What is .99 repeating as a fraction?

Quote: weaselman

So, you do know that most real numbers take an infinite number of digits to be represented in base-10 notation. What is it that's so special about 0.999... to you, or 0.333 ... for that matter, that you deny them the right to be multiplied by an integer?

In fact, it's "almost all" real numbers that have an infinite (non-terminating) representation. It's also "almost all" rational numbers, and almost all reals are irrational, so it's very few numbers which have a terminating decimal representation. In fact, the only numbers with a terminating decimal representation are the set of numbers
x / 2^a*5^b for all integers x, a, b. The rational number 1/3 doesn't fit that formula, so it has a non-terminating but repeating decimal representation 0.333... But 0.333... still equals 1/3, and multiplying both sides by 3 still works.

Quote: MathExtremist

In fact, it's "almost all" real numbers

Yes. More over, almost all real number cannot actually be notated or otherwise described in any way whatsoever at all :)

Quote: MathExtremist

It's also "almost all" rational numbers,

This I don't agree with. Rationals are countable (aleph-null), so, for terminating decimals to be a null-subset, there would have to be a finite number of them, which is obviously not so. I would say, there are "equal amounts" of terminating and non-terminating decimals among rational numbers (both subsets are aleph-null)

I thought "almost all" meant "all but a countably-infinite subset". Does it require that the larger set be uncountable?

Rational numbers x / 2^a * 5^b are obviously a subset of
rational numbers x / 2^a * 3^b * 5^c * 7^d * 11^e ... ad infinitum for all prime factors

but only the former have terminating decimal representations. Is it nevertheless proper to say those sets are equal in cardinality because both are countable?

Quote: MathExtremist

I thought "almost all" meant "all but a countably-infinite subset".

If that was true, you could argue that "almost all integers are less than 10" :)

Quote:

Does it require that the larger set be uncountable?

"Almost all" means all but, perhaps, a zero-measured subset.
Now, a zero-measured subset can be defined more or less formally, depending on the context, but, in simple terms, for infinite sets, you can think of it as a subset of a smaller cardinality.
So, when talking about a continuum, "almost all" means "all except, maybe, finite or countable number of elements", as you said, but in application to countable sets, "almost all" means "all, but, maybe, a finite number".

Quote:

Rational numbers x / 2^a * 5^b are obviously a subset of
rational numbers x / 2^a * 3^b * 5^c * 7^d * 11^e ... ad infinitum for all prime factors

but only the former have terminating decimal representations.
Is it nevertheless proper to say those sets are equal in cardinality because both are countable?

Yes. Like all countable sets, they have the cardinality Aleph-null.
BTW, having a (proper) subset of equal cardinality is the definition of an infinite set.

Thanks - I forgot that context matters. I suppose it would be accurate to say: "Almost all integers are not equal to 0.999..." :)

I was just waiting until we got to countably and non-countably infinite in this thread. I'm excited.

Quote: MathExtremist

".999... does not equal 1 for all fractions". I can't even begin to fathom what that means, and as a result I can't possibly refute it. Nice work, sir!

a != b for all values of x.

I'll take what I can get.

Quote: MathExtremist

"

a != b for all values of x."

You got it! Now try it with x = unknown = 2. Please don't respond to this post. I'll go with my previous post, I'll take what I can get.

Drinking again tonight.

This is how I understand it:

The natural numbers are a countable infinite set, which are a subset of the real numbers, an uncountable infinite set. The difference of which is directly related to this question.

With natural numbers I can pick some number k1, and another number k2, in which there is exactly one natural number between them, or similarly no natural number.

However, I can pick some real number r1 and another number r2. Because real numbers are uncountably infinite, I can find another number r3 between r1 and r2. Then I can find another, and another, and through induction I can ALWAYS find another number between r1 and the last r I picked. Thus I can always find a real number between 1 and .999... and thus 1 does not equal .999 repeating.