Mixing
What is the order-type of the set of natural numbers, when written in alphabetical order?
$begingroup$
We are all familiar with the standard nomenclature for the smallish
natural numbers, such as
one, two, three, ..., one hundred, one hundred one, ..., fifteen
thousand two hundred forty-nine.
I have in mind the simple American number naming
conventions,
together with the names for large
numbers. (Update Names of large numbers seems to be more thorough. Note to Wikipedians: should probably merge those two pages somehow.)
Preliminary question. Is there a sensible naming system that
provides a canonical name for every natural number?
That is, I want a naming system that extends the current naming
system sensibly in such a way that every number gets a unique name. Please provide a system and explain why it is sensible.
For example, if there were some natural way to extend the Latin naming convention indefinitely, that would be great.
Let me assume that some of you will be able to provide such a
naming system.
Main Question. What is the order-type of the set of natural
numbers, when written in alphabetical order?
For example, the order will not be the same as the order $omega$
of the natural number themselves, since presumably there will be
infinitely many numbers starting with "o", as in one hundred, one
million, one thousand, and so on, and these will all be
alphabetically preceding two hundred, two million, two thousand and
so on.
So the order type will probably be related naturally $Ltimes 26$
for some order $L$, or actually, less than $26$, since probably not
every letter will be a legitimate first letter of a number name.
It is conceivable that the order type will depend on syntactic features of the naming convention.
Here is a part of the order, for numbers up to 100: (from hervé
graumann
1988)
1) eight
2) eighteen
3) eighty
4) eighty-eight
5) eighty-five
6) eighty-four
7) eighty-nine
8) eighty-one
9) eighty-seven
10) eighty-six
11) eighty-three
12) eighty-two
13) eleven
14) fifteen
15) fifty
16) fifty-eight
17) fifty-five
18) fifty-four
19) fifty-nine
20) fifty-one
21) fifty-seven
22) fifty-six
23) fifty-three
24) fifty-two
25) five
26) forty
27) forty-eight
28) forty-five
29) forty-four
30) forty-nine
31) forty-one
32) forty-seven
33) forty-six
34) forty-three
35) forty-two
36) four
37) fourteen
38) hundred
39) nine
40) nineteen
41) ninety
42) ninety-eight
43) ninety-five
44) ninety-four
45) ninety-nine
46) ninety-one
47) ninety-seven
48) ninety-six
49) ninety-three
50) ninety-two
51) one
52) seven
53) seventeen
54) seventy
55) seventy-eight
56) seventy-five
57) seventy-four
58) seventy-nine
59) seventy-one
60) seventy-seven
61) seventy-six
62) seventy-three
63) seventy-two
64) six
65) sixteen
66) sixty
67) sixty-eight
68) sixty-five
69) sixty-four
70) sixty-nine
71) sixty-one
72) sixty-seven
73) sixty-six
74) sixty-three
75) sixty-two
76) ten
77) thirteen
78) thirty
79) thirty-eight
80) thirty-five
81) thirty-four
82) thirty-nine
83) thirty-one
84) thirty-seven
85) thirty-six
86) thirty-three
87) thirty-two
88) three
89) twelve
90) twenty
91) twenty-eight
92) twenty-five
93) twenty-four
94) twenty-nine
95) twenty-one
96) twenty-seven
97) twenty-six
98) twenty-three
99) twenty-two
100) two
101) zero
Let me add that I don't necessarily expect that the order is a well-order. For example, if we have a naming convention whereby $10^k$ is represented for large $k$ simply by repeating "penpenpenpen$cdots$pen", then we could make a descending sequence via penpenpenpen$cdots$pen twelve, which would descend as the number of pen's increased, since we would be replacing t with p.
elementary-number-theory elementary-set-theory logic order-theory puzzle
edited Dec 21 '18 at 15:45
Aloizio Macedo♦
23.6k23887
asked Dec 20 '18 at 13:24
JDHJDH
32.6k680145
$endgroup$
|
show 17 more comments
$begingroup$
We are all familiar with the standard nomenclature for the smallish
natural numbers, such as
one, two, three, ..., one hundred, one hundred one, ..., fifteen
thousand two hundred forty-nine.
I have in mind the simple American number naming
conventions,
together with the names for large
numbers. (Update Names of large numbers seems to be more thorough. Note to Wikipedians: should probably merge those two pages somehow.)
Preliminary question. Is there a sensible naming system that
provides a canonical name for every natural number?
That is, I want a naming system that extends the current naming
system sensibly in such a way that every number gets a unique name. Please provide a system and explain why it is sensible.
For example, if there were some natural way to extend the Latin naming convention indefinitely, that would be great.
Let me assume that some of you will be able to provide such a
naming system.
Main Question. What is the order-type of the set of natural
numbers, when written in alphabetical order?
For example, the order will not be the same as the order $omega$
of the natural number themselves, since presumably there will be
infinitely many numbers starting with "o", as in one hundred, one
million, one thousand, and so on, and these will all be
alphabetically preceding two hundred, two million, two thousand and
so on.
So the order type will probably be related naturally $Ltimes 26$
for some order $L$, or actually, less than $26$, since probably not
every letter will be a legitimate first letter of a number name.
It is conceivable that the order type will depend on syntactic features of the naming convention.
Here is a part of the order, for numbers up to 100: (from hervé
graumann
1988)
1) eight
2) eighteen
3) eighty
4) eighty-eight
5) eighty-five
6) eighty-four
7) eighty-nine
8) eighty-one
9) eighty-seven
10) eighty-six
11) eighty-three
12) eighty-two
13) eleven
14) fifteen
15) fifty
16) fifty-eight
17) fifty-five
18) fifty-four
19) fifty-nine
20) fifty-one
21) fifty-seven
22) fifty-six
23) fifty-three
24) fifty-two
25) five
26) forty
27) forty-eight
28) forty-five
29) forty-four
30) forty-nine
31) forty-one
32) forty-seven
33) forty-six
34) forty-three
35) forty-two
36) four
37) fourteen
38) hundred
39) nine
40) nineteen
41) ninety
42) ninety-eight
43) ninety-five
44) ninety-four
45) ninety-nine
46) ninety-one
47) ninety-seven
48) ninety-six
49) ninety-three
50) ninety-two
51) one
52) seven
53) seventeen
54) seventy
55) seventy-eight
56) seventy-five
57) seventy-four
58) seventy-nine
59) seventy-one
60) seventy-seven
61) seventy-six
62) seventy-three
63) seventy-two
64) six
65) sixteen
66) sixty
67) sixty-eight
68) sixty-five
69) sixty-four
70) sixty-nine
71) sixty-one
72) sixty-seven
73) sixty-six
74) sixty-three
75) sixty-two
76) ten
77) thirteen
78) thirty
79) thirty-eight
80) thirty-five
81) thirty-four
82) thirty-nine
83) thirty-one
84) thirty-seven
85) thirty-six
86) thirty-three
87) thirty-two
88) three
89) twelve
90) twenty
91) twenty-eight
92) twenty-five
93) twenty-four
94) twenty-nine
95) twenty-one
96) twenty-seven
97) twenty-six
98) twenty-three
99) twenty-two
100) two
101) zero
Let me add that I don't necessarily expect that the order is a well-order. For example, if we have a naming convention whereby $10^k$ is represented for large $k$ simply by repeating "penpenpenpen$cdots$pen", then we could make a descending sequence via penpenpenpen$cdots$pen twelve, which would descend as the number of pen's increased, since we would be replacing t with p.
elementary-number-theory elementary-set-theory logic order-theory puzzle
edited Dec 21 '18 at 15:45
Aloizio Macedo♦
23.6k23887
asked Dec 20 '18 at 13:24
JDHJDH
32.6k680145
$endgroup$
2
$begingroup$
I have a method for writing the names of numbers in alphabet: For $n$ just write a string of $n+1$ times the letter "a". So $0$ is "a", and "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" is $345$. Now the order type is again $omega$!
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 13:35
1
$begingroup$
This exact question was asked two weeks ago on r/math subreddit.
$endgroup$
– Wojowu
Dec 20 '18 at 13:48
1
$begingroup$
@Wojowu Great! I guess the puzzle is in the air.
$endgroup$
– JDH
Dec 20 '18 at 13:53
1
$begingroup$
@Holo: It will be, the string will just be indexed with a non-standard integer.
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 20:48
3
$begingroup$
Actually, not even all native English speakers are "familiar with" the naming scheme of your first example: in British English, 15,249 is read out as "fifteen thousand two hundred and forty nine". Aside from that the form of your question amounts to: "please invent something nice and then tell me what its properties are". This is not a good MSE question.
$endgroup$
– Rob Arthan
Dec 20 '18 at 22:54
|
show 17 more comments
$begingroup$
We are all familiar with the standard nomenclature for the smallish
natural numbers, such as
one, two, three, ..., one hundred, one hundred one, ..., fifteen
thousand two hundred forty-nine.
I have in mind the simple American number naming
conventions,
together with the names for large
numbers. (Update Names of large numbers seems to be more thorough. Note to Wikipedians: should probably merge those two pages somehow.)
Preliminary question. Is there a sensible naming system that
provides a canonical name for every natural number?
That is, I want a naming system that extends the current naming
system sensibly in such a way that every number gets a unique name. Please provide a system and explain why it is sensible.
For example, if there were some natural way to extend the Latin naming convention indefinitely, that would be great.
Let me assume that some of you will be able to provide such a
naming system.
Main Question. What is the order-type of the set of natural
numbers, when written in alphabetical order?
For example, the order will not be the same as the order $omega$
of the natural number themselves, since presumably there will be
infinitely many numbers starting with "o", as in one hundred, one
million, one thousand, and so on, and these will all be
alphabetically preceding two hundred, two million, two thousand and
so on.
So the order type will probably be related naturally $Ltimes 26$
for some order $L$, or actually, less than $26$, since probably not
every letter will be a legitimate first letter of a number name.
It is conceivable that the order type will depend on syntactic features of the naming convention.
Here is a part of the order, for numbers up to 100: (from hervé
graumann
1988)
1) eight
2) eighteen
3) eighty
4) eighty-eight
5) eighty-five
6) eighty-four
7) eighty-nine
8) eighty-one
9) eighty-seven
10) eighty-six
11) eighty-three
12) eighty-two
13) eleven
14) fifteen
15) fifty
16) fifty-eight
17) fifty-five
18) fifty-four
19) fifty-nine
20) fifty-one
21) fifty-seven
22) fifty-six
23) fifty-three
24) fifty-two
25) five
26) forty
27) forty-eight
28) forty-five
29) forty-four
30) forty-nine
31) forty-one
32) forty-seven
33) forty-six
34) forty-three
35) forty-two
36) four
37) fourteen
38) hundred
39) nine
40) nineteen
41) ninety
42) ninety-eight
43) ninety-five
44) ninety-four
45) ninety-nine
46) ninety-one
47) ninety-seven
48) ninety-six
49) ninety-three
50) ninety-two
51) one
52) seven
53) seventeen
54) seventy
55) seventy-eight
56) seventy-five
57) seventy-four
58) seventy-nine
59) seventy-one
60) seventy-seven
61) seventy-six
62) seventy-three
63) seventy-two
64) six
65) sixteen
66) sixty
67) sixty-eight
68) sixty-five
69) sixty-four
70) sixty-nine
71) sixty-one
72) sixty-seven
73) sixty-six
74) sixty-three
75) sixty-two
76) ten
77) thirteen
78) thirty
79) thirty-eight
80) thirty-five
81) thirty-four
82) thirty-nine
83) thirty-one
84) thirty-seven
85) thirty-six
86) thirty-three
87) thirty-two
88) three
89) twelve
90) twenty
91) twenty-eight
92) twenty-five
93) twenty-four
94) twenty-nine
95) twenty-one
96) twenty-seven
97) twenty-six
98) twenty-three
99) twenty-two
100) two
101) zero
Let me add that I don't necessarily expect that the order is a well-order. For example, if we have a naming convention whereby $10^k$ is represented for large $k$ simply by repeating "penpenpenpen$cdots$pen", then we could make a descending sequence via penpenpenpen$cdots$pen twelve, which would descend as the number of pen's increased, since we would be replacing t with p.
elementary-number-theory elementary-set-theory logic order-theory puzzle
edited Dec 21 '18 at 15:45
Aloizio Macedo♦
23.6k23887
asked Dec 20 '18 at 13:24
JDHJDH
32.6k680145
$endgroup$
We are all familiar with the standard nomenclature for the smallish
natural numbers, such as
one, two, three, ..., one hundred, one hundred one, ..., fifteen
thousand two hundred forty-nine.
I have in mind the simple American number naming
conventions,
together with the names for large
numbers. (Update Names of large numbers seems to be more thorough. Note to Wikipedians: should probably merge those two pages somehow.)
Preliminary question. Is there a sensible naming system that
provides a canonical name for every natural number?
That is, I want a naming system that extends the current naming
system sensibly in such a way that every number gets a unique name. Please provide a system and explain why it is sensible.
For example, if there were some natural way to extend the Latin naming convention indefinitely, that would be great.
Let me assume that some of you will be able to provide such a
naming system.
Main Question. What is the order-type of the set of natural
numbers, when written in alphabetical order?
For example, the order will not be the same as the order $omega$
of the natural number themselves, since presumably there will be
infinitely many numbers starting with "o", as in one hundred, one
million, one thousand, and so on, and these will all be
alphabetically preceding two hundred, two million, two thousand and
so on.
So the order type will probably be related naturally $Ltimes 26$
for some order $L$, or actually, less than $26$, since probably not
every letter will be a legitimate first letter of a number name.
It is conceivable that the order type will depend on syntactic features of the naming convention.
Here is a part of the order, for numbers up to 100: (from hervé
graumann
1988)
1) eight
2) eighteen
3) eighty
4) eighty-eight
5) eighty-five
6) eighty-four
7) eighty-nine
8) eighty-one
9) eighty-seven
10) eighty-six
11) eighty-three
12) eighty-two
13) eleven
14) fifteen
15) fifty
16) fifty-eight
17) fifty-five
18) fifty-four
19) fifty-nine
20) fifty-one
21) fifty-seven
22) fifty-six
23) fifty-three
24) fifty-two
25) five
26) forty
27) forty-eight
28) forty-five
29) forty-four
30) forty-nine
31) forty-one
32) forty-seven
33) forty-six
34) forty-three
35) forty-two
36) four
37) fourteen
38) hundred
39) nine
40) nineteen
41) ninety
42) ninety-eight
43) ninety-five
44) ninety-four
45) ninety-nine
46) ninety-one
47) ninety-seven
48) ninety-six
49) ninety-three
50) ninety-two
51) one
52) seven
53) seventeen
54) seventy
55) seventy-eight
56) seventy-five
57) seventy-four
58) seventy-nine
59) seventy-one
60) seventy-seven
61) seventy-six
62) seventy-three
63) seventy-two
64) six
65) sixteen
66) sixty
67) sixty-eight
68) sixty-five
69) sixty-four
70) sixty-nine
71) sixty-one
72) sixty-seven
73) sixty-six
74) sixty-three
75) sixty-two
76) ten
77) thirteen
78) thirty
79) thirty-eight
80) thirty-five
81) thirty-four
82) thirty-nine
83) thirty-one
84) thirty-seven
85) thirty-six
86) thirty-three
87) thirty-two
88) three
89) twelve
90) twenty
91) twenty-eight
92) twenty-five
93) twenty-four
94) twenty-nine
95) twenty-one
96) twenty-seven
97) twenty-six
98) twenty-three
99) twenty-two
100) two
101) zero
Let me add that I don't necessarily expect that the order is a well-order. For example, if we have a naming convention whereby $10^k$ is represented for large $k$ simply by repeating "penpenpenpen$cdots$pen", then we could make a descending sequence via penpenpenpen$cdots$pen twelve, which would descend as the number of pen's increased, since we would be replacing t with p.
elementary-number-theory elementary-set-theory logic order-theory puzzle
elementary-number-theory elementary-set-theory logic order-theory puzzle
edited Dec 21 '18 at 15:45
Aloizio Macedo♦
23.6k23887
asked Dec 20 '18 at 13:24
JDHJDH
32.6k680145
edited Dec 21 '18 at 15:45
Aloizio Macedo♦
23.6k23887
asked Dec 20 '18 at 13:24
JDHJDH
32.6k680145
edited Dec 21 '18 at 15:45
Aloizio Macedo♦
23.6k23887
edited Dec 21 '18 at 15:45
Aloizio Macedo♦
23.6k23887
edited Dec 21 '18 at 15:45
Aloizio Macedo♦
23.6k23887
23.6k23887
asked Dec 20 '18 at 13:24
JDHJDH
32.6k680145
asked Dec 20 '18 at 13:24
JDHJDH
32.6k680145
asked Dec 20 '18 at 13:24
JDHJDH
32.6k680145
32.6k680145
2
$begingroup$
I have a method for writing the names of numbers in alphabet: For $n$ just write a string of $n+1$ times the letter "a". So $0$ is "a", and "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" is $345$. Now the order type is again $omega$!
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 13:35
1
$begingroup$
This exact question was asked two weeks ago on r/math subreddit.
$endgroup$
– Wojowu
Dec 20 '18 at 13:48
1
$begingroup$
@Wojowu Great! I guess the puzzle is in the air.
$endgroup$
– JDH
Dec 20 '18 at 13:53
1
$begingroup$
@Holo: It will be, the string will just be indexed with a non-standard integer.
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 20:48
3
$begingroup$
Actually, not even all native English speakers are "familiar with" the naming scheme of your first example: in British English, 15,249 is read out as "fifteen thousand two hundred and forty nine". Aside from that the form of your question amounts to: "please invent something nice and then tell me what its properties are". This is not a good MSE question.
$endgroup$
– Rob Arthan
Dec 20 '18 at 22:54
|
show 17 more comments
2
$begingroup$
I have a method for writing the names of numbers in alphabet: For $n$ just write a string of $n+1$ times the letter "a". So $0$ is "a", and "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" is $345$. Now the order type is again $omega$!
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 13:35
1
$begingroup$
This exact question was asked two weeks ago on r/math subreddit.
$endgroup$
– Wojowu
Dec 20 '18 at 13:48
1
$begingroup$
@Wojowu Great! I guess the puzzle is in the air.
$endgroup$
– JDH
Dec 20 '18 at 13:53
1
$begingroup$
@Holo: It will be, the string will just be indexed with a non-standard integer.
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 20:48
3
$begingroup$
Actually, not even all native English speakers are "familiar with" the naming scheme of your first example: in British English, 15,249 is read out as "fifteen thousand two hundred and forty nine". Aside from that the form of your question amounts to: "please invent something nice and then tell me what its properties are". This is not a good MSE question.
$endgroup$
– Rob Arthan
Dec 20 '18 at 22:54
2
2
$begingroup$
I have a method for writing the names of numbers in alphabet: For $n$ just write a string of $n+1$ times the letter "a". So $0$ is "a", and "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" is $345$. Now the order type is again $omega$!
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 13:35
$begingroup$
I have a method for writing the names of numbers in alphabet: For $n$ just write a string of $n+1$ times the letter "a". So $0$ is "a", and "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" is $345$. Now the order type is again $omega$!
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 13:35
1
1
$begingroup$
This exact question was asked two weeks ago on r/math subreddit.
$endgroup$
– Wojowu
Dec 20 '18 at 13:48
$begingroup$
This exact question was asked two weeks ago on r/math subreddit.
$endgroup$
– Wojowu
Dec 20 '18 at 13:48
1
1
$begingroup$
@Wojowu Great! I guess the puzzle is in the air.
$endgroup$
– JDH
Dec 20 '18 at 13:53
$begingroup$
@Wojowu Great! I guess the puzzle is in the air.
$endgroup$
– JDH
Dec 20 '18 at 13:53
1
1
$begingroup$
@Holo: It will be, the string will just be indexed with a non-standard integer.
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 20:48
$begingroup$
@Holo: It will be, the string will just be indexed with a non-standard integer.
$endgroup$
– Asaf Karagila♦
Dec 20 '18 at 20:48
3
3
$begingroup$
Actually, not even all native English speakers are "familiar with" the naming scheme of your first example: in British English, 15,249 is read out as "fifteen thousand two hundred and forty nine". Aside from that the form of your question amounts to: "please invent something nice and then tell me what its properties are". This is not a good MSE question.
$endgroup$
– Rob Arthan
Dec 20 '18 at 22:54
$begingroup$
Actually, not even all native English speakers are "familiar with" the naming scheme of your first example: in British English, 15,249 is read out as "fifteen thousand two hundred and forty nine". Aside from that the form of your question amounts to: "please invent something nice and then tell me what its properties are". This is not a good MSE question.
$endgroup$
– Rob Arthan
Dec 20 '18 at 22:54
|
show 17 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let us consider the digit-pronunciation naming system, by which
one simply pronounces the digits of a number in order, so that
$7216$ is pronounced "seven two one six" and so on for any number.
Thus, we obtain a naming system of the numbers, and while it does
not extend the standard nomenclature, nevertheless I find it to be
perfectly sensible, providing a definite unique name for every
natural number. This naming system is sometimes actually used for
very large numbers, such as reading off the number on a credit
card, and it is also commonly used to help disambiguate small
numbers, such as $50$ and $15$. So I find it to be a reasonable
naming system.
Let us place the natural numbers in alphabetical order with respect
to this naming system. Thus, $882746$ appears alphabetically before
$87$, which appears before $8734$. Note that any prefix of a word
appears earlier in the alphabetical order.
Theorem. The order type of the natural numbers, in alphabetical
order with respect to the digit-pronunciation naming system, is
exactly $$omegacdot(1+mathbb{Q})+1.$$
Proof. That is, we have $1+mathbb{Q}$ many copies of $omega$, with a
final point on top.
I will analyze the naming system with respect to base ten, but a
similar analysis works regardless of the base.
Consider first the alphabetical order of the ten digits themselves:
eight, five, four, nine, one, seven, six, three, two, zero
Notice that these digit names are prefix-free — none of them
is an initial segment of another. Thus, when comparing the names of
two numbers, we will never be in a situation where part of one
digit is combined with part of another in order to make the
alphabetical comparison. Rather, the alphabetical order is the same
as the lexical order on the strings of digits themselves,
considered in the alphabetical digit order above.
The largest number of all, in the alphabetical order, is zero,
since no other number starts with the letter "z", and so this
number will appear as the very last entry alphabetically. This
explains the final $+1$ in the theorem claim.
The smallest number in alphabetical order, in contrast, is $8$,
since it begins with "e", and the only other numbers beginning with
"e" also begin with $8$, followed possibly by additional digits,
and thus will appear after the single-digit $8$.
The next number after $8$, alphabetically, is $88$ and then $888$
and $8888$ and so on. I claim that every number (except $0$) has an
alphabetical successor, which is simply to add a digit $8$ at the
end of the decimal representation of the number. For example, the
next number after $532876$ is $5328768$, because any other digit
sequence above the first number must either extend it or deviate
from one of those digits. But $5328768$ will be below any other
higher deviation or extension, and so it is a successor. Similarly,
$53287688$ and $532876888$ are the next few numbers, simply adding
more $8$'s at the end.
Thus, every number except $0$ in the alphabetical order is followed
by a sequence of order type $omega$, which is obtained by simply
tacking on additional $8$s. And so the order will be a number of
copies of $omega$, plus one more point $0$ at the top.
Let me argue that those copies of $omega$ are themselves densely
ordered. If one number $m$ precedes another $n$ alphabetically, but
$n$ is not just adding $8$'s to the end of the decimal
representation of $m$, then either there is some alphabetically
upward deviation in the digits of $m$ to form $n$, or else $n$
extends the digits of $m$, but eventually using some digits other
than $8$. It is easy to see that we can find another number in
between, which also won't be just adding $8$s.
Perhaps it is easiest to see this by example. The number $7536$ is
alphabetically prior to $752$, since "three" is alphabetically
earlier than "two". In between these numbers, we can find $75366$,
which has it own copy of $omega$ arising from $753668$, $7536688$,
$75366888$ and so on.
Thus, the blocks of $omega$ obtained by appending $8$'s are
themselves densely ordered: between any two of them we can find
another.
Notice that there is a very first such block of $omega$ in the
alphabetical order the numbers, namely, the block consisting of
$8$, $88$, $888$ and so on, which appears at the very beginning of
the numbers in alphabetical order.
There is in contrast no largest block, before the final $0$,
because if we are given any number $n$, we can append some other
digits other than $8$ to the end of the decimal representation, and
thereby find another copy of $omega$ above $n$ in the alphabetical
order.
Thus, the $omega$ blocks arising from appending $8$'s are
themselves densely ordered, with a first such block and no last
such block. Since there are only countably many numbers, we must
have exactly $1+mathbb{Q}$ many such blocks of size $omega$. And
with the final point $0$ at the very top, it follows that the order
type of the natural numbers in the digit-pronunciation naming
system is precisely $$omegacdot(1+mathbb{Q})+1,$$ as claimed.
$Box$.
Several of us had discussed this problem over beers last night in
Münster, including Stefan Hoffelner and Stefan Mesken, following my talk at the Münster Logic
Oberseminar. Stefan Hoffelner had suggested that we consider the digit-pronunciation naming system.
Let me say finally that it seems to me that the features of the
digit-pronunciation naming system will appear essentially in all
the naming systems, and so I expect this kind of analysis to be
able to extend to the other nomenclatures, with perhaps slightly
different endpoint effects.
answered Jan 12 at 8:49
JDHJDH
32.6k680145
$endgroup$
add a comment |
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$begingroup$
Let us consider the digit-pronunciation naming system, by which
one simply pronounces the digits of a number in order, so that
$7216$ is pronounced "seven two one six" and so on for any number.
Thus, we obtain a naming system of the numbers, and while it does
not extend the standard nomenclature, nevertheless I find it to be
perfectly sensible, providing a definite unique name for every
natural number. This naming system is sometimes actually used for
very large numbers, such as reading off the number on a credit
card, and it is also commonly used to help disambiguate small
numbers, such as $50$ and $15$. So I find it to be a reasonable
naming system.
Let us place the natural numbers in alphabetical order with respect
to this naming system. Thus, $882746$ appears alphabetically before
$87$, which appears before $8734$. Note that any prefix of a word
appears earlier in the alphabetical order.
Theorem. The order type of the natural numbers, in alphabetical
order with respect to the digit-pronunciation naming system, is
exactly $$omegacdot(1+mathbb{Q})+1.$$
Proof. That is, we have $1+mathbb{Q}$ many copies of $omega$, with a
final point on top.
I will analyze the naming system with respect to base ten, but a
similar analysis works regardless of the base.
Consider first the alphabetical order of the ten digits themselves:
eight, five, four, nine, one, seven, six, three, two, zero
Notice that these digit names are prefix-free — none of them
is an initial segment of another. Thus, when comparing the names of
two numbers, we will never be in a situation where part of one
digit is combined with part of another in order to make the
alphabetical comparison. Rather, the alphabetical order is the same
as the lexical order on the strings of digits themselves,
considered in the alphabetical digit order above.
The largest number of all, in the alphabetical order, is zero,
since no other number starts with the letter "z", and so this
number will appear as the very last entry alphabetically. This
explains the final $+1$ in the theorem claim.
The smallest number in alphabetical order, in contrast, is $8$,
since it begins with "e", and the only other numbers beginning with
"e" also begin with $8$, followed possibly by additional digits,
and thus will appear after the single-digit $8$.
The next number after $8$, alphabetically, is $88$ and then $888$
and $8888$ and so on. I claim that every number (except $0$) has an
alphabetical successor, which is simply to add a digit $8$ at the
end of the decimal representation of the number. For example, the
next number after $532876$ is $5328768$, because any other digit
sequence above the first number must either extend it or deviate
from one of those digits. But $5328768$ will be below any other
higher deviation or extension, and so it is a successor. Similarly,
$53287688$ and $532876888$ are the next few numbers, simply adding
more $8$'s at the end.
Thus, every number except $0$ in the alphabetical order is followed
by a sequence of order type $omega$, which is obtained by simply
tacking on additional $8$s. And so the order will be a number of
copies of $omega$, plus one more point $0$ at the top.
Let me argue that those copies of $omega$ are themselves densely
ordered. If one number $m$ precedes another $n$ alphabetically, but
$n$ is not just adding $8$'s to the end of the decimal
representation of $m$, then either there is some alphabetically
upward deviation in the digits of $m$ to form $n$, or else $n$
extends the digits of $m$, but eventually using some digits other
than $8$. It is easy to see that we can find another number in
between, which also won't be just adding $8$s.
Perhaps it is easiest to see this by example. The number $7536$ is
alphabetically prior to $752$, since "three" is alphabetically
earlier than "two". In between these numbers, we can find $75366$,
which has it own copy of $omega$ arising from $753668$, $7536688$,
$75366888$ and so on.
Thus, the blocks of $omega$ obtained by appending $8$'s are
themselves densely ordered: between any two of them we can find
another.
Notice that there is a very first such block of $omega$ in the
alphabetical order the numbers, namely, the block consisting of
$8$, $88$, $888$ and so on, which appears at the very beginning of
the numbers in alphabetical order.
There is in contrast no largest block, before the final $0$,
because if we are given any number $n$, we can append some other
digits other than $8$ to the end of the decimal representation, and
thereby find another copy of $omega$ above $n$ in the alphabetical
order.
Thus, the $omega$ blocks arising from appending $8$'s are
themselves densely ordered, with a first such block and no last
such block. Since there are only countably many numbers, we must
have exactly $1+mathbb{Q}$ many such blocks of size $omega$. And
with the final point $0$ at the very top, it follows that the order
type of the natural numbers in the digit-pronunciation naming
system is precisely $$omegacdot(1+mathbb{Q})+1,$$ as claimed.
$Box$.
Several of us had discussed this problem over beers last night in
Münster, including Stefan Hoffelner and Stefan Mesken, following my talk at the Münster Logic
Oberseminar. Stefan Hoffelner had suggested that we consider the digit-pronunciation naming system.
Let me say finally that it seems to me that the features of the
digit-pronunciation naming system will appear essentially in all
the naming systems, and so I expect this kind of analysis to be
able to extend to the other nomenclatures, with perhaps slightly
different endpoint effects.
answered Jan 12 at 8:49
JDHJDH
32.6k680145
$endgroup$
add a comment |
$begingroup$
Let us consider the digit-pronunciation naming system, by which
one simply pronounces the digits of a number in order, so that
$7216$ is pronounced "seven two one six" and so on for any number.
Thus, we obtain a naming system of the numbers, and while it does
not extend the standard nomenclature, nevertheless I find it to be
perfectly sensible, providing a definite unique name for every
natural number. This naming system is sometimes actually used for
very large numbers, such as reading off the number on a credit
card, and it is also commonly used to help disambiguate small
numbers, such as $50$ and $15$. So I find it to be a reasonable
naming system.
Let us place the natural numbers in alphabetical order with respect
to this naming system. Thus, $882746$ appears alphabetically before
$87$, which appears before $8734$. Note that any prefix of a word
appears earlier in the alphabetical order.
Theorem. The order type of the natural numbers, in alphabetical
order with respect to the digit-pronunciation naming system, is
exactly $$omegacdot(1+mathbb{Q})+1.$$
Proof. That is, we have $1+mathbb{Q}$ many copies of $omega$, with a
final point on top.
I will analyze the naming system with respect to base ten, but a
similar analysis works regardless of the base.
Consider first the alphabetical order of the ten digits themselves:
eight, five, four, nine, one, seven, six, three, two, zero
Notice that these digit names are prefix-free — none of them
is an initial segment of another. Thus, when comparing the names of
two numbers, we will never be in a situation where part of one
digit is combined with part of another in order to make the
alphabetical comparison. Rather, the alphabetical order is the same
as the lexical order on the strings of digits themselves,
considered in the alphabetical digit order above.
The largest number of all, in the alphabetical order, is zero,
since no other number starts with the letter "z", and so this
number will appear as the very last entry alphabetically. This
explains the final $+1$ in the theorem claim.
The smallest number in alphabetical order, in contrast, is $8$,
since it begins with "e", and the only other numbers beginning with
"e" also begin with $8$, followed possibly by additional digits,
and thus will appear after the single-digit $8$.
The next number after $8$, alphabetically, is $88$ and then $888$
and $8888$ and so on. I claim that every number (except $0$) has an
alphabetical successor, which is simply to add a digit $8$ at the
end of the decimal representation of the number. For example, the
next number after $532876$ is $5328768$, because any other digit
sequence above the first number must either extend it or deviate
from one of those digits. But $5328768$ will be below any other
higher deviation or extension, and so it is a successor. Similarly,
$53287688$ and $532876888$ are the next few numbers, simply adding
more $8$'s at the end.
Thus, every number except $0$ in the alphabetical order is followed
by a sequence of order type $omega$, which is obtained by simply
tacking on additional $8$s. And so the order will be a number of
copies of $omega$, plus one more point $0$ at the top.
Let me argue that those copies of $omega$ are themselves densely
ordered. If one number $m$ precedes another $n$ alphabetically, but
$n$ is not just adding $8$'s to the end of the decimal
representation of $m$, then either there is some alphabetically
upward deviation in the digits of $m$ to form $n$, or else $n$
extends the digits of $m$, but eventually using some digits other
than $8$. It is easy to see that we can find another number in
between, which also won't be just adding $8$s.
Perhaps it is easiest to see this by example. The number $7536$ is
alphabetically prior to $752$, since "three" is alphabetically
earlier than "two". In between these numbers, we can find $75366$,
which has it own copy of $omega$ arising from $753668$, $7536688$,
$75366888$ and so on.
Thus, the blocks of $omega$ obtained by appending $8$'s are
themselves densely ordered: between any two of them we can find
another.
Notice that there is a very first such block of $omega$ in the
alphabetical order the numbers, namely, the block consisting of
$8$, $88$, $888$ and so on, which appears at the very beginning of
the numbers in alphabetical order.
There is in contrast no largest block, before the final $0$,
because if we are given any number $n$, we can append some other
digits other than $8$ to the end of the decimal representation, and
thereby find another copy of $omega$ above $n$ in the alphabetical
order.
Thus, the $omega$ blocks arising from appending $8$'s are
themselves densely ordered, with a first such block and no last
such block. Since there are only countably many numbers, we must
have exactly $1+mathbb{Q}$ many such blocks of size $omega$. And
with the final point $0$ at the very top, it follows that the order
type of the natural numbers in the digit-pronunciation naming
system is precisely $$omegacdot(1+mathbb{Q})+1,$$ as claimed.
$Box$.
Several of us had discussed this problem over beers last night in
Münster, including Stefan Hoffelner and Stefan Mesken, following my talk at the Münster Logic
Oberseminar. Stefan Hoffelner had suggested that we consider the digit-pronunciation naming system.
Let me say finally that it seems to me that the features of the
digit-pronunciation naming system will appear essentially in all
the naming systems, and so I expect this kind of analysis to be
able to extend to the other nomenclatures, with perhaps slightly
different endpoint effects.
answered Jan 12 at 8:49
JDHJDH
32.6k680145
$endgroup$
add a comment |
$begingroup$
Let us consider the digit-pronunciation naming system, by which
one simply pronounces the digits of a number in order, so that
$7216$ is pronounced "seven two one six" and so on for any number.
Thus, we obtain a naming system of the numbers, and while it does
not extend the standard nomenclature, nevertheless I find it to be
perfectly sensible, providing a definite unique name for every
natural number. This naming system is sometimes actually used for
very large numbers, such as reading off the number on a credit
card, and it is also commonly used to help disambiguate small
numbers, such as $50$ and $15$. So I find it to be a reasonable
naming system.
Let us place the natural numbers in alphabetical order with respect
to this naming system. Thus, $882746$ appears alphabetically before
$87$, which appears before $8734$. Note that any prefix of a word
appears earlier in the alphabetical order.
Theorem. The order type of the natural numbers, in alphabetical
order with respect to the digit-pronunciation naming system, is
exactly $$omegacdot(1+mathbb{Q})+1.$$
Proof. That is, we have $1+mathbb{Q}$ many copies of $omega$, with a
final point on top.
I will analyze the naming system with respect to base ten, but a
similar analysis works regardless of the base.
Consider first the alphabetical order of the ten digits themselves:
eight, five, four, nine, one, seven, six, three, two, zero
Notice that these digit names are prefix-free — none of them
is an initial segment of another. Thus, when comparing the names of
two numbers, we will never be in a situation where part of one
digit is combined with part of another in order to make the
alphabetical comparison. Rather, the alphabetical order is the same
as the lexical order on the strings of digits themselves,
considered in the alphabetical digit order above.
The largest number of all, in the alphabetical order, is zero,
since no other number starts with the letter "z", and so this
number will appear as the very last entry alphabetically. This
explains the final $+1$ in the theorem claim.
The smallest number in alphabetical order, in contrast, is $8$,
since it begins with "e", and the only other numbers beginning with
"e" also begin with $8$, followed possibly by additional digits,
and thus will appear after the single-digit $8$.
The next number after $8$, alphabetically, is $88$ and then $888$
and $8888$ and so on. I claim that every number (except $0$) has an
alphabetical successor, which is simply to add a digit $8$ at the
end of the decimal representation of the number. For example, the
next number after $532876$ is $5328768$, because any other digit
sequence above the first number must either extend it or deviate
from one of those digits. But $5328768$ will be below any other
higher deviation or extension, and so it is a successor. Similarly,
$53287688$ and $532876888$ are the next few numbers, simply adding
more $8$'s at the end.
Thus, every number except $0$ in the alphabetical order is followed
by a sequence of order type $omega$, which is obtained by simply
tacking on additional $8$s. And so the order will be a number of
copies of $omega$, plus one more point $0$ at the top.
Let me argue that those copies of $omega$ are themselves densely
ordered. If one number $m$ precedes another $n$ alphabetically, but
$n$ is not just adding $8$'s to the end of the decimal
representation of $m$, then either there is some alphabetically
upward deviation in the digits of $m$ to form $n$, or else $n$
extends the digits of $m$, but eventually using some digits other
than $8$. It is easy to see that we can find another number in
between, which also won't be just adding $8$s.
Perhaps it is easiest to see this by example. The number $7536$ is
alphabetically prior to $752$, since "three" is alphabetically
earlier than "two". In between these numbers, we can find $75366$,
which has it own copy of $omega$ arising from $753668$, $7536688$,
$75366888$ and so on.
Thus, the blocks of $omega$ obtained by appending $8$'s are
themselves densely ordered: between any two of them we can find
another.
Notice that there is a very first such block of $omega$ in the
alphabetical order the numbers, namely, the block consisting of
$8$, $88$, $888$ and so on, which appears at the very beginning of
the numbers in alphabetical order.
There is in contrast no largest block, before the final $0$,
because if we are given any number $n$, we can append some other
digits other than $8$ to the end of the decimal representation, and
thereby find another copy of $omega$ above $n$ in the alphabetical
order.
Thus, the $omega$ blocks arising from appending $8$'s are
themselves densely ordered, with a first such block and no last
such block. Since there are only countably many numbers, we must
have exactly $1+mathbb{Q}$ many such blocks of size $omega$. And
with the final point $0$ at the very top, it follows that the order
type of the natural numbers in the digit-pronunciation naming
system is precisely $$omegacdot(1+mathbb{Q})+1,$$ as claimed.
$Box$.
Several of us had discussed this problem over beers last night in
Münster, including Stefan Hoffelner and Stefan Mesken, following my talk at the Münster Logic
Oberseminar. Stefan Hoffelner had suggested that we consider the digit-pronunciation naming system.
Let me say finally that it seems to me that the features of the
digit-pronunciation naming system will appear essentially in all
the naming systems, and so I expect this kind of analysis to be
able to extend to the other nomenclatures, with perhaps slightly
different endpoint effects.
answered Jan 12 at 8:49
JDHJDH
32.6k680145
$endgroup$
Let us consider the digit-pronunciation naming system, by which
one simply pronounces the digits of a number in order, so that
$7216$ is pronounced "seven two one six" and so on for any number.
Thus, we obtain a naming system of the numbers, and while it does
not extend the standard nomenclature, nevertheless I find it to be
perfectly sensible, providing a definite unique name for every
natural number. This naming system is sometimes actually used for
very large numbers, such as reading off the number on a credit
card, and it is also commonly used to help disambiguate small
numbers, such as $50$ and $15$. So I find it to be a reasonable
naming system.
Let us place the natural numbers in alphabetical order with respect
to this naming system. Thus, $882746$ appears alphabetically before
$87$, which appears before $8734$. Note that any prefix of a word
appears earlier in the alphabetical order.
Theorem. The order type of the natural numbers, in alphabetical
order with respect to the digit-pronunciation naming system, is
exactly $$omegacdot(1+mathbb{Q})+1.$$
Proof. That is, we have $1+mathbb{Q}$ many copies of $omega$, with a
final point on top.
I will analyze the naming system with respect to base ten, but a
similar analysis works regardless of the base.
Consider first the alphabetical order of the ten digits themselves:
eight, five, four, nine, one, seven, six, three, two, zero
Notice that these digit names are prefix-free — none of them
is an initial segment of another. Thus, when comparing the names of
two numbers, we will never be in a situation where part of one
digit is combined with part of another in order to make the
alphabetical comparison. Rather, the alphabetical order is the same
as the lexical order on the strings of digits themselves,
considered in the alphabetical digit order above.
The largest number of all, in the alphabetical order, is zero,
since no other number starts with the letter "z", and so this
number will appear as the very last entry alphabetically. This
explains the final $+1$ in the theorem claim.
The smallest number in alphabetical order, in contrast, is $8$,
since it begins with "e", and the only other numbers beginning with
"e" also begin with $8$, followed possibly by additional digits,
and thus will appear after the single-digit $8$.
The next number after $8$, alphabetically, is $88$ and then $888$
and $8888$ and so on. I claim that every number (except $0$) has an
alphabetical successor, which is simply to add a digit $8$ at the
end of the decimal representation of the number. For example, the
next number after $532876$ is $5328768$, because any other digit
sequence above the first number must either extend it or deviate
from one of those digits. But $5328768$ will be below any other
higher deviation or extension, and so it is a successor. Similarly,
$53287688$ and $532876888$ are the next few numbers, simply adding
more $8$'s at the end.
Thus, every number except $0$ in the alphabetical order is followed
by a sequence of order type $omega$, which is obtained by simply
tacking on additional $8$s. And so the order will be a number of
copies of $omega$, plus one more point $0$ at the top.
Let me argue that those copies of $omega$ are themselves densely
ordered. If one number $m$ precedes another $n$ alphabetically, but
$n$ is not just adding $8$'s to the end of the decimal
representation of $m$, then either there is some alphabetically
upward deviation in the digits of $m$ to form $n$, or else $n$
extends the digits of $m$, but eventually using some digits other
than $8$. It is easy to see that we can find another number in
between, which also won't be just adding $8$s.
Perhaps it is easiest to see this by example. The number $7536$ is
alphabetically prior to $752$, since "three" is alphabetically
earlier than "two". In between these numbers, we can find $75366$,
which has it own copy of $omega$ arising from $753668$, $7536688$,
$75366888$ and so on.
Thus, the blocks of $omega$ obtained by appending $8$'s are
themselves densely ordered: between any two of them we can find
another.
Notice that there is a very first such block of $omega$ in the
alphabetical order the numbers, namely, the block consisting of
$8$, $88$, $888$ and so on, which appears at the very beginning of
the numbers in alphabetical order.
There is in contrast no largest block, before the final $0$,
because if we are given any number $n$, we can append some other
digits other than $8$ to the end of the decimal representation, and
thereby find another copy of $omega$ above $n$ in the alphabetical
order.
Thus, the $omega$ blocks arising from appending $8$'s are
themselves densely ordered, with a first such block and no last
such block. Since there are only countably many numbers, we must
have exactly $1+mathbb{Q}$ many such blocks of size $omega$. And
with the final point $0$ at the very top, it follows that the order
type of the natural numbers in the digit-pronunciation naming
system is precisely $$omegacdot(1+mathbb{Q})+1,$$ as claimed.
$Box$.
Several of us had discussed this problem over beers last night in
Münster, including Stefan Hoffelner and Stefan Mesken, following my talk at the Münster Logic
Oberseminar. Stefan Hoffelner had suggested that we consider the digit-pronunciation naming system.
Let me say finally that it seems to me that the features of the
digit-pronunciation naming system will appear essentially in all
the naming systems, and so I expect this kind of analysis to be
able to extend to the other nomenclatures, with perhaps slightly
different endpoint effects.
answered Jan 12 at 8:49
JDHJDH
32.6k680145
edited Jan 12 at 16:15
answered Jan 12 at 8:49
JDHJDH
32.6k680145
answered Jan 12 at 8:49
JDHJDH
32.6k680145
answered Jan 12 at 8:49
JDHJDH
32.6k680145
32.6k680145
add a comment |
add a comment |
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I have a method for writing the names of numbers in alphabet: For $n$ just write a string of $n+1$ times the letter "a". So $0$ is "a", and "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" is $345$. Now the order type is again $omega$!
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– Asaf Karagila♦
Dec 20 '18 at 13:35
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This exact question was asked two weeks ago on r/math subreddit.
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– Wojowu
Dec 20 '18 at 13:48
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@Wojowu Great! I guess the puzzle is in the air.
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– JDH
Dec 20 '18 at 13:53
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@Holo: It will be, the string will just be indexed with a non-standard integer.
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– Asaf Karagila♦
Dec 20 '18 at 20:48
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Actually, not even all native English speakers are "familiar with" the naming scheme of your first example: in British English, 15,249 is read out as "fifteen thousand two hundred and forty nine". Aside from that the form of your question amounts to: "please invent something nice and then tell me what its properties are". This is not a good MSE question.
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– Rob Arthan
Dec 20 '18 at 22:54