(9) + 3 = 1 + 3
3.(9) = 4

Other arguments through real analysis say that if there doesn't exist a real number inbetween a and b, then a = b.  Now no number can exist between 3.(9) and 4, so they must be the same.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: 2+2 = 3.99 recurring??

...if there doesn't exist a real number inbetween a and b, then a = b.

Yes.

I've managed to convince myself that 3.999... does in fact equal 4 by the following method:

Now from basic mathematics we know that this infinite sum is a geometric series and converges to

so

This method satisfies me completely, I wouldn't dare argue against this now. What I quoted you on is also very logical and convincing, as well is the rest of your post. Good work.

Edit:

Another way to look at it is this:

4.000000...
-3.999999...
0.000000...

Somebody could argue that there's a 1 at the end of it, but there are an infinite amount of zeros before it, so the 1 will never be reached. I don't know, this way of thinking isn't as formal as the rest of this post, but it's just another view of the problem.

Last edited by Zhylliolom (2006-07-24 16:36:59)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: 2+2 = 3.99 recurring??

4.000000...
-3.999999...
-------------
0.000000...

Interesting argument.  Perhaps we can take it a few steps further.

The real numbers must have a countable number of decimal digits (I'm not entirely sure if this is true, although it seems to be).  Now understanding countable is hard enough, but basically if you put the digits into this form:

{1st digit, 2nd digit, 3rd digit, 4th digit...}

Then you can name what position any certain digit in.  For example, pi:

{1, 4, 1, 5, 9, 2, 6, 5, ...}

1 is at places 1 and 3.  4 is at place 2.  5 is at places 4 and 8.  And so on.  And this can be done for any digit in pi.

4.0 - 3.(9) = 0.00000....1 as you said.  But in a countable set:

{0, 0, 0, ..., 1}

The digit 1 doesn't have a position.  Thus, it is not in the countable set.  That is:

{0, 0, 0, ..., 1} = {0, 0, 0, ...}, when talking about countable sets

So 0.0....1 = 0

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: 2+2 = 3.99 recurring??

Can you help me figure out what it is?

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X'(y-Xβ)=0

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: 2+2 = 3.99 recurring??

If you construct the 3.999...'s as a sequence of 9's, then you can take a limit and it diverges to infinity.  Otherwise, it's undefined.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Zhylliolom

Real Member

Registered: 2005-09-05

Posts: 412

Re: 2+2 = 3.99 recurring??

I'd imagine that the expression you posted is undefined, George.

On a slightly related note to this thread (I didn't want to start a new one... I don't know if it would be worth it), here's an article I was reading where 0.999... comes up as something that makes the real numbers contradictory and flawed (as well as trichotomy, pure blasphemy!). Anyway, here is the article if you want to read the ideas of a deranged mathematician:

http://falseproofs.blogspot.com/2006/06 … ltura.html

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: 2+2 = 3.99 recurring??

From the article ('blog'):

For any set of positive numbers, there exists one number which is the smallest.

S = {1/n : n is natural} S is a set of positive numbers with no smallest number.

Fermat's Last Theorem is a statement about integers. It says nothing about real numbers so there is no reason that Fermat's Last Theorem needs to be reformulated.

It is about the integers.  But I believe (as I have only heard a summary of it) that Wiles actually uses real numbers to make his conclusions on integers.  Again, not sure about this.

It seems that Escultura likes to just go on intuition rather than axioms.

Edit: great site by the way, thanks for the link.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: 2+2 = 3.99 recurring??

Thank you very much, Zhylliolom.

But server not found

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X'(y-Xβ)=0

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: 2+2 = 3.99 recurring??

I got it just fine George.  Try it again, maybe it was just down for a while.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: 2+2 = 3.99 recurring??

Still unavailable. I don't know what goes wrong- whether it be the internet jam or my government monitoring. Can you recommand some good proxy softwares ( free better) ?

Besides, I never managed to enter  wikipedia, but I heard it has a secure access starting with https://, yet  don't know the detail.

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X'(y-Xβ)=0