linear combinations of of irrational number with integer coefficients.
Let ${x_1, dots,x_k}$ be real numbers, under which conditions do I have that the set
$$
langle x_1, dots,x_krangle
=
{sum a_i x_i: a_i in mathbb{Z} }
$$
is discrete set?
Well, this is clearly the case if $x_i$'s are all integers. On the other hand, if I take $x_1=1$ and $x_2$ to be the Liouville Constant, I have a dense set. So what would be necessary and/or sufficient conditions?
real-analysis analysis
asked Dec 7 at 8:00
Kernel
772521
add a comment |
Let ${x_1, dots,x_k}$ be real numbers, under which conditions do I have that the set
$$
langle x_1, dots,x_krangle
=
{sum a_i x_i: a_i in mathbb{Z} }
$$
is discrete set?
Well, this is clearly the case if $x_i$'s are all integers. On the other hand, if I take $x_1=1$ and $x_2$ to be the Liouville Constant, I have a dense set. So what would be necessary and/or sufficient conditions?
real-analysis analysis
asked Dec 7 at 8:00
Kernel
772521
add a comment |
Let ${x_1, dots,x_k}$ be real numbers, under which conditions do I have that the set
$$
langle x_1, dots,x_krangle
=
{sum a_i x_i: a_i in mathbb{Z} }
$$
is discrete set?
Well, this is clearly the case if $x_i$'s are all integers. On the other hand, if I take $x_1=1$ and $x_2$ to be the Liouville Constant, I have a dense set. So what would be necessary and/or sufficient conditions?
real-analysis analysis
asked Dec 7 at 8:00
Kernel
772521
Let ${x_1, dots,x_k}$ be real numbers, under which conditions do I have that the set
$$
langle x_1, dots,x_krangle
=
{sum a_i x_i: a_i in mathbb{Z} }
$$
is discrete set?
Well, this is clearly the case if $x_i$'s are all integers. On the other hand, if I take $x_1=1$ and $x_2$ to be the Liouville Constant, I have a dense set. So what would be necessary and/or sufficient conditions?
real-analysis analysis
real-analysis analysis
asked Dec 7 at 8:00
Kernel
772521
asked Dec 7 at 8:00
Kernel
772521
asked Dec 7 at 8:00
Kernel
772521
asked Dec 7 at 8:00
Kernel
772521
asked Dec 7 at 8:00
Kernel
772521
772521
add a comment |
add a comment |
1 Answer
1
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If $frac x y$ is irrational then ${nx+my:n,m in mathbb Z}$ is dense. If $frac x y$ is rational then ${nx+my:n,m in mathbb Z}$ is discrete.
answered Dec 7 at 8:06
Kavi Rama Murthy
48.6k31854
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
If $frac x y$ is irrational then ${nx+my:n,m in mathbb Z}$ is dense. If $frac x y$ is rational then ${nx+my:n,m in mathbb Z}$ is discrete.
answered Dec 7 at 8:06
Kavi Rama Murthy
48.6k31854
add a comment |
If $frac x y$ is irrational then ${nx+my:n,m in mathbb Z}$ is dense. If $frac x y$ is rational then ${nx+my:n,m in mathbb Z}$ is discrete.
answered Dec 7 at 8:06
Kavi Rama Murthy
48.6k31854
add a comment |
If $frac x y$ is irrational then ${nx+my:n,m in mathbb Z}$ is dense. If $frac x y$ is rational then ${nx+my:n,m in mathbb Z}$ is discrete.
answered Dec 7 at 8:06
Kavi Rama Murthy
48.6k31854
If $frac x y$ is irrational then ${nx+my:n,m in mathbb Z}$ is dense. If $frac x y$ is rational then ${nx+my:n,m in mathbb Z}$ is discrete.
answered Dec 7 at 8:06
Kavi Rama Murthy
48.6k31854
answered Dec 7 at 8:06
Kavi Rama Murthy
48.6k31854
answered Dec 7 at 8:06
Kavi Rama Murthy
48.6k31854
answered Dec 7 at 8:06
Kavi Rama Murthy
48.6k31854
48.6k31854
add a comment |
add a comment |
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