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Can ratio of smooth numbers approach 1?
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Can the ratio of $q$-smooth numbers ever approach one? Asked more rigorously, can we ever have
$$1in overlinea/b; :; a,b ;text are qtext-smooth=overlineprod_p_ile qp_i^alpha_i; :; (a_1, dots, a_pi(q))inmathbbZ^pi(q);?$$
If $q=2$, the answer is obviously no because
$$1notin overline2^alpha; :; alphainmathbbZ=0cup 2^alpha; :; alphainmathbbZ$$
however beyond this the problem is not easy. The reason I would like to know this is because this would imply that
$$G(q)=inf a/b; :; a>b,; a,b ; qtext-smooth>1.$$
I suspect that we may be able to say even further that
$$overlineprod_p_ile qp_i^alpha_i; :; (a_1, dots, a_pi(q))inmathbbZ^pi(q)=0cup prod_p_ile qp_i^alpha_i; :; (a_1, dots, a_pi(q))inmathbbZ^pi(q)$$
which would clearly imply the result.
number-theory limits elementary-number-theory
asked 38 mins ago
Will Fisher
3,477629
add a comment |Â
up vote
2
down vote
favorite
Can the ratio of $q$-smooth numbers ever approach one? Asked more rigorously, can we ever have
$$1in overlinea/b; :; a,b ;text are qtext-smooth=overlineprod_p_ile qp_i^alpha_i; :; (a_1, dots, a_pi(q))inmathbbZ^pi(q);?$$
If $q=2$, the answer is obviously no because
$$1notin overline2^alpha; :; alphainmathbbZ=0cup 2^alpha; :; alphainmathbbZ$$
however beyond this the problem is not easy. The reason I would like to know this is because this would imply that
$$G(q)=inf a/b; :; a>b,; a,b ; qtext-smooth>1.$$
I suspect that we may be able to say even further that
$$overlineprod_p_ile qp_i^alpha_i; :; (a_1, dots, a_pi(q))inmathbbZ^pi(q)=0cup prod_p_ile qp_i^alpha_i; :; (a_1, dots, a_pi(q))inmathbbZ^pi(q)$$
which would clearly imply the result.
number-theory limits elementary-number-theory
asked 38 mins ago
Will Fisher
3,477629
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Can the ratio of $q$-smooth numbers ever approach one? Asked more rigorously, can we ever have
$$1in overlinea/b; :; a,b ;text are qtext-smooth=overlineprod_p_ile qp_i^alpha_i; :; (a_1, dots, a_pi(q))inmathbbZ^pi(q);?$$
If $q=2$, the answer is obviously no because
$$1notin overline2^alpha; :; alphainmathbbZ=0cup 2^alpha; :; alphainmathbbZ$$
however beyond this the problem is not easy. The reason I would like to know this is because this would imply that
$$G(q)=inf a/b; :; a>b,; a,b ; qtext-smooth>1.$$
I suspect that we may be able to say even further that
$$overlineprod_p_ile qp_i^alpha_i; :; (a_1, dots, a_pi(q))inmathbbZ^pi(q)=0cup prod_p_ile qp_i^alpha_i; :; (a_1, dots, a_pi(q))inmathbbZ^pi(q)$$
which would clearly imply the result.
number-theory limits elementary-number-theory
asked 38 mins ago
Will Fisher
3,477629
Can the ratio of $q$-smooth numbers ever approach one? Asked more rigorously, can we ever have
$$1in overlinea/b; :; a,b ;text are qtext-smooth=overlineprod_p_ile qp_i^alpha_i; :; (a_1, dots, a_pi(q))inmathbbZ^pi(q);?$$
If $q=2$, the answer is obviously no because
$$1notin overline2^alpha; :; alphainmathbbZ=0cup 2^alpha; :; alphainmathbbZ$$
however beyond this the problem is not easy. The reason I would like to know this is because this would imply that
$$G(q)=inf a/b; :; a>b,; a,b ; qtext-smooth>1.$$
I suspect that we may be able to say even further that
$$overlineprod_p_ile qp_i^alpha_i; :; (a_1, dots, a_pi(q))inmathbbZ^pi(q)=0cup prod_p_ile qp_i^alpha_i; :; (a_1, dots, a_pi(q))inmathbbZ^pi(q)$$
which would clearly imply the result.
number-theory limits elementary-number-theory
number-theory limits elementary-number-theory
asked 38 mins ago
Will Fisher
3,477629
asked 38 mins ago
Will Fisher
3,477629
asked 38 mins ago
Will Fisher
3,477629
asked 38 mins ago
Will Fisher
3,477629
asked 38 mins ago
Will Fisher
3,477629
3,477629
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Yes, if $qge 3$.
This is because $log_2(3)$ is irrational, so the fractional parts of $log_2(3^n)$ lie densely in $[0,1)$. In particular they can become arbitrarily small.
answered 31 mins ago
Henning Makholm
234k16299531
add a comment |Â
up vote
3
down vote
Yes. We can just consider fractions of the form $frac 3^a2^b$. The base $2$ log of this is $alog_2(3)-b$. Because $log_2(3)$ is irrational, we can choose $a,b$ to make this difference arbitrarily close to $0$, which means the ratio can be arbitrarily close to $1$.
answered 30 mins ago
Ross Millikan
285k23195363
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Yes, if $qge 3$.
This is because $log_2(3)$ is irrational, so the fractional parts of $log_2(3^n)$ lie densely in $[0,1)$. In particular they can become arbitrarily small.
answered 31 mins ago
Henning Makholm
234k16299531
add a comment |Â
up vote
3
down vote
accepted
Yes, if $qge 3$.
This is because $log_2(3)$ is irrational, so the fractional parts of $log_2(3^n)$ lie densely in $[0,1)$. In particular they can become arbitrarily small.
answered 31 mins ago
Henning Makholm
234k16299531
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Yes, if $qge 3$.
This is because $log_2(3)$ is irrational, so the fractional parts of $log_2(3^n)$ lie densely in $[0,1)$. In particular they can become arbitrarily small.
answered 31 mins ago
Henning Makholm
234k16299531
Yes, if $qge 3$.
This is because $log_2(3)$ is irrational, so the fractional parts of $log_2(3^n)$ lie densely in $[0,1)$. In particular they can become arbitrarily small.
answered 31 mins ago
Henning Makholm
234k16299531
edited 13 mins ago
answered 31 mins ago
Henning Makholm
234k16299531
answered 31 mins ago
Henning Makholm
234k16299531
answered 31 mins ago
Henning Makholm
234k16299531
234k16299531
add a comment |Â
add a comment |Â
up vote
3
down vote
Yes. We can just consider fractions of the form $frac 3^a2^b$. The base $2$ log of this is $alog_2(3)-b$. Because $log_2(3)$ is irrational, we can choose $a,b$ to make this difference arbitrarily close to $0$, which means the ratio can be arbitrarily close to $1$.
answered 30 mins ago
Ross Millikan
285k23195363
add a comment |Â
up vote
3
down vote
Yes. We can just consider fractions of the form $frac 3^a2^b$. The base $2$ log of this is $alog_2(3)-b$. Because $log_2(3)$ is irrational, we can choose $a,b$ to make this difference arbitrarily close to $0$, which means the ratio can be arbitrarily close to $1$.
answered 30 mins ago
Ross Millikan
285k23195363
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Yes. We can just consider fractions of the form $frac 3^a2^b$. The base $2$ log of this is $alog_2(3)-b$. Because $log_2(3)$ is irrational, we can choose $a,b$ to make this difference arbitrarily close to $0$, which means the ratio can be arbitrarily close to $1$.
answered 30 mins ago
Ross Millikan
285k23195363
Yes. We can just consider fractions of the form $frac 3^a2^b$. The base $2$ log of this is $alog_2(3)-b$. Because $log_2(3)$ is irrational, we can choose $a,b$ to make this difference arbitrarily close to $0$, which means the ratio can be arbitrarily close to $1$.
answered 30 mins ago
Ross Millikan
285k23195363
answered 30 mins ago
Ross Millikan
285k23195363
answered 30 mins ago
Ross Millikan
285k23195363
answered 30 mins ago
Ross Millikan
285k23195363
285k23195363
add a comment |Â
add a comment |Â
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