I heard that .9 repeated is equal to 1. Is that true?

Question

Can you explain it to me? I was told this and it doesn't make any sense!

Answer 1

Yes.

1/9 (one ninth)=0.1_ (the _ stands for repeating). Therefore 9*(1/9) (nine ninths)=9*0.1_=0.9_ (each one just turns into a nine, so there's no carrying). But of course, nine ninths is one. So 0.9_=1.

Answer 2

Blizzard Entertainment® Announces .999~ (Repeating) = 1


IRVINE, Calif. - APRIL 1, 2004 - At a press conference earlier today, Blizzard Entertainment® announced that .999~ does in fact equal 1. For seven and a half years, enthusiastic forum-goers have fervently debated the issue on the Battle.net® forums. Does .999~ equal 1? Does .999~ not equal 1? The popular forum topic reached its epic conclusion today, as Blizzard revealed the proof that finally brought an end to the discussion and thus resolved the conundrum that has plagued visitors to the forums for so long. "We are very excited to close the book on this subject once and for all," stated Mike Morhaime, Blizzard Entertainment president and co-founder. "We've witnessed the heartache and concern over whether .999~ does or does not equal 1, and we're proud that the following proof finally and conclusively addresses the issue for our customers." Consider:


Proof:

lim(m --> ∞) sum(n = 1)^m (9)/(10^n) = 1
0.9999... = 1

Thus x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.



Elaborate and yet crafted with deceptive simplicity and elegance, Blizzard's proof that .999~ = 1 is a true wonder of modern mathematics. More information on how the proof was derived will be revealed in the months ahead. Please stay tuned to http://www.blizzard.com for further announcements regarding this and other upcoming proofs from Blizzard Entertainment.

About Blizzard Entertainment, Inc.
Best known for blockbuster hits including World of Warcraft® and the Warcraft®, StarCraft®, and Diablo® series, Blizzard Entertainment, Inc. (www.blizzard.com), a division of Vivendi Games, is a premier developer and publisher of entertainment software renowned for creating many of the industry's most critically acclaimed games. Blizzard's track record includes nine #1-selling games and multiple Game of the Year awards. The company's free Internet gaming service, Battle.net®, reigns as the largest in the world, with millions of active users.

Answer 3

.9 repeated does equal 1 because 1/3 = .3 repeated and 1/3 x 3 = 3/3 = 1
.3 repeated x 3 = .9 repeated. Get it?

Here's another way:
let n = .9 repeated
10n = 9.9 repeated
10n - n = 9n 9.9 repeated - n = 9.0
Therefore 9n = 9
9n / 9 = n 9 / 9 = 1
n = 1I think .9 repeated does equal 1 because 1/3 = .3 repeated and 1/3 x 3 = 3/3 = 1
.3 repeated x 3 = .9 repeated. Get it?

Here's another way:
let n = .9 repeated
10n = 9.9 repeated
10n - n = 9n 9.9 repeated - n = 9.0
Therefore 9n = 9
9n / 9 = n 9 / 9 = 1
n = 1

Answer 4

Rounded off to the nearest decimal it equals 1 When repeated 3 X .999 = 1

Answer 5

Yes, this is true. This is, in fact, a necessary feature of any positional system of representing the real numbers. Permit me to explain: suppose that we wish to invent some relation mapping real numbers onto strings of symbols selected from a finite alphabet, with the property that these strings can be lexicographically totally ordered and that this ordering correspond to the order of the real numbers (that is, if S1 and S2 are strings corresponding to real numbers R1 and R2, then S1 ≤ S2 implies R1 ≤ R2 and S1 ≥ S2 implies R1 ≥ R2). Suppose also that we wish every real number to be representable by at least one such string. We can immediately conclude two things: first, that we must allow for strings of infinite length, since the set of finite-length strings is countable and the set of reals isn't. And second, that given any pair of strings A and B such that A and B agree up to the nth position, A_n is the symbol that comes immediately after B_n, every element of A after the nth is the symbol considered lowest in the ordering, and every element of B after the nth is the symbol considered highest in the ordering, then either at least one of A or B is an invalid string, or else A and B correspond to the same real number.

Since the second conclusion may not be immediately obvious, permit me to prove it for you: suppose the contrary - that A and B correspond to real numbers a and b such that a≠b. Since the reals are dense, there exists a third number c such that a > c > b. Since every real number is mapped to at least one string, c must have some string representing it. It cannot be mapped to any string ≥ A, because that would imply that c ≥ A, which is false. Similarly, it cannot be mapped to any string ≤ B. Therefore it must be mapped to a new string C such that A > C > B. However, any string that differs from B only in positions after the nth would be strictly less than B, so this means that the nth position at least must be changed. The smallest change to this would be to have the nth position be the symbol that comes immediately after B_n, and then have every element thereafter be minimal. But this is string A, and we already established that C ≠ A. Therefore it is impossible to find C such that A > C > B, and our initial assumption, that A and B map to distinct real numbers, is false. Q.E.D.

Now, relating this to the decimal system, we can consider the decimal mapping of real numbers as the subset of strings from the alphabet {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} that are infinite in both directions, such that at most finitely many of the elements in negative positions are nonzero (okay, technically this only is only a mapping of the nonnegative reals, since to indicate a negative number we must use the symbol '-' which doesn't appear in this alphabet, but that doesn't affect the basic properties of the theorem). By convention, we don't actually write any of the infinite prefix of zeros, or any of the infinite suffix of zeros where one exists, preferring instead to have the decimal simply terminate. The zero position may be considered to the the one to the immediate left of the decimal point. The decimals can represent any real number (they are designed to), and are also designed so that the order of the reals does correspond to the order of the decimal expansions (we wouldn't bother using them otherwise). Therefore, by this theorem, at least one of the following MUST be true:

a) 1.00000... is an invalid decimal expansion
b) 0.99999... is an invalid decimal expansion
c) 0.99999... = 1

a is, I think you will agree, completely absurd, since 1 was around long before we thought to make infinite decimal expansions. b is possibly sensible, although it does pose some problems for anyone attempting to find the decimal expansion of a*b by formally multiplying the decimal expansions of a and b (suppose we wished to find 9*1/9 from the decimal expansions of 9 and 1/9 respectively - quite the conundrum, as 9 * .11111... = .999999..., which wouldn't be a valid result). Therefore, mathematicians have decided that usefulness and consistency are best preserved by allowing them both to be valid decimal expansions, which means that we are forced, by the above theorem, to have .99999... equal 1. And in a similar fashion: 2.999999.... = 3, 1.482799999... = 1.4828, 754999.9999... = 755000, and so on. Every decimal expansion that terminates has another equivalent decimal expansion that ends in an infinite series of nines. And this is true in any base - for instance, in binary, 1/4 has the expansions 0.01 and 0.001111111... . Also, it can be shown that ONLY the numbers with terminating expansions have multiple decimal expansions.

You will see a lot of proofs, most of them valid, relating .9999... to 1 by a series of arithmetical operations (such as the 1*1/9 above), or perhaps by recasting them in terms of infinite series. The purpose of the above is to show you that the issue is more fundamental than that, and you really can't have anything like the decimal system without at least a few redundant expansions.

- Pascal

Answer 6

Yep, .99999... is equal to 1.

One explanation would be relatively simple...
1/3 is the same as .3333333....
2/3 is the same as .6666666....
If you add the fractions you get 3/3 which is 1.
If you add the decimals you get .99999999...

There is another way to do the same problem.
If you say that
.9999.. = x,

Then multiply both sides by 10 you get
9.9999... = 10x

Subtract the two equations to get rid of the repeating decimals
9.999999... = 10x
-.999999... = x
---------------------
9 = 9x

Divide both sides by 9, and you get x = 1,
so .9999999.... = 1.

Hope this makes sense.

Answer 7

When you add a 9 to the end, you converge to 1. An infinite converging leads to 1.

Answer 8

yes. u r right.

let us say we have 999 as the number of coins in a bag. it will be easy to say that the bag contains 1000 coins than saying the exact and correct answer of 999. similarly for 99 we say 100.

by the same token, when we go below the decimal,
0.99 can be rounded off as 1
1.99 can be rounded off as 2
2.01 can also be rounded off to 2
23.097389 is almost equal to 23.1
and so on.

i should tell u that this kind of correction came in to the world because of the laziness of the human to calculate and write down.

this as not an issue with computers. they can handle with out any laziness.

Answer 9

No. Even if you repeat .9 a hundred or a thousand times, it would still be .9. Rephrase your question into something that gives a reason for an answer.

Answer 10

.9 repeated is not equal to 1.
It's very, very, VERY close to 1.
In fact, it really wants to be 1, but it will never get there.

Actually, it's like saying .9(infinity 9's -1)8 is equal to .9 repeated.

If .9 repeated _was_ equal to 1, then .9(infinity 9's - 1)8 is also equal to 1.

Continuing along that line an infinite number of times, if .9 repeated was equal to 1, then 0 is also equal to 1.

*cough*