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.










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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.










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      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.










      share|cite|improve this question













      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






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      asked 34 mins ago









      Will Fisher

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          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.






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            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 $1$.






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              2 Answers
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              active

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              2 Answers
              2






              active

              oldest

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              active

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              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.






              share|cite|improve this answer


























                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.






                share|cite|improve this answer
























                  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.






                  share|cite|improve this answer














                  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.







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                  edited 8 mins ago

























                  answered 26 mins ago









                  Henning Makholm

                  234k16299531




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                      up vote
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                      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 $1$.






                      share|cite|improve this answer
























                        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 $1$.






                        share|cite|improve this answer






















                          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 $1$.






                          share|cite|improve this answer












                          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 $1$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 26 mins ago









                          Ross Millikan

                          285k23195363




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