is 0.9 reccuring the same as 1?

Question

if 0.3 reccuring is 1/3, and three times 1/3 is one, but three times 0.3recurring is 0.9 recurring, is 1 the same as 0.9recurring?

thanx evr1 who answered, I came accross it in my gcse revision and it got me thinking( gcse = british school exam at 14-16years old)

Answer 1

Only if you round up :-)

Answer 2

Yes. This is because 0.3 recurring is 1/3, so three times that would be 0.9 recurring = 3/3 = 1.

Answer 3

I agree with Piopo -- the answers to mathematical questions often depend on the exact definitions which people use. For this example, it depends on what one means by the decimal representation of a number. One might expect when one first learns about decimals that every positive number x has a unique decimal expression
A_1 A_2 ... A_s . a_1 a_2 a_3 ... where all of these digits are between 0 and 9, the A_i's are the numbers before the decimal sign, the a_i's are the numbers after the decimal sign, and not all of the digits are 0.

In fact, this is not true. It turns out (see Hardy and Wright, Introduction to Number Theory) that this is only true if you further infinitely many of the a_i's are not 9, i.e., that the digits aren't all 9 after some point. In other words, for what we think of as terminating decimals, corresponding to rational numbers whose denominators are some power of 2 times some power of 5, there is an ambiguity in trying to express them as a decimal. They can either be expressed as a number whose digits are eventually all 0, or they could reasonably be expressed as a number whose digits are eventually all 9. This is what happens with your example, where 1.00000... can also be expressed as 0.999999...

If what is important to you is that all possibly decimal expansions you right down be considered to be numbers, then you would consider both of these decimal expansions as being the same number, so then in that case, yes, 1 is the same as 0.9 recurring.

However, if what is important to you is that every positive number correspond to EXACTLY one decimal expansion, then you would probably define decimal expansions by a procedure to compute them, which eventually assigns one decimal expansion to each positive number. This is the approach Hardy and Wright takes, and that procedure always assigns terminating decimals the expansion ending in all 0's. Thus, by leaving out some decimal expansions (i.e., ones with recurring 9's), the decimal expansions really do correspond to real numbers, while otherwise, the only way to think of them as corresponding directly with real numbers is by taking all instances of ambiguity and bundling them together as "equivalent", which is somewhat of an awkward construction. I personally prefer just leaving out the numbers with recurring 9's.

Answer 4

Yes and No and Al is correct.

There is a problem with your question, 0.3 is not = 1/3 in the world of infinite recurring numbers. It is 0.333... just ast 1/9 is not equal to 0.9 but rather 0.999... which would only be equal to 1 if you round up.

In the world of fractions, of course you could write it as equal to 1 but you can't suddenly jump to fractions then decimals then back to fractions to prove a proof.

For example, you can't say
1/3 = 0.3
3 x 1/3 = 1
therefore 0.3 must = 1
In a proof of this fashion you would have to say 1/3 = 0.333...

Answer 5

0.9 recurring is indeed the decimal representation of 0.3 recurring multiplied by 3, but strictly speaking, 0.9 recurring is NOT 1. it comes infinitessimally close to 1, and in terms of mathematical limits, it "approaches" 1, but it never reaches 1 (we call this "asymptotic to 1"). as you add more "9"s to the decimal, we say that it "approaches 1" and that it's "limit" is equivalent to 1.

the essentially same thing can be observed with the function f(x) = 1/x. if you continue on increasing the value of x, 1/x then becomes closer and closer to zero. but it will NEVER be zero, no matter how large a number you slap into its denominator. we say that this function's limit is zero, and that it is asymptotic to zero. by this we mean that it comes inifinitessimally close to zero, but it really never does.

Answer 6

mathbear77 is right. .99999.... is equal to 1. No rounding is involved. .99999.... is not a sequence, so talking of limits is not correct. You could represent it as a series, in which cause it would be .9 + .09 + .009 ....

Then, you would get the sum of that (infinite) series, which would be 1. For any real number x such that x < 1, there would be an N such that the sum of the first N terms of that series would be greater than x.

Answer 7

Yes; based on your premise.

Under different circumstances, 0.9 recurring approaches 1 but never quite reaches it. For approximations, it may be considered as equal to 1.

Answer 8

With 0.9 recurring long enough; they will eventually equal out.
And eventually,give the same result when you use them for calculating other figures.

Answer 9

Yes... look this question up in yahoo answers, others have answered it better than anyone here. or look here: http://en.wikipedia.org/wiki/Proof_that_0.999..._equals_1

The answer is YES although there will always be people who argue it (there are people who truly still believe the earth is flat). Most of the no answers here are completely wrong and most of the people who answer no just got through pre-calc or calc 1 and now think they know everything about math.

Answer 10

yes. 0.3 recurring is the decimal representation of 1/3 just like 0.50........ is of 1/2

Answer 11

Yes. No rounding is required if you understand what .9 recurring actually means.