.9999999.... = 1

.... I agree.... for very large values of .9999999~

Chris

You wrote:

".... I agree.... for very large values of .9999999~"

I'm not sure what you mean by very large values. .99999.... is a single value, an infinite number of 9s after the decimal point, and is exactly equal to 1.

just a joke.

someone on here has a byline that says "2 + 2 = 5 for very large values of 2"

I've always loved repeating decimals. A couple of fun facts:

X / 9 = .XXXXXXX...

ABC / 999 = .ABCABCABC...

ABCDEFGHIJK / 99999999999 = .ABCDEFGHIJKABCDEFGHIJK... you get the idea

(where x, a,b,c etc are each a digit between 0 and 9)

That's really cool, thanks.

... another fun fact (not entirely related, but in a similar vein):

Any number of the form ABCABC is divisible by 7, 11, 13 and ABC

7 x 11 x 13 x ABC = 1001 x ABC = ABCABC

Just an observation, in the original post it is stated that:

0.3333333... = 1/3

This is just as much of an assumption as 0.99999999... = 1 and hence proves nothing!

(although i do agree that it equals one of course)

I believe 1 divided by 3 IS 1, no assumption. Do the "long" division yourself -

1.0 / 3 = .3 remainder 1

10/3 = 3 remainder 1

10/3 = 3 remainder 1

10/3 = 3 remainder 1, to infinity, you get the picture...

.333333333....

Does that mean anything at all? Afterall, = 1 or not only matters depending on application precision required~ ;)

It means our number system is redundant. Worth knowing. Whether there is a practical application that this effects, I doubt it.

It actually has some pretty incredible philosophical implications if you ask me.

I would agree only if the equal sign was squiggly. Just because our base 10 decimal numbering system has difficulty with fractions with multiples of 3 in the denominator doesn't make nearly equal numbers equal. They are equal only because of numerical semantics.

Actually, it is exactly equal to one, a squiggly line, which suggests an approximation would be incorrect.

Just as π=3.1415.....

1=.999999.....

If you multiply .999999.... X 2 It doesn't equal 2.

I forgot until I read my post what my byline was,kind of ironic. :)

Actually, .999999.... x 2 = 2 .

I don't deny it's hard to believe, I had to see the proofs to believe it myself, but it's true, .99999.... = 1, so whatever 1 does, .999999..... does.

x/.9999..... = x

x + .9999.....= x + 1

x * .9999.....= x

If you are having trouble believing this you should probably read the Wikipedia article Rodger put up, it has a couple more proofs as well as some explainations about why people have trouble understanding this concept.

If you don't want to do that, you can do this: Find out the degree of precision of your calculator (mine is 1E-10). Put in 0.99... with as many 9's as it takes to be outside of the range of your calculators precision (so for me, 11 of them). Now, as far as your calculator is concerned you have just put in an infinite number of 9's. Hit enter, multiply it by 2, whatever you need to do. What comes out?

Just because your calculator rounds doesn't make it equal....

The limit of that expression when b_x (x>=1) = 0.9 is one(b_0=0), meaning it approches 1 asymptotically as x approaches infinity. However an asymptote is never reached only approached so 0.99999999.......approaches but does not reach one.

The number .999... does not "approach infinity," it IS infinte in length. I think that's where your confusion arises.

You wrote: "However an asymptote is never reached only approached so 0.99999999.......approaches but does not reach one."

No, this is an infinite series that converges, not a limit, so .9999.... and 1 are exactly equal, nothing approaches anything, they are equivalent.

OK...if .9999 repeating = 1 then .3333 repeating = .34 and .34*3 = 1.02

Actually, if you use that same proof Rodger used with the geometric series then .333... equals 1/3 (kind of expected I guess). Unless I'm misunderstanding it that is.

OK...if .9999 repeating = 1 then .3333 repeating = .34 and .34*3 = 1.02

This doesn't work because you can't express .3333.... as an infinite geometric series (try to and you'll see it doesn't work), so you have no mathematical reason to say that .333333...... = .34

Since your premise is incorrect, your answer of 1.02 not equal to 1 only proves that your premise is incorrect, not that .99999.... is not equal to one.

No....... a little dense are we? 0.3999.... = 0.34

How on earth did you go from 0.999... = 1 to 0.333... = 0.34?

Heres a fun repeating decimal math problem/brain teaser for you all:

given the equation

X = (.ABCABC... + .ACBACB... + .BACBAC... +.BCABCA... +.CABCAB... +.CBACBA...) / (.AAA... + .BBB... + .CCC...)

where A,B,C are digits (0-9), for how many sets of A, B, and C does the resulting X become an integer?

(this was once given to me on a math competition ages ago. I'll post the answer sometime in the future)

(this was once given to me on a math competition ages ago. I'll post the answer sometime in the future)

Please do, I can't even venture a guess as to the answer at this point but I'd be interested to know.

I figured it out. Given by some information from RHANSEN (thanks for pointing that out!!!), ABC / 999 = .ABCABCABC...

You can make that formula work, but only if you have an infinite repeating decimal, which the above equation gives you.

X = (.ABCABC... + .ACBACB... + .BACBAC... +.BCABCA... +.CABCAB... +.CBACBA...) / (.AAA... + .BBB... + .CCC...)

I tried it in mathematica 5.0, and it converges to the integer 2, no matter which set of numbers you use, and it makes perfect sense.

Does it look like I'm right? There must be some way to use an improper integral with the upper bound at infinity, to solve this, but I don't see a way at the moment. Any ideas?

I added an image of the formula from mathematica, I hope it shows up.

Your right Nickjd, I checked your answer with some numbers and got two as well (nice work, I didn't think to use the trick in the earlier post which is definitely the way to go).

In general terms I get the following equation:

x= (abc + acb + bac + bca + cab + cba)/(aaa + bbb + ccc)

and Nickjd has showed that x = 2 so I know now;

(abc + acb + bac + bca + cab + cba)/(aaa + bbb + ccc) = 2

(I used Rhansens trick of abc/999 = abcabc.... and cancelled out the 999 with the 1/999 to get the equation above)

I just don't see how the left hand side can be equal to 2. Can anyone help me simplify the left hand side into 2? Am I missing something obvious, or is it a number theory trick?

You are close to correct. There is a trick here. Remember what I asked. I didn't ask WHAT x is, but for HOW MANY combinations of A,B,C is x an integer.

See if you can go the one extra step :)

because a, b, and c are digits, ABC is actually (A*100 + B*10 + C)

does that help?

Ha! I think I've got it now, thanks to your hint:

abc + acb + bca + bca + cab + cba / aaa + bbb + ccc =

=2 [(ax100)+(bx100)+(cx100)+(ax10)+(bx10)+(cx10)+a+b+c]/[(ax100)+(bx100)+(cx100)+(ax10)+(bx10)+(cx10)+a+b+c]

=2(1)=2

Is that right?

Your logic is correct. Now what is the answer to my original question: for how many combinations of a,b,c is x an integer?

Unless I'm misunderstanding your question, we've already answered it. For all values of a, b, and c the answer is 2, an integer. What's more, this is independent of the counting system used, so it's true of binary, hexadecimal, base 20, base 60, whatever. Always equals two.

AHA... I got you. the question is a trick one. Not all combinations return a valid result.

The correct answer is 999 possible combinations of a,b,c. This is because all results return x=2, so it would be 10*10*10 = 1000 different combinations. BUT you forgot that we are doing some division here, and if A,B,C are all 0, then X is undefined because it is 0/0.

Don't feel bad, I got this question wrong in the competition (I forgot the 0 case), but it was such a fun one that I still remember it. Hope you enjoyed it also!

Rhansen,

You got me. That's true. So it's true for all combinations except 000. Cool puzzle, thanks.

Roger

This is easy. .99999999... does not equal 1. It approaches 1, gets infinitely close to 1, but is never, will never, and can never be 1. .9999999... is exactly equal to [1-(1x10^(-infinity))] It doesn't matter if your mathematical knowledge stops at elementary arithmetic or quantum mechanics, the answer is still the same: .9999999 = 1 is exactly the same as saying 1=2.

No, .99999...... is exactly equal to one. That's the whole point of this post. Proofs are provided in the original post. If you really can't accept this as fact, take a look at the proofs.

The point here is that .999999.... in a nonterminating number and 1 is a terminating number and both are ways of representing the same quantity, as counter-intuitive as it may seem, the proofs don't lie.

You wrote: " and the more vigorous proof with converging geometric series:"

----------

I like it! Not only is the proof rigorous, it's vigorous!

Nothin' better than a vigorous proof on a saturday night!

--Europium

Ok, time to step back for a minute. Say you have a candy bar, and magically cut it into 3 equal pieces (assume your cutting utensil has no thickness so you don't loose any material due to the cutting kerf). You now have a total of 3 pieces representing 1/3 of the total candy bar. The decimal representation of this is .333333 (repeating infinitely). Yet you still have a total of 1 candy bar on your desk (3 * 1/3) = 1. So the although .33333 (infinite) * 3 = .999999 (infinite), the .333333 (infinite) is actually just a representation of 1/3 in decimal form. In the same way .999999 is a representation of 1 if you multiply in the decimal rather than the fraction form. (I'm not a math major if you can tell, but it makes sense)

Then your co-worker walks by your desk and eats your candy bar which is sitting out while you are typing and you realize that you should not become distracted when you have a candy bar sitting on your desk.

"Apparently .9999999...... is exactly equal to one."

Then 1 - 0.9999999...... is exactly equal to nothing.

To make a long story short, "Much Ado About Nothing."

Who thinks it's not?
Send me a personal message.