# million, billion, trillion... then what?

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If you want the definition of countability, here it is:

A nonempty set "A" is said to be countable if and only if there exists a function f: N->A such that it is surjective (onto). N refers to the natural numbers, N = {1, 2, 3, 4, ... }

This means in the direction we are going (If there exists a function f, the A is countable), we have to assign every number in A to at least one natural number and that every natural number must be assigned one value in A.

The definition of countable infinite is this: A is CI if there exists a one-to-one and onto function f: N -> A.

So, for one natural number, there must be one value of A.. and every value of A must have one value in N, all unique.

I mean, you can skip the binary bit and go straight to decimal using the same method...

0.1, 0.2, ..., 0.01, 0.11, ..., 0.02, ... to infinity.
The union of all the values does in fact equal every number in from zero to one (non inclusive) but we still cannot count every number, only terminating rationals.

Hi Fraggle Rocker,
Fraggle Rocker said:
This list contains all numbers if the integrity of our numbering system is taken for granted.
I don't see it. I don't know what you mean by "the integrity of our numbering system", but I do see that the list does not contain all numbers.

Any number can be expressed in binary notation. (I could haved done it in decimal but I assume that we're all scientists here and binary was easier.) I think the problem you're driving at is that some finite numbers have an infinite number of digits.
That's right... If your list is to contain all numbers, then your list must include numbers with infinite digits. Does it?

I can describe to you with complete precision how to navigate down my list to the number 1/3, which is .01010101... in binary and therefore has a sequence number of 10101010... And if you take my instructions you will indeed find it. Except of course for the fact that finding it will take an infinite amount of time.
Hmm... If I will find it, then I must be able to find it in a finite time. If it takes an infinite time, then I will never find it, I think.

The fact that my way of sorting fractions gives a one-to-one mapping between integers and fractions troubles me.
Now we're on to something.
I think that it does not, in fact, give a one-to-one mapping between integers and fractions.
I think that one-to-one mapping means that for each finite integer, you must be able to find a matching finite fraction, and for each finite fraction, you must be able to find a matching finite integer..
If I can count integers, then why doesn't this automatically imply that I can also count anything they can map to?
You're correct. You can count anything that map to the integers.

I guess it's because an integer with an infinite number of digits has an infinite value
This could be a problem - is there such a thing as an infinite integer? I think that integers might be finite by definition.
whereas all fractions have finite values.
But all integers are fractions, therefore if all fractions are finite then all integers are also finite, right?

Does it violate the definition of countability to say that you have to count to infinity to count 1/3, much less pi?
I think it violates the definition of one-to-one mapping.

Word of the day - surjective.

Thanks Absane!

Pete said:
but I do see that the list does not contain all numbers.

Well, as I said in my previous post, his list DOES include every number from zero to one, non-inclusive.

That's right... If your list is to contain all numbers, then your list must include numbers with infinite digits. Does it?

It does.

Hmm... If I will find it, then I must be able to find it in a finite time. If it takes an infinite time, then I will never find it, I think.

You must be able to find the N that corresponds with it in a finite amount of time, since the value in N is in fact finite.

Now we're on to something.
I think that it does not, in fact, give a one-to-one mapping between integers and fractions.
I think that one-to-one mapping means that for each finite integer, you must be able to find a matching finite fraction, and for each finite fraction, you must be able to find a matching finite integer..

His list is in fact countable, as I stated. It is just not countable in the way he wants or thinks it is.

Well, as I said in my previous post, his list DOES include every number from zero to one, non-inclusive.
Can you prove it?

Pete said:
That's right... If your list is to contain all numbers, then your list must include numbers with infinite digits. Does it?
It does
I'll have to defer to you authority on this one, since you obviously have more mathematical education than I. But I'm not comfortable with it!

You must be able to find the N that corresponds with it in a finite amount of time, since the value in N is in fact finite.
Then what is its value? What finite value in N corresponds with 1/3 in Fraggle's list?

His list is in fact countable, as I stated. It is just not countable in the way he wants or thinks it is.
His list is certainly countable, but I think that there is no surjective function that maps the items in his list onto the rationals, therefore I think that the list is irrelevant to the question of whether the rationals are countable.

Pete said:
I do see that the list does not contain all numbers.
.1 - that's all one-bit numbers.
.01, .11 - that's all two-bit numbers.
.001, .101, .011, .111 - that's all three-bit numbers.
.0001, .1001, .0101, .1101, .0011, .1011, .0111, .1111 - that's all four-bit numbers.
Etc.

It's just counting from one to infinity and reversing the order of the bits. It contains all numbers (between zero and one), it is mappable to integers (with the pesky problem of finite fractions mapping to integers of finite length), and it's ordered in the same sequence as the mapped integers.
If your list is to contain all numbers, then your list must include numbers with infinite digits. Does it?
Yes. They're all at the bottom of the list. But you can't actually see them because the list is infinitely long. Then again I suppose you couldn't "see" one of them anyway because you could only see the portion of it transcribed in a string of digits of arbitrary finite length.
I think that it does not, in fact, give a one-to-one mapping between integers and fractions. I think that one-to-one mapping means that for each finite integer, you must be able to find a matching finite fraction, and for each finite fraction, you must be able to find a matching finite integer.
Yes, I understand that now.
This could be a problem - is there such a thing as an infinite integer? I think that integers might be finite by definition.
I'm not sure what the definition of an integer is. I'm still struggling through the Wikipedia definition of "surjective." But I thought the definition hinged on not having a fractional part. A "cardinal number" as we say in linguistics. I just don't know how to say that in proper mathematical terms.
But all integers are fractions, therefore if all fractions are finite then all integers are also finite, right?
No, for the purpose of this discussion an integer has the the part to the right of the decimal (or binary) point equal zero and a fraction has the part to the left equal zero. Using standard definitions, though, an integer is indeed a fraction with the denominator equal one.

To sum up, I see the error of my ways. I stick by my guns and maintain that my series is an ordered list of all possible fractions if it is allowed to go to infinity. But I agree that it does not satisfy the proper definition of "countable."

The definition of "surjective" is still in question.

Surjective means this:

Well, a function is a mapping of set A to set B, such that every element in A has a value in B and that each value in A cannot have more than one value in B.

When the function is said to be surjective, every element in B must have a value in A, but it does not have to be unique.

For example, Let A = {1, 2, 3} and B = {4, 5, 6}

A function f:A->B might take the values (1,4) (2, 6) (3, 6). This is a function, as every value in A is defined in B and each value is unique to that in A. However, it is not surjective because 5 is not defined in f as a value for an element in A.

An example of a surjective function g:A->B could be (1,4) (2,6) (3,5).

Understand? maybe now you can reply to my post I hope!

It's fairly interesting to note that if the cardinality (that is, the "size") of the codomain (B in the case) is greater than the cardinality of the domain (A in this case), then an onto function cannot exist. So if A = N (all the natural numbers) and we cannot create an onto function to all R (or (0,1) in your case) then it seems the codomain's cardinality is in fact "bigger" than the domain. cadrinality of N = infinity and cadrinality of R is infinity.. seems R is bigger One infinity greater than the other. However, this is nothing new. Cantor showed this many many years ago.

I have a lot of fun telling people this when I am drunk... they think I am not serious. hehe.

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you may also want to refer to this system

number of zeros
(10 to the power of)

3 thousand
6 million
9 billion
12 trillion
18 quintillion
21 sextillion
24 septillion
27 octillion
30 nonillion
33 decillion
36 undecillion
39 duodecillion
42 tredecillion
45 quattuordecillion
48 quindecillion
51 sexdecillion
54 septdecillion
57 octodecillion
60 novemdecillion
100 googol
googol googolplex

dauthus said:
you may also want to refer to this system. . . 12 trillion. . . 21 sextillion. . . 54 septdecillion
Bearing in mind that it is not universal, as has been mentioned in the earlier postings on this thread over the years. Much of the world refers to this as the American system, although it arguably originated in France. Many other countries use a power-of-six paradigm. Million=6, billion=12, trillion=18, etc. It's a little more intuitive because "all" you have to do is translate the Latin prefix into your national language and multiply by six to get the number of zeros, instead of performing the extra step of adding one before multiplying by three. But of course it's more cumbersome because 10^22, for example, must be pronounced "ten thousand trillion." Which is a real mouthful for languages whose grammar requires this to be rendered as "ten thousands of trillions of..." (Some countries call this a "trilliard," which probably makes them wonder what we're doing in all of our gaudily advertised "billiard parlors.")

Also bearing in mind that this is still a bit of an elite exercise. There appears to be no formal consensus on the precise form of some of those higher prefixes. You're more likely to run into "septemdecillion," which is proper Latin, than "septdecillion." And "sexillion" vies with "sextillion" since we can't seem to make up our minds whether we're drawing from the Latin cardinal or ordinal numeral series.

Wow! I had never known it was all this confusing... I guess this is why it's better to use scientific notation.

Wow! I had never known it was all this confusing... I guess this is why it's better to use scientific notation.

Until you got 3 significant digits and you need something bigger than trillion

Thats why when I saw \$50 Billion worth of bonds it wasn't worth any more than £35 Thousand Million's (which is are also called Billion's)

I always thought that it was much more logical say thousand million instead of billion. It is more easily distinguishable in vocal communications. I have seen many times where people mistaken billion for million, whereas they would not if it was thousand-million.

I always thought that it was much more logical to say thousand million instead of billion. It is more easily distinguishable in vocal communications. I have seen many times where people mistaken billion for million, whereas they would not if it was thousand-million.
It's logical if you only have one or two significant digits. If someone said to me, "The GDP of Earth is nineteen thousand seven hundred eighty-eight billion six hundred seventy-two thousand four hundred thirteen million nine hundred eight thousand five hundred twenty dollars," I would not be able to parse it.

Nineteen quadrillion seven hundred eighty-eight trillion six hundred seventy two billion four hundred thirteen million nine hundred eight thousand five hundred twenty dollars is easier.

Eight bumps...

Nineteen quadrillion seven hundred eighty-eight trillion six hundred seventy two billion four hundred thirteen million nine hundred eight thousand five hundred twenty dollars is easier.
Is that the US debt now?

million = 10<sup>6</sup>
billion = 10<sup>9</sup>
trillion = 10<sup>12</sup>
quintillion = 10<sup>18</sup>
hexillion = 10<sup>21</sup>
heptillion = 10<sup>24</sup>
octillion = 10<sup>27</sup>
nonillion = 10<sup>30</sup>
decillion = 10<sup>33</sup>
unodecillion = 10<sup>36</sup>
duodecillion = 10<sup>39</sup>
etc.

hexillion and heptillion are incorrect. It's sextillion and septillion

You dredged up a 15-month old thread for that?

You dredged up a 15-month old thread for that?
Particulary since post #128 shows the correct names.

Mick, if you're going to commit thread necromancy at least do it right and read it before you post. Don't Google one post and respond to it without checking the rest of the thread. Chances are, somebody already said what you're about to say and that's why the discussion went dormant.

-- Note from a Moderator (although not the moderator of this subforum).

Hmm actually I mistated it, I just checked and it is ((10)^10)^100 not (10^100)^(10^100). But irregardless it is supposed to be larger than the number of atoms in the universe and thus it is unwritable in long form base ten.

"irregardless" is not a word.

I want my own big number. It isn't really very big vs. the really big numbers mentioned in this thread, but still it is pretty big and the name hasn't been used because I searched Google and SciForums and it is not there, and it doesn't show up in the ieSpell dictionary.

I'm going to call it a bezillion. A bezillion is a googol divided by 10. A googol is a 1 followed by 100 zeros, and a bezillion is a 1 followed by 99 zeros.

You may use it when the occasion comes up .

I'll check Google later and see if I have establised a bezillion on the web.

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