Hopping to infinity along a string of digitswhat is the new order of the digits here ? Both the numbers $144$ and $441$ consists of the same digits?Deterministic Turing machine for a duplicate concatenation of a stringprove that language is not decidable (string and reverse)Sequences of consecutive digits of arbitrary length in irrational ternary numbers (take two).Can I guess an irrational number formula from its digits?Reversing the digits of an infinite decimalIs each string decideble?Conjecture about the digits of $pi$Cardinality of number of digitsNumbers such that they equal the product of their own digits

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Hopping to infinity along a string of digits

what is the new order of the digits here ? Both the numbers $144$ and $441$ consists of the same digits?Deterministic Turing machine for a duplicate concatenation of a stringprove that language is not decidable (string and reverse)Sequences of consecutive digits of arbitrary length in irrational ternary numbers (take two).Can I guess an irrational number formula from its digits?Reversing the digits of an infinite decimalIs each string decideble?Conjecture about the digits of $pi$Cardinality of number of digitsNumbers such that they equal the product of their own digits

Let $s$ be an infinite string of decimal digits, for example:
beginarraycccccccccc
s = 3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 & cdots
endarray
Consider a marker, the head, pointing to the first digit, $3$ in the above example. Interpret the digit under
the head as an instruction to move the head $3$ digits to the right, i.e., to the $4$th digit. Now the head is pointing to $1$. Interpret this as an instruction to move $1$ place to the left. Continue in this manner, hopping through the string, alternately moving right and left. Think of the head as akin to the head of a Turing machine, and $s$ as the tape of instructions.

There are three possible behaviors.
(1) The head moves off the left end of $s$:

beginarraycccccccccc
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text $^wedge$ & text & text & text & text & text & text & text & text & text \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text & text & text & text $^wedge$ & text & text & text & text & text & text \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text & text & text $^wedge$ & text & text & text & text & text & text & text \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text & text & text & text & text & text & text $^wedge$ & text & text & text \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text & text & text & text & text $^wedge$ & text & text & text & text & text \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text & text & text & text & text & text & text & text & text & text $^wedge$ \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text & text & text & text & text & text & text $^wedge$ & text & text & text \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text & text & text & text & text & text & text & text & text $^wedge$ & text \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text & text & text & text $^wedge$ & text & text & text & text & text & text \
3 & 1 & 4 & 1 & 5 & 9 & 2 & 6 & 5 & 3 \
text & text & text & text & text $^wedge$ & text & text & text & text & text
endarray

(2) The head goes into a cycle, e.g., when the head hits $0$:

beginarraycccccccccccccc
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text $^wedge$ & text & text & text & text & text & text & text & text & text & text & text & text & text \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text & text & text & text & text & text & text $^wedge$ & text & text & text & text & text & text & text \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text & text & text & text $^wedge$ & text & text & text & text & text & text & text & text & text & text \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text & text & text & text & text & text & text & text & text & text & text $^wedge$ & text & text & text \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text & text & text & text & text $^wedge$ & text & text & text & text & text & text & text & text & text \
6 & 4 & 5 & 7 & 5 & 1 & 3 & 1 & 1 & 0 & 6 & 4 & 5 & 9 \
text & text & text & text & text & text & text & text & text & text $^wedge$ & text & text & text & text
endarray

(3) The head moves off rightward to infnity:

beginarrayccccccccccccc
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text $^wedge$ & text & text & text & text & text & text & text & text & text & text & text & text \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text & text & text & text $^wedge$ & text & text & text & text & text & text & text & text & text \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text & text & text $^wedge$ & text & text & text & text & text & text & text & text & text & text \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text & text & text & text & text & text $^wedge$ & text & text & text & text & text & text & text \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text & text & text & text & text $^wedge$ & text & text & text & text & text & text & text & text \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text & text & text & text & text & text & text & text $^wedge$ & text & text & text & text & text \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text & text & text & text & text & text & text $^wedge$ & text & text & text & text & text & text \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text & text & text & text & text & text & text & text & text & text $^wedge$ & text & text & text \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text & text & text & text & text & text & text & text & text $^wedge$ & text & text & text & text \
3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 & 1 & 3 \
text & text & text & text & text & text & text & text & text & text & text & text $^wedge$ & text \
endarray

This last string could be viewed as the decimal expansion of $31/99 = 0.3131313131313cdots$.

Q1. What is an example of an irrational number
$0.d_1 d_2 d_3 cdots$ whose string $s=d_1 d_2 d_3 cdots$ causes the head to hop rightward to infinity?

Q1.5. (Added). Is there an explicit irrational algebraic number with the hop-to-$infty$ property?

I'm thinking of something like $sqrt7-2$, the
2nd example above (which cycles).

Q2. More generally, which strings
cause the head to hop rightward to infinity?

Update (summarizing answers, 13Apr2019).
Q1. There are irrationals with the hop-to-$infty$ property
(@EthanBolker, @TheSimpliFire),
but explicit construction requires using, e.g., the
Thue-Morse sequence (@Wojowu).
Q1.5. @EthanBolker suggests this may be difficult, and @Wojowu suggests it may be false (b/c: nine consecutive zeros): Perhaps no algebraic irrational has the hop-to-$infty$ property.
Q2. A partial algorithmic characterization by @TheSimpliFire.

edited Apr 14 at 0:51

asked Apr 13 at 12:04

Joseph O'Rourke

18.5k351117

1

$begingroup$
You want a string such that the $D_1=(1+d_1)$th digit is less than $d_1$, that the $D_2=(1+d_1-D_1)$th digit is greater than $D_1$, that the $D_3=(1+d_1-D_1+D_2)$th digit is less than $D_2$, etc. I think it is possible to generate an algorithm and you can try for some simulations.
$endgroup$
– TheSimpliFire
Apr 13 at 12:32

2

$begingroup$
Another question. For sequences that don't hop to infinity behavior is determined by a (finite) initial subsequence. There are only countably many of those. What are they?
$endgroup$
– Ethan Bolker
Apr 13 at 12:42

3

$begingroup$
I think Q1.5 is hard since digit sequences for algebraic numbers are hard adamczewski.perso.math.cnrs.fr/Siauliai.pdf
$endgroup$
– Ethan Bolker
Apr 13 at 13:58

1

$begingroup$
Followup question: what are the measures of the three sets in the interval $(0,1)$?
$endgroup$
– eyeballfrog
Apr 13 at 16:30

2

$begingroup$
@eyeballfrog The set of numbers with hop to infinity property has measure zero, because you can't hop past a string of nine zeros. This also strongly suggests no algebraic irrational has this property.
$endgroup$
– Wojowu
Apr 13 at 18:21

|
show 3 more comments

There are three possible behaviors.
(1) The head moves off the left end of $s$:

(2) The head goes into a cycle, e.g., when the head hits $0$:

(3) The head moves off rightward to infnity:

This last string could be viewed as the decimal expansion of $31/99 = 0.3131313131313cdots$.

Q1. What is an example of an irrational number
$0.d_1 d_2 d_3 cdots$ whose string $s=d_1 d_2 d_3 cdots$ causes the head to hop rightward to infinity?

Q1.5. (Added). Is there an explicit irrational algebraic number with the hop-to-$infty$ property?

I'm thinking of something like $sqrt7-2$, the
2nd example above (which cycles).

Q2. More generally, which strings
cause the head to hop rightward to infinity?

edited Apr 14 at 0:51

asked Apr 13 at 12:04

Joseph O'Rourke

18.5k351117

1

$begingroup$
You want a string such that the $D_1=(1+d_1)$th digit is less than $d_1$, that the $D_2=(1+d_1-D_1)$th digit is greater than $D_1$, that the $D_3=(1+d_1-D_1+D_2)$th digit is less than $D_2$, etc. I think it is possible to generate an algorithm and you can try for some simulations.
$endgroup$
– TheSimpliFire
Apr 13 at 12:32

2

$begingroup$
Another question. For sequences that don't hop to infinity behavior is determined by a (finite) initial subsequence. There are only countably many of those. What are they?
$endgroup$
– Ethan Bolker
Apr 13 at 12:42

3

$begingroup$
I think Q1.5 is hard since digit sequences for algebraic numbers are hard adamczewski.perso.math.cnrs.fr/Siauliai.pdf
$endgroup$
– Ethan Bolker
Apr 13 at 13:58

1

$begingroup$
Followup question: what are the measures of the three sets in the interval $(0,1)$?
$endgroup$
– eyeballfrog
Apr 13 at 16:30

2

$begingroup$
@eyeballfrog The set of numbers with hop to infinity property has measure zero, because you can't hop past a string of nine zeros. This also strongly suggests no algebraic irrational has this property.
$endgroup$
– Wojowu
Apr 13 at 18:21

|
show 3 more comments

There are three possible behaviors.
(1) The head moves off the left end of $s$:

(2) The head goes into a cycle, e.g., when the head hits $0$:

(3) The head moves off rightward to infnity:

This last string could be viewed as the decimal expansion of $31/99 = 0.3131313131313cdots$.

Q1. What is an example of an irrational number
$0.d_1 d_2 d_3 cdots$ whose string $s=d_1 d_2 d_3 cdots$ causes the head to hop rightward to infinity?

Q1.5. (Added). Is there an explicit irrational algebraic number with the hop-to-$infty$ property?

I'm thinking of something like $sqrt7-2$, the
2nd example above (which cycles).

Q2. More generally, which strings
cause the head to hop rightward to infinity?

edited Apr 14 at 0:51

asked Apr 13 at 12:04

Joseph O'Rourke

18.5k351117

There are three possible behaviors.
(1) The head moves off the left end of $s$:

(2) The head goes into a cycle, e.g., when the head hits $0$:

(3) The head moves off rightward to infnity:

This last string could be viewed as the decimal expansion of $31/99 = 0.3131313131313cdots$.

Q1. What is an example of an irrational number
$0.d_1 d_2 d_3 cdots$ whose string $s=d_1 d_2 d_3 cdots$ causes the head to hop rightward to infinity?

Q1.5. (Added). Is there an explicit irrational algebraic number with the hop-to-$infty$ property?

I'm thinking of something like $sqrt7-2$, the
2nd example above (which cycles).

Q2. More generally, which strings
cause the head to hop rightward to infinity?

sequences-and-series recreational-mathematics irrational-numbers decimal-expansion turing-machines

edited Apr 14 at 0:51

asked Apr 13 at 12:04

Joseph O'Rourke

18.5k351117

edited Apr 14 at 0:51

asked Apr 13 at 12:04

Joseph O'Rourke

18.5k351117

edited Apr 14 at 0:51

asked Apr 13 at 12:04

Joseph O'Rourke

18.5k351117

asked Apr 13 at 12:04

Joseph O'Rourke

18.5k351117

asked Apr 13 at 12:04

Joseph O'Rourke

18.5k351117

1

$begingroup$
You want a string such that the $D_1=(1+d_1)$th digit is less than $d_1$, that the $D_2=(1+d_1-D_1)$th digit is greater than $D_1$, that the $D_3=(1+d_1-D_1+D_2)$th digit is less than $D_2$, etc. I think it is possible to generate an algorithm and you can try for some simulations.
$endgroup$
– TheSimpliFire
Apr 13 at 12:32

2

$begingroup$
Another question. For sequences that don't hop to infinity behavior is determined by a (finite) initial subsequence. There are only countably many of those. What are they?
$endgroup$
– Ethan Bolker
Apr 13 at 12:42

3

$begingroup$
I think Q1.5 is hard since digit sequences for algebraic numbers are hard adamczewski.perso.math.cnrs.fr/Siauliai.pdf
$endgroup$
– Ethan Bolker
Apr 13 at 13:58

1

$begingroup$
Followup question: what are the measures of the three sets in the interval $(0,1)$?
$endgroup$
– eyeballfrog
Apr 13 at 16:30

2

$begingroup$
@eyeballfrog The set of numbers with hop to infinity property has measure zero, because you can't hop past a string of nine zeros. This also strongly suggests no algebraic irrational has this property.
$endgroup$
– Wojowu
Apr 13 at 18:21

|
show 3 more comments

1

$begingroup$
You want a string such that the $D_1=(1+d_1)$th digit is less than $d_1$, that the $D_2=(1+d_1-D_1)$th digit is greater than $D_1$, that the $D_3=(1+d_1-D_1+D_2)$th digit is less than $D_2$, etc. I think it is possible to generate an algorithm and you can try for some simulations.
$endgroup$
– TheSimpliFire
Apr 13 at 12:32

2

$begingroup$
Another question. For sequences that don't hop to infinity behavior is determined by a (finite) initial subsequence. There are only countably many of those. What are they?
$endgroup$
– Ethan Bolker
Apr 13 at 12:42

3

$begingroup$
I think Q1.5 is hard since digit sequences for algebraic numbers are hard adamczewski.perso.math.cnrs.fr/Siauliai.pdf
$endgroup$
– Ethan Bolker
Apr 13 at 13:58

1

$begingroup$
Followup question: what are the measures of the three sets in the interval $(0,1)$?
$endgroup$
– eyeballfrog
Apr 13 at 16:30

2

$begingroup$
@eyeballfrog The set of numbers with hop to infinity property has measure zero, because you can't hop past a string of nine zeros. This also strongly suggests no algebraic irrational has this property.
$endgroup$
– Wojowu
Apr 13 at 18:21

You want a string such that the $D_1=(1+d_1)$th digit is less than $d_1$, that the $D_2=(1+d_1-D_1)$th digit is greater than $D_1$, that the $D_3=(1+d_1-D_1+D_2)$th digit is less than $D_2$, etc. I think it is possible to generate an algorithm and you can try for some simulations.

– TheSimpliFire
Apr 13 at 12:32

Another question. For sequences that don't hop to infinity behavior is determined by a (finite) initial subsequence. There are only countably many of those. What are they?

– Ethan Bolker
Apr 13 at 12:42

I think Q1.5 is hard since digit sequences for algebraic numbers are hard adamczewski.perso.math.cnrs.fr/Siauliai.pdf

– Ethan Bolker
Apr 13 at 13:58

Followup question: what are the measures of the three sets in the interval $(0,1)$?

– eyeballfrog
Apr 13 at 16:30

@eyeballfrog The set of numbers with hop to infinity property has measure zero, because you can't hop past a string of nine zeros. This also strongly suggests no algebraic irrational has this property.

– Wojowu
Apr 13 at 18:21

|
show 3 more comments

1 Answer
1

active

oldest

votes

$$
x 1^x-2 y 1^y-2 z1^z-2 ldots
$$
moves off to infinity for any sequence of digits $xyzldots$ between $3$ and $9$. Select a sequence that defines an irrational number.

More generally

$$
x 1 ?^x-1 y 1 ?^y-1 z 1 ?^z-1 ldots
$$
works, where $?^n$ is an arbitrary string of $n$ digits, since those spots will never be hopped on.

edited Apr 13 at 13:14

answered Apr 13 at 12:24

Ethan Bolker

47.6k556123

$begingroup$
(+1) I was thinking exactly the same thing just as you posted your answer!
$endgroup$
– TheSimpliFire
Apr 13 at 12:25

$begingroup$
Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.
$endgroup$
– Joseph O'Rourke
Apr 13 at 14:17

4

$begingroup$
Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.
$endgroup$
– Ethan Bolker
Apr 13 at 14:47

$begingroup$
I worry about explicitly specifying $t=0.xyzcdots$, excluding digits $ 0,1,2 $, guaranteeing $t$ is irrational.
$endgroup$
– Joseph O'Rourke
Apr 13 at 16:46

4

$begingroup$
@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.
$endgroup$
– Wojowu
Apr 13 at 18:28

|
show 1 more comment

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1 Answer
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1 Answer
1

active

oldest

votes

$$
x 1^x-2 y 1^y-2 z1^z-2 ldots
$$
moves off to infinity for any sequence of digits $xyzldots$ between $3$ and $9$. Select a sequence that defines an irrational number.

More generally

$$
x 1 ?^x-1 y 1 ?^y-1 z 1 ?^z-1 ldots
$$
works, where $?^n$ is an arbitrary string of $n$ digits, since those spots will never be hopped on.

edited Apr 13 at 13:14

answered Apr 13 at 12:24

Ethan Bolker

47.6k556123

$begingroup$
(+1) I was thinking exactly the same thing just as you posted your answer!
$endgroup$
– TheSimpliFire
Apr 13 at 12:25

$begingroup$
Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.
$endgroup$
– Joseph O'Rourke
Apr 13 at 14:17

4

$begingroup$
Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.
$endgroup$
– Ethan Bolker
Apr 13 at 14:47

$begingroup$
I worry about explicitly specifying $t=0.xyzcdots$, excluding digits $ 0,1,2 $, guaranteeing $t$ is irrational.
$endgroup$
– Joseph O'Rourke
Apr 13 at 16:46

4

$begingroup$
@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.
$endgroup$
– Wojowu
Apr 13 at 18:28

|
show 1 more comment

$$
x 1^x-2 y 1^y-2 z1^z-2 ldots
$$
moves off to infinity for any sequence of digits $xyzldots$ between $3$ and $9$. Select a sequence that defines an irrational number.

More generally

$$
x 1 ?^x-1 y 1 ?^y-1 z 1 ?^z-1 ldots
$$
works, where $?^n$ is an arbitrary string of $n$ digits, since those spots will never be hopped on.

edited Apr 13 at 13:14

answered Apr 13 at 12:24

Ethan Bolker

47.6k556123

$begingroup$
(+1) I was thinking exactly the same thing just as you posted your answer!
$endgroup$
– TheSimpliFire
Apr 13 at 12:25

$begingroup$
Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.
$endgroup$
– Joseph O'Rourke
Apr 13 at 14:17

4

$begingroup$
Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.
$endgroup$
– Ethan Bolker
Apr 13 at 14:47

$begingroup$
I worry about explicitly specifying $t=0.xyzcdots$, excluding digits $ 0,1,2 $, guaranteeing $t$ is irrational.
$endgroup$
– Joseph O'Rourke
Apr 13 at 16:46

4

$begingroup$
@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.
$endgroup$
– Wojowu
Apr 13 at 18:28

|
show 1 more comment

$$
x 1^x-2 y 1^y-2 z1^z-2 ldots
$$
moves off to infinity for any sequence of digits $xyzldots$ between $3$ and $9$. Select a sequence that defines an irrational number.

More generally

$$
x 1 ?^x-1 y 1 ?^y-1 z 1 ?^z-1 ldots
$$
works, where $?^n$ is an arbitrary string of $n$ digits, since those spots will never be hopped on.

edited Apr 13 at 13:14

answered Apr 13 at 12:24

Ethan Bolker

47.6k556123

$$
x 1^x-2 y 1^y-2 z1^z-2 ldots
$$
moves off to infinity for any sequence of digits $xyzldots$ between $3$ and $9$. Select a sequence that defines an irrational number.

More generally

$$
x 1 ?^x-1 y 1 ?^y-1 z 1 ?^z-1 ldots
$$
works, where $?^n$ is an arbitrary string of $n$ digits, since those spots will never be hopped on.

edited Apr 13 at 13:14

answered Apr 13 at 12:24

Ethan Bolker

47.6k556123

edited Apr 13 at 13:14

answered Apr 13 at 12:24

Ethan Bolker

47.6k556123

answered Apr 13 at 12:24

Ethan Bolker

47.6k556123

answered Apr 13 at 12:24

Ethan Bolker

47.6k556123

$begingroup$
(+1) I was thinking exactly the same thing just as you posted your answer!
$endgroup$
– TheSimpliFire
Apr 13 at 12:25

$begingroup$
Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.
$endgroup$
– Joseph O'Rourke
Apr 13 at 14:17

4

$begingroup$
Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.
$endgroup$
– Ethan Bolker
Apr 13 at 14:47

$begingroup$
I worry about explicitly specifying $t=0.xyzcdots$, excluding digits $ 0,1,2 $, guaranteeing $t$ is irrational.
$endgroup$
– Joseph O'Rourke
Apr 13 at 16:46

4

$begingroup$
@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.
$endgroup$
– Wojowu
Apr 13 at 18:28

|
show 1 more comment

$begingroup$
(+1) I was thinking exactly the same thing just as you posted your answer!
$endgroup$
– TheSimpliFire
Apr 13 at 12:25

$begingroup$
Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.
$endgroup$
– Joseph O'Rourke
Apr 13 at 14:17

4

$begingroup$
Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.
$endgroup$
– Ethan Bolker
Apr 13 at 14:47

$begingroup$
I worry about explicitly specifying $t=0.xyzcdots$, excluding digits $ 0,1,2 $, guaranteeing $t$ is irrational.
$endgroup$
– Joseph O'Rourke
Apr 13 at 16:46

4

$begingroup$
@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.
$endgroup$
– Wojowu
Apr 13 at 18:28

(+1) I was thinking exactly the same thing just as you posted your answer!

– TheSimpliFire
Apr 13 at 12:25

Nice! Could you be more explicit about how you know the resulting digit string is irrational? Thanks.

– Joseph O'Rourke
Apr 13 at 14:17

Clearly it will be irrational if $t = 0.xyzldots$ is, since then periodicity is impossible. It can be even when $t$ is rational because the $?$'s can force that.

– Ethan Bolker
Apr 13 at 14:47

I worry about explicitly specifying $t=0.xyzcdots$, excluding digits $ 0,1,2 $, guaranteeing $t$ is irrational.

– Joseph O'Rourke
Apr 13 at 16:46

@JosephO'Rourke Take the Thue-Morse sequence, add 3 to each term and take that as a binary sequence. Don't hope for a better sort of answer - our understanding of decimal expansions of "natural" constants is really bad, so you won't be able to exclude 0,1,2 without artificially constructing the decimal expansion.

– Wojowu
Apr 13 at 18:28

|
show 1 more comment

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