Digits in an algebraic irrational number
$begingroup$
I am trying to solve a problem and I got a conditional result related to normality of algebraic irrational numbers (Borel conjecture).
I know that by using Ridout theorem or Schmidt subspace theorem is possible to find a good lower bound for the number of nonzero "digits" in the $g$-ary expansion of an algebraic irrational number (for any basis $ggeq 2$).
However, my question is in the opposite direction:
Is it possible to prove that every algebraic irrational number has at least one 0 in its $g$-ary expansion, for all sufficiently large $ggeq 2$?
Of course, if this statement is true, then it is possible to prove that, in fact, there are infinitely many $0$'s in its $g$-ary expansion (by multiplying the algebraic number for some convenient power of $10$).
Any help will be welcomed.
nt.number-theory measure-theory diophantine-approximation transcendental-number-theory equidistribution
asked 12 hours ago
JeanJean
843
$endgroup$
add a comment |
$begingroup$
I am trying to solve a problem and I got a conditional result related to normality of algebraic irrational numbers (Borel conjecture).
I know that by using Ridout theorem or Schmidt subspace theorem is possible to find a good lower bound for the number of nonzero "digits" in the $g$-ary expansion of an algebraic irrational number (for any basis $ggeq 2$).
However, my question is in the opposite direction:
Is it possible to prove that every algebraic irrational number has at least one 0 in its $g$-ary expansion, for all sufficiently large $ggeq 2$?
Of course, if this statement is true, then it is possible to prove that, in fact, there are infinitely many $0$'s in its $g$-ary expansion (by multiplying the algebraic number for some convenient power of $10$).
Any help will be welcomed.
nt.number-theory measure-theory diophantine-approximation transcendental-number-theory equidistribution
asked 12 hours ago
JeanJean
843
$endgroup$
add a comment |
$begingroup$
I am trying to solve a problem and I got a conditional result related to normality of algebraic irrational numbers (Borel conjecture).
I know that by using Ridout theorem or Schmidt subspace theorem is possible to find a good lower bound for the number of nonzero "digits" in the $g$-ary expansion of an algebraic irrational number (for any basis $ggeq 2$).
However, my question is in the opposite direction:
Is it possible to prove that every algebraic irrational number has at least one 0 in its $g$-ary expansion, for all sufficiently large $ggeq 2$?
Of course, if this statement is true, then it is possible to prove that, in fact, there are infinitely many $0$'s in its $g$-ary expansion (by multiplying the algebraic number for some convenient power of $10$).
Any help will be welcomed.
nt.number-theory measure-theory diophantine-approximation transcendental-number-theory equidistribution
asked 12 hours ago
JeanJean
843
$endgroup$
I am trying to solve a problem and I got a conditional result related to normality of algebraic irrational numbers (Borel conjecture).
I know that by using Ridout theorem or Schmidt subspace theorem is possible to find a good lower bound for the number of nonzero "digits" in the $g$-ary expansion of an algebraic irrational number (for any basis $ggeq 2$).
However, my question is in the opposite direction:
Is it possible to prove that every algebraic irrational number has at least one 0 in its $g$-ary expansion, for all sufficiently large $ggeq 2$?
Of course, if this statement is true, then it is possible to prove that, in fact, there are infinitely many $0$'s in its $g$-ary expansion (by multiplying the algebraic number for some convenient power of $10$).
Any help will be welcomed.
nt.number-theory measure-theory diophantine-approximation transcendental-number-theory equidistribution
nt.number-theory measure-theory diophantine-approximation transcendental-number-theory equidistribution
asked 12 hours ago
JeanJean
843
asked 12 hours ago
JeanJean
843
asked 12 hours ago
JeanJean
843
asked 12 hours ago
JeanJean
843
asked 12 hours ago
JeanJean
843
843
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$.
Taking an even-numbered convergent of the continued fraction of $x$ gives us a rational $p/q$ such that
$$frac{p}{q} < x < frac{p}{q} + frac{1}{q^2}$$
so that the first two digits in the base-$q$ expansion of $x$ are $p$ and $0$.
Of course, this is a far cry from all sufficiently large $g$ (which is surely true, but I very much doubt it's provable in the current state of the art).
answered 11 hours ago
Robert IsraelRobert Israel
42.6k51122
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1 Answer
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1 Answer
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$begingroup$
What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$.
Taking an even-numbered convergent of the continued fraction of $x$ gives us a rational $p/q$ such that
$$frac{p}{q} < x < frac{p}{q} + frac{1}{q^2}$$
so that the first two digits in the base-$q$ expansion of $x$ are $p$ and $0$.
Of course, this is a far cry from all sufficiently large $g$ (which is surely true, but I very much doubt it's provable in the current state of the art).
answered 11 hours ago
Robert IsraelRobert Israel
42.6k51122
$endgroup$
add a comment |
$begingroup$
What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$.
Taking an even-numbered convergent of the continued fraction of $x$ gives us a rational $p/q$ such that
$$frac{p}{q} < x < frac{p}{q} + frac{1}{q^2}$$
so that the first two digits in the base-$q$ expansion of $x$ are $p$ and $0$.
Of course, this is a far cry from all sufficiently large $g$ (which is surely true, but I very much doubt it's provable in the current state of the art).
answered 11 hours ago
Robert IsraelRobert Israel
42.6k51122
$endgroup$
add a comment |
$begingroup$
What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$.
Taking an even-numbered convergent of the continued fraction of $x$ gives us a rational $p/q$ such that
$$frac{p}{q} < x < frac{p}{q} + frac{1}{q^2}$$
so that the first two digits in the base-$q$ expansion of $x$ are $p$ and $0$.
Of course, this is a far cry from all sufficiently large $g$ (which is surely true, but I very much doubt it's provable in the current state of the art).
answered 11 hours ago
Robert IsraelRobert Israel
42.6k51122
$endgroup$
What is known is that every real irrational has a $0$ in its $g$-ary expansion for infinitely many $g$. WLOG take $0 < x < 1$.
Taking an even-numbered convergent of the continued fraction of $x$ gives us a rational $p/q$ such that
$$frac{p}{q} < x < frac{p}{q} + frac{1}{q^2}$$
so that the first two digits in the base-$q$ expansion of $x$ are $p$ and $0$.
Of course, this is a far cry from all sufficiently large $g$ (which is surely true, but I very much doubt it's provable in the current state of the art).
answered 11 hours ago
Robert IsraelRobert Israel
42.6k51122
answered 11 hours ago
Robert IsraelRobert Israel
42.6k51122
answered 11 hours ago
Robert IsraelRobert Israel
42.6k51122
answered 11 hours ago
Robert IsraelRobert Israel
42.6k51122
42.6k51122
add a comment |
add a comment |
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