2.999 (repeating) equals 3 not rounded?

Question

I saw a math proof once that said 2.9999 (repeating) equal 3, if you don't round it. I can't it anymore. Can anyone clue me in?

Answer 1

the reason this is well i cant really explain it but here is an example.

.33333 repeated is 1/3
.66666666 repeated is2/3
and .999999 repeated is 3/3

3/3 is = to 1

so 2.99999repeated is equal to 3

Answer 2

Well, the proof is in the .999 ... > repeating.

You could phrase the problem as 1 + 1 + .9999 repeating.

Consider: the fraction 1/3 or 1 divided by three. The approximation of this number is .33333333 ... >

So, 1/3 + 1/3 + 1/3 could be written:

.333... + .333... + .333...

which is equivalent to .999... > repeating.

What's more, one can basically say that regardless of how far one keeps subdividing one keeps getting the same remainder because one continues to divide the same number by the same number infinitely. So no matter how many time you keep dividing it you will keep getting the same answer. Or, one could also work backward and say:

What is the difference between .999 repeating and 1?

So let's try this:

What is 1 minus .9? It's 0.1
1 - .99 = .01
1 - .999 = .001
1 - .9999 = .0001
1 - .9999999999999999 = .0000000000000001

You see that the more 9's you add to the end the smaller and smaller the difference between 1 and .999... > repeating.

So, one could express this as a limit (in calculus) and say that as the number of 9's tacked on to the end -> approaches infinity the difference between .999... > and 1 approaches zero. IE, there is ostensibly not a remainder of the addition of 1/3 + 1/3 + 1/3.

We could also express the relationship of .999 repeating to 1 as a fraction or a division problem:

1/0.9 = 1.111... >
1/0.99 = 1.010101... >
1/0.999 = 1.001001001... >
1/0.9999 = 1.000100010001... >

So, the number of 0's between 1. and the next 1 (in the decimal) is x-1 where x is the number of 9's after the decimal in the bottom of the fraction. So, as x -> infinity, the number of 0's approaches infinity, and .999 gets infinitely close to 1 because it has an infinitely small or basically 0 remainder. (in reality .333...> repeating is just an >>approximation<< of 1/3 and 3/3 DOES equal 1).

Essentially therre would be a remainder if you could ever get to it, but the more 9's you add, the more 0's there are between you an the remainder. In fact there are infinitely many zeros since we're approaching infinity. IE you will NEVER got to a remainder simply because you keep adding zeroes between you and the 1 for each 9 you add. So you will NEVER under any circumstance reach that elusive 1 remainder. ;o)

Hopefully somewhere in the examples was a cogent argument for why it works. ;o) I had a better argument on maths.org. Hey, I even found it! w00t! This was like last year or the year before. But a good argument. I'm in there somewhere a page or two down...

https://nrich.maths.org/discus/messages/67613/67006.html

Answer 3

The short answer is that a repeating single digit decimal .xxxxx.... repeating is defined as x/9, so .9999.... (repeating) is defined to be 9/9=1, just as .6666.... is defined as 6/9=2/3. And 1+2=3, so 2.9999....=3.

The full proof requires Limit Theory, which is part of Calculus and/or Real Analysis. No numerical proof exists, because the proof requires continuity and equality concepts that can't be expressed as numbers.

A sketch of the proof without the advanced stuff goes like this:

if 3 and 2.99999.... were, not equal, then 3 - 2.99999.... > 0.
It turns out that you can't find any number (say, c) such that c = 2.99999... and c>0, so 3 and 2.99999.... are equal.

Not sure if that helps much, but that's how the proof goes.

Answer 4

Although many of the previous posts are a little easier to understand, sometimes they just don't feel that thorough:
for example 1/9=0.111111. . . so 1=9/9=9•0.1111 . . . =0.999. . .
and x=9.9999. . ., 10x=9.999. . . 9x=9, and x=1.
I just feel like there might be some hidden rounding in there.

Remember that we are working base 10, so 2=2, 2.9=2+9/10, 2.99=2+9/10+9/100, 2.999 = 2+9/10+9/100+9/1000 = 2+9(1/10+(1/10)^2+(1/10)^3) . . . and so on.

Here is a proof using this property of 2.999999 . . .

let x=9/10. Then 2.99999. . . =2+lim (n->∞) 9/10•(0≤i≤n)∑(1/10)^i = 2+lim (n->∞) 9/10•[(1/10)^n-1]/[1/10-1]=2+ 9/10•[0-1]/(-9/10) = 2+9/10•(10/9)=3

Answer 5

Here is a simple answer.
If you accept that 2.999... (repeating) represents a real number, then it can't be anything other than 3. That's all there is to it. If you
accept that 2.99999... represents a number, then if it is anything els but 3 it will break the rest of your arithmetic.

2.9999... = 2 + .9 + .09 +.009 +...

When you use the value 2.999... you are accepting the validity of the infinite sum above. The value of the infinite sum is the limit of the sequence of partial sums. That limit is 3.

Answer 6

There are some good answers here.
I'm fond of the multiply both sides by 10
solution. However, there are some very wrong
statements also.

2.99 (repeating) does equal 3 exactly.
People saying things about it approaches
3 in some limit are not correct. The repeating 9 is not
some sort of infinite series, it is an actual number that
is not approaching anything. The mathematical proof
PROVES they are the same:

x = 2.9999...
10x = 29.999 ....
subtract
9x = 27
x = 3

NOT A LIMIT. There is no limit to approach here,
because there is no inifinite process being done! It is
a number.

Answer 7

Let x = 2.9999.... (forever)

Then, 10x=29.99999.... (forever)

Subtracting x from 10 x gives us

10x-x=9x

and subtracting 2.9999... (forever) from 29.999... (forever) gives us

9x=27 so x=3.

Here is how it might look on a blackboard:

10x=29.999....
x=2.9.....

9x=27

x=3

It is interesting to note that decimal representation of numbers is not unique. So, in the same way you can show that 0.9999...... (forever)=1.

This a consequence of the Completeness Axiom for the Real Numbers which states that between any two distinct real numbers, there is a real number. With a lot of math work, we can show that there are no real numbers between 0.9999...... (forever) and 1, so they must be equal.

There are other ways of looking at this problem that involve infinite series. (.99999.... (forever) is a geometric series with a=r=9/10, using notation that would be encountered in a standard calculus text.)

Unless you are in a real analysis course, I hope the above helps answer your question! :)

Answer 8

Yup, 2.999... (repeating forever) = 3 (not rounded).

8/3 = 2.66666666... (repeating forever)
1/3 = 0.33333333... (repeating forever)
Add both lines and you have
9/3 = 2.99999999... (repeating forever).
9/3 = 3, so 2.9999999... (repeating) = 3.

Another way to see it is like this:
x = 2.99999999... (repeating)
10x = 29.9999999... (repeating) [Multiplying by 10 merely moves the decimal point.]
Subtract line one from line two.
10x - x = 29.9999999... - 2.99999999...
9x = 27
x = 3.

Answer 9

2.999 recurring (this means that the 9's go on forever) does equal 3.

Answer 10

This is a good one.

1/3 = 0.333...
1/3+1/3+1/3 = 0.999...
but 1/3+1/3+1/3=1
therefore 1=0.999...

You can add the 2 in to get your answer but the principal is the same.

Answer 11

This is the same as 0.99999... = 1, just adding two to each side. The reason for the first equation is that 0.11111... = 1/9, and 0.999999.... = 9 * 0.1111111 = 9 * 1/9 = 1.