2+2 = 3.99 recurring??

jd · 2006-07-20 09:24:55

I heard somewhere this was true.

mikau · 2006-07-20 09:55:29

thats why they say..

n = 3.99...

10n = 39.99... multiply both sides by 10
9n = 36 subtracted n from the left side, and 3.99... from the right.
n = 4. divide both sides by 9. QED.

numen · 2006-07-20 11:26:19

Is (...) supposed to mean limit? Now that's new.

Besides, I think this would be a valid argument against it:

Edit: 3,999... is not irrational by definition so it has to be rational, so I take the above statement back.

Last edited by numen (2006-07-22 11:09:37)

MathsIsFun · 2006-07-20 12:18:26

Ahhh ... this one again. Have a read here: Does 0.999... equal 1?

jd · 2006-07-20 17:10:23

I heard it on QI something about when you add 2 and 2 using log it turns out to be 3.99 recurring.

luca-deltodesco · 2006-07-20 21:46:18

well 3.99 recurring equals 4 anyways, so it doesnt matter, the only way it matters, is that it could change a result by a very small amount due to calculators, computers whatever not being able to calculate with infinately long fractional parts

Kurre · 2006-07-21 01:16:21

well the difference between 3,9999...and 4 is 0.00............001, but since the 1 is at the end of an endless number of 0s, 3,99999 = 4, right?

ben · 2006-07-21 03:29:11

No! Please don't write 0.0.......1, it has no meaning. Rather write 3.9..... or 3.99....... etc. The elipses imply an infinite number of nines. As we really don't have time to write out 9 an infinite number of time by convention we call that number 4.

Or, to be cute. If 3.99.... is not equal to 4, you have two choices. Either tell us what lies between these two numbers, or you assert that the real line is not continuous. I promise you, either option will result in a very serious headache.

ben · 2006-07-21 03:53:20

numen wrote:

Is (...) supposed to mean limit? Now that's new.
Besides, I think this would be a valid argument against it:

Certainly not! Here's your homework.

Is 1/3 rational or not?

Is 1/3 + 1/3 +1/3 equal to 1?

Is 1 rational or not?

What is the decimal represenation of 1/3?

Of 1/3 + 1/3 + 1/3?

See where it goes?

Kurre · 2006-07-21 05:29:26

ben wrote:

Either tell us what lies between these two numbers

for me it sounds like the number that is between those numbers are 10^(-∞)

John E. Franklin · 2006-07-21 05:33:52

Who was the smart guy who proved you couldn't have something on both ends of an infinite number of things, ben?
I've heard you can't make a set that way, but who proved it? What mathematician?

Didn't you know infinity isn't really everything? It's just everything about something.

Last edited by John E. Franklin (2006-07-21 05:35:13)

ben · 2006-07-21 06:12:45

John E. Franklin wrote:

Who was the smart guy who proved you couldn't have something on both ends of an infinite number of things, ben?

Dunno, Cantor maybe? He was interested in different sorts of infininty (!). He also went bonkers......

I've heard you can't make a set that way

You mean you can't have an infinite set? Of course you can

Kurre wrote:

for me it sounds like the number that is between those numbers are 10^(-∞)

I realize this was tongue in cheek, but it does show a common difficulty. Infinity can not be treated like "just some hugely big number". It is an abstract concept.

Like, infinity is what there is when you run out of counting numbers. Huh? You simply cannot manipulate it as though it were an element of R. So 10^(-∞) has no meaning. Sorry to appear tough!

(Actually, there is a real line, called, I think, the projective real line, or something like that, which does precisely that. Try googling, as I know FA about it)

numen · 2006-07-21 06:36:03

ben,

1/3+1/3+1/3=1.

But note that 0,333... is not 1/3, it's an approximation.
It's just slightly less than 1/3. It approaches 1/3 like a limit, but never becomes it. 4 is in Q, 3,999... isn't.

Keep in mind that an infinitely large amount of zero's adds up to zero, while infinitely small is larger than 0. The difference is infinitely small, thus not equal to zero, so they're strictly not the same.

Ricky · 2006-07-21 06:57:36

If you are talking about infinity without talking about limits, then you aren't talking about the real numbers. The only way to even have the concept of infinity with a real number is to use limits.

It can be proven that any number with repeating decimals is rational, in the reals. I'll have to find this proof later on and post it here. But 3.999... certainly has repeating decimals, so it must be rational. So there has to exist and a and b such that a/b = 3.9999...

Further more, there aren't any real numbers between 3.999... and 4. But between every two distinct real numbers lies another real number, specifically their average. So it must be the case that 3.999... and 4 are not distinct.

When you deal with things such as infinitesimals, then 3.999... does not equal 4.

ben · 2006-07-21 07:04:31

numen wrote:

But note that 0,333... is not 1/3, it's an approximation.

No, it's the best our notation can render. Everyone except you understands that. The onus is on you to show which is an approximation. 1/3 or 0.33.... Mathematically, that is.

It's just slightly less than 1/3.

Watch your logic here. If you want to argue that 1/3 in not equal to 0.33.. you are not allowed to use that as an assumption.

It approaches 1/3 like a limit, but never becomes it.

It has absolutely nothing, nothing to do with limits.

Anyway, I'm running away from this thread. The like get me cross, and I have no wish to fall out with anyone here.

numen · 2006-07-21 07:29:17

If that proof is true, then 3,999... is in Q.

If 1,999... = 2, then we could multiply both by 10000...
thus, 1999... = 2000... right?

Actually, I don't think I'll continue from here, I'm really confused now...

luca-deltodesco · 2006-07-21 09:00:36

numen wrote:

If that proof is true, then 3,999... is in Q.
If 1,999... = 2, then we could multiply both by 10000...
thus, 1999... = 2000... right?
Actually, I don't think I'll continue from here, I'm really confused now...

1.999.... = 2
1999.999... = 2000

remember the infinate number of decimals

George,Y · 2006-07-22 17:26:29

Infinite digits is tricky. The only way to deny 3.999...=4 is to say infinite digits and infinite division is illegal, meaningless or unpratical.

Humans can set up a real axis, but nature wouldn't bother to do the trick.

krassi_holmz · 2006-07-22 19:45:20

3.99... is a number, like the others, and it has some interestion properties, which allow to "calculate" it:
I'm writing 3.(9) instead of 3.999... , because it doesn't use this "...".
For example:
1.101010(234)=1.101010234234234...,

Call 3.(9) C:
C=3.(9)

Now C has an infinite number of digits, but it also have a property, given with the definition:
C 10^n has the same fractional part as C, so we may exclude it, by substracting:
For example:
10C-C=39.(9)-3.(9)=36.(0)=36
C=36/9=4.
That kind of method is used for evaluating different kinds of numbers:
For example

Last edited by krassi_holmz (2006-07-22 19:48:29)

Heldensheld · 2006-07-24 02:36:42

My first contribution .

One number can simply not equal another!

All the numbers are different!

Is that not right?

numen · 2006-07-24 04:45:01

Yes, but these are the same number (yes, I've realized that now, sorry!)
They're just presented differently, 4 or 7-3 or 3,999... but if a number like 3 would equal another number 4 you'd have big problems!

Ricky · 2006-07-24 07:24:18

3 = 4 in the integers mod 1. In other words, they both have the same remainder when divided by 1, so they are equal.

In the same sense, 2 = 5 in the integers mod 3, because when divided by 3 they both have a remainder of 2.

But overall, what you should understand that whether or not numbers are equal depends on why number system you are talking about.

ben · 2006-07-24 07:48:50

numen wrote:

Yes, but these are the same number (yes, I've realized that now, sorry!)
They're just presented differently, 4 or 7-3 or 3,999... but if a number like 3 would equal another number 4 you'd have big problems!

Hey, don't be sorry. This is still a deep discussion in the philosophy of math! In what sense is 2 + 3 the same as 4 + 1? They share only one element, the = sign. In philosophy, it is by no means a trivial question whether 0.99... = 1 or not.

In math, we just define it thus, and move on to the really good stuff

Zhylliolom · 2006-07-24 14:40:35

The only explaination for the phenomenon that comes to my mind immediately is this:

For one of the steps, you multiply n by 10, and assume you have a new digit from nowhere in saying 10n = 39.999... In reality, when we multiply a number by 10, we lose a decimal place, like so:

10 × 3.99 = 39.9.

So with this thinking, in our problem here we would have 10n = 39.999...9, where this 9 at the end is one decimal place closer to the ones place than the original "final" 9. (of course this is a foolish thing to say, how can there be a "final" 9 in an infinite sequence of 9's? But this is all for the sake of argument, and in the end it seems to be a valid explaination, so follow me still) Now this is a different number than the 39.999... erroneously declared earlier. If we subtracted 3.999... from this number, we'd get 35.999...91. (once again, I realize that it's foolish to assume this can be done with an infinite decimal, but I feel that it shows the real workings of this problem) Divide this by 9 and you get n = 3.999..., and everyone is happy and the world of mathematics is safe and sound once again. Now, the procedure I described is something we aren't sure works for an infinite amount of digits (or maybe it does and I don't know that is does for sure), so let's see how this works for a finite amount of digits. Let m = 3.99. Then 10m = 39.9, and subtracting m, 9m = 35.91. Divide both sides by 9 and you get m = 3.99, as expected. No matter how many decimal places of 9's we have after the 3, we can still work this out and m doesn't magically round up, unless you're using a calculator that rounds up after so many digits. Assuming that this trend continues into the realm of infinite 9's after the decimal is a lot safer than assuming 3.999... = 4. I think this is the correct way to look at the problem.

Ricky · 2006-07-24 16:13:08

Assuming that this trend continues into the realm of infinite

There are very few, if any, mathimatical properties that continue when you go from finite to infinite. This is not one of them.

Math Is Fun Forum

#1 2006-07-20 09:24:55

2+2 = 3.99 recurring??

#2 2006-07-20 09:55:29

Re: 2+2 = 3.99 recurring??

#3 2006-07-20 11:26:19

Re: 2+2 = 3.99 recurring??

#4 2006-07-20 12:18:26

Re: 2+2 = 3.99 recurring??

#5 2006-07-20 17:10:23

Re: 2+2 = 3.99 recurring??

#6 2006-07-20 21:46:18

Re: 2+2 = 3.99 recurring??

#7 2006-07-21 01:16:21

Re: 2+2 = 3.99 recurring??

#8 2006-07-21 03:29:11

Re: 2+2 = 3.99 recurring??

#9 2006-07-21 03:53:20

Re: 2+2 = 3.99 recurring??

#10 2006-07-21 05:29:26

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#11 2006-07-21 05:33:52

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#12 2006-07-21 06:12:45

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#13 2006-07-21 06:36:03

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#14 2006-07-21 06:57:36

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#15 2006-07-21 07:04:31

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#16 2006-07-21 07:29:17

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#17 2006-07-21 09:00:36

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#18 2006-07-22 17:26:29

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#19 2006-07-22 19:45:20

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#20 2006-07-24 02:36:42

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#21 2006-07-24 04:45:01

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#22 2006-07-24 07:24:18

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#23 2006-07-24 07:48:50

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#24 2006-07-24 14:40:35

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#25 2006-07-24 16:13:08

Re: 2+2 = 3.99 recurring??

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