Let $a_1< cdots< a_nleq x$, where no $a_i$ divides product of others, show that $nleq pi(x)$.
$begingroup$
Let $a_1< a_2< cdots< a_nleq x$ be a set of positive integers such that no $a_i$ divides the product of the others. Prove that $nleq pi(x)$.
I have tried to argue by contradiction, which I assume $n> pi(x)$, then $a_1, a_2, cdots,a_n$ cannot be mutually relatively prime, so some of $a_i$ must divides product of other $a$'s, but I cannot give a very accurate explanation about it. Any suggestion?
prime-numbers analytic-number-theory
asked Jan 25 at 11:23
kelvin hong 方kelvin hong 方
80018
$endgroup$
add a comment |
$begingroup$
Let $a_1< a_2< cdots< a_nleq x$ be a set of positive integers such that no $a_i$ divides the product of the others. Prove that $nleq pi(x)$.
I have tried to argue by contradiction, which I assume $n> pi(x)$, then $a_1, a_2, cdots,a_n$ cannot be mutually relatively prime, so some of $a_i$ must divides product of other $a$'s, but I cannot give a very accurate explanation about it. Any suggestion?
prime-numbers analytic-number-theory
asked Jan 25 at 11:23
kelvin hong 方kelvin hong 方
80018
$endgroup$
add a comment |
$begingroup$
Let $a_1< a_2< cdots< a_nleq x$ be a set of positive integers such that no $a_i$ divides the product of the others. Prove that $nleq pi(x)$.
I have tried to argue by contradiction, which I assume $n> pi(x)$, then $a_1, a_2, cdots,a_n$ cannot be mutually relatively prime, so some of $a_i$ must divides product of other $a$'s, but I cannot give a very accurate explanation about it. Any suggestion?
prime-numbers analytic-number-theory
asked Jan 25 at 11:23
kelvin hong 方kelvin hong 方
80018
$endgroup$
Let $a_1< a_2< cdots< a_nleq x$ be a set of positive integers such that no $a_i$ divides the product of the others. Prove that $nleq pi(x)$.
I have tried to argue by contradiction, which I assume $n> pi(x)$, then $a_1, a_2, cdots,a_n$ cannot be mutually relatively prime, so some of $a_i$ must divides product of other $a$'s, but I cannot give a very accurate explanation about it. Any suggestion?
prime-numbers analytic-number-theory
prime-numbers analytic-number-theory
asked Jan 25 at 11:23
kelvin hong 方kelvin hong 方
80018
asked Jan 25 at 11:23
kelvin hong 方kelvin hong 方
80018
asked Jan 25 at 11:23
kelvin hong 方kelvin hong 方
80018
asked Jan 25 at 11:23
kelvin hong 方kelvin hong 方
80018
asked Jan 25 at 11:23
kelvin hong 方kelvin hong 方
80018
80018
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since no $a_i$ divides the product of the others, then for each fixed index $i$ there must be some prime $p_i$ such that $ineq jimplies v_{p_i}(a_i)>v_{p_i}(a_j)$. There may, of course, be several such primes...if so, let $p_i$ denote the least of them. Clearly $ineq jimplies p_ineq p_j$. In this way we have produced $n$ distinct primes, all less than $x$ so we are done.
Note: here, as usual, for a prime $p$ and a natural number $a$ we define $v_p(a)$ to be the order to which $p$ divides $a$. Thus, for example, $v_3(18)=2$ and $v_5(18)=0$.
answered Jan 25 at 11:29
lulululu
42.9k25080
$endgroup$
$begingroup$
I have thought a while, but why $ineq j$ implies $p_ineq p_j$? Is the statement $ineq j$ implies $v_{p_i}(a_i)> v_{p_i}(a_j)$ true for all $jneq i$ or just some $jneq i$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:51
1
$begingroup$
Suppose $p_i=p_j=p$. Then $v_p(a_i)>v_p(a_j)$ and $v_p(a_j)>v_p(a_i)$.
$endgroup$
– lulu
Jan 25 at 11:52
$begingroup$
thank, now I know how to do this.
$endgroup$
– kelvin hong 方
Jan 25 at 11:55
add a comment |
$begingroup$
Assume to the contrary that there exists $x$ and $a_1<a_2<...<a_nleq x$ and no $a_i$ divides the product of the others and $n>pi(x)$. We pick such $a_1,a_2,...,a_n$ and $x$ such that the sum $a_1+a_2+...+a_n$ is minimum. Then we can show that $a_{i}$'s are pairwise coprime.(Suppose that $(a_i,a_j)=dneq 1$ Then replace (a_i,a_j) by (a_i/d,a_j/d) and consider the minimality of the sum $a_1+a_2+...+a_n$.) Then each $a_i$ has a prime divisor(as $a_ineq 1$) which does not divide any other $a_j$'s. Therefore each $a_i$ has a distinct prime divisor. Hence there exists at least $n$ primes that are less than or equal to $x$. This is a contradiction. Hence the conclusion follows.
answered Jan 25 at 11:30
NotoriousJuanGNotoriousJuanG
843
$endgroup$
$begingroup$
Is it possible after we make the sum to be minimum, then some of $a_i$ becomes $1$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:48
1
$begingroup$
In that case, 1 divides the product of the others.
$endgroup$
– NotoriousJuanG
Jan 25 at 17:06
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Since no $a_i$ divides the product of the others, then for each fixed index $i$ there must be some prime $p_i$ such that $ineq jimplies v_{p_i}(a_i)>v_{p_i}(a_j)$. There may, of course, be several such primes...if so, let $p_i$ denote the least of them. Clearly $ineq jimplies p_ineq p_j$. In this way we have produced $n$ distinct primes, all less than $x$ so we are done.
Note: here, as usual, for a prime $p$ and a natural number $a$ we define $v_p(a)$ to be the order to which $p$ divides $a$. Thus, for example, $v_3(18)=2$ and $v_5(18)=0$.
answered Jan 25 at 11:29
lulululu
42.9k25080
$endgroup$
$begingroup$
I have thought a while, but why $ineq j$ implies $p_ineq p_j$? Is the statement $ineq j$ implies $v_{p_i}(a_i)> v_{p_i}(a_j)$ true for all $jneq i$ or just some $jneq i$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:51
1
$begingroup$
Suppose $p_i=p_j=p$. Then $v_p(a_i)>v_p(a_j)$ and $v_p(a_j)>v_p(a_i)$.
$endgroup$
– lulu
Jan 25 at 11:52
$begingroup$
thank, now I know how to do this.
$endgroup$
– kelvin hong 方
Jan 25 at 11:55
add a comment |
$begingroup$
Since no $a_i$ divides the product of the others, then for each fixed index $i$ there must be some prime $p_i$ such that $ineq jimplies v_{p_i}(a_i)>v_{p_i}(a_j)$. There may, of course, be several such primes...if so, let $p_i$ denote the least of them. Clearly $ineq jimplies p_ineq p_j$. In this way we have produced $n$ distinct primes, all less than $x$ so we are done.
Note: here, as usual, for a prime $p$ and a natural number $a$ we define $v_p(a)$ to be the order to which $p$ divides $a$. Thus, for example, $v_3(18)=2$ and $v_5(18)=0$.
answered Jan 25 at 11:29
lulululu
42.9k25080
$endgroup$
$begingroup$
I have thought a while, but why $ineq j$ implies $p_ineq p_j$? Is the statement $ineq j$ implies $v_{p_i}(a_i)> v_{p_i}(a_j)$ true for all $jneq i$ or just some $jneq i$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:51
1
$begingroup$
Suppose $p_i=p_j=p$. Then $v_p(a_i)>v_p(a_j)$ and $v_p(a_j)>v_p(a_i)$.
$endgroup$
– lulu
Jan 25 at 11:52
$begingroup$
thank, now I know how to do this.
$endgroup$
– kelvin hong 方
Jan 25 at 11:55
add a comment |
$begingroup$
Since no $a_i$ divides the product of the others, then for each fixed index $i$ there must be some prime $p_i$ such that $ineq jimplies v_{p_i}(a_i)>v_{p_i}(a_j)$. There may, of course, be several such primes...if so, let $p_i$ denote the least of them. Clearly $ineq jimplies p_ineq p_j$. In this way we have produced $n$ distinct primes, all less than $x$ so we are done.
Note: here, as usual, for a prime $p$ and a natural number $a$ we define $v_p(a)$ to be the order to which $p$ divides $a$. Thus, for example, $v_3(18)=2$ and $v_5(18)=0$.
answered Jan 25 at 11:29
lulululu
42.9k25080
$endgroup$
Since no $a_i$ divides the product of the others, then for each fixed index $i$ there must be some prime $p_i$ such that $ineq jimplies v_{p_i}(a_i)>v_{p_i}(a_j)$. There may, of course, be several such primes...if so, let $p_i$ denote the least of them. Clearly $ineq jimplies p_ineq p_j$. In this way we have produced $n$ distinct primes, all less than $x$ so we are done.
Note: here, as usual, for a prime $p$ and a natural number $a$ we define $v_p(a)$ to be the order to which $p$ divides $a$. Thus, for example, $v_3(18)=2$ and $v_5(18)=0$.
answered Jan 25 at 11:29
lulululu
42.9k25080
edited Jan 25 at 13:24
answered Jan 25 at 11:29
lulululu
42.9k25080
answered Jan 25 at 11:29
lulululu
42.9k25080
answered Jan 25 at 11:29
lulululu
42.9k25080
42.9k25080
$begingroup$
I have thought a while, but why $ineq j$ implies $p_ineq p_j$? Is the statement $ineq j$ implies $v_{p_i}(a_i)> v_{p_i}(a_j)$ true for all $jneq i$ or just some $jneq i$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:51
1
$begingroup$
Suppose $p_i=p_j=p$. Then $v_p(a_i)>v_p(a_j)$ and $v_p(a_j)>v_p(a_i)$.
$endgroup$
– lulu
Jan 25 at 11:52
$begingroup$
thank, now I know how to do this.
$endgroup$
– kelvin hong 方
Jan 25 at 11:55
add a comment |
$begingroup$
I have thought a while, but why $ineq j$ implies $p_ineq p_j$? Is the statement $ineq j$ implies $v_{p_i}(a_i)> v_{p_i}(a_j)$ true for all $jneq i$ or just some $jneq i$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:51
1
$begingroup$
Suppose $p_i=p_j=p$. Then $v_p(a_i)>v_p(a_j)$ and $v_p(a_j)>v_p(a_i)$.
$endgroup$
– lulu
Jan 25 at 11:52
$begingroup$
thank, now I know how to do this.
$endgroup$
– kelvin hong 方
Jan 25 at 11:55
$begingroup$
I have thought a while, but why $ineq j$ implies $p_ineq p_j$? Is the statement $ineq j$ implies $v_{p_i}(a_i)> v_{p_i}(a_j)$ true for all $jneq i$ or just some $jneq i$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:51
$begingroup$
I have thought a while, but why $ineq j$ implies $p_ineq p_j$? Is the statement $ineq j$ implies $v_{p_i}(a_i)> v_{p_i}(a_j)$ true for all $jneq i$ or just some $jneq i$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:51
1
1
$begingroup$
Suppose $p_i=p_j=p$. Then $v_p(a_i)>v_p(a_j)$ and $v_p(a_j)>v_p(a_i)$.
$endgroup$
– lulu
Jan 25 at 11:52
$begingroup$
Suppose $p_i=p_j=p$. Then $v_p(a_i)>v_p(a_j)$ and $v_p(a_j)>v_p(a_i)$.
$endgroup$
– lulu
Jan 25 at 11:52
$begingroup$
thank, now I know how to do this.
$endgroup$
– kelvin hong 方
Jan 25 at 11:55
$begingroup$
thank, now I know how to do this.
$endgroup$
– kelvin hong 方
Jan 25 at 11:55
add a comment |
$begingroup$
Assume to the contrary that there exists $x$ and $a_1<a_2<...<a_nleq x$ and no $a_i$ divides the product of the others and $n>pi(x)$. We pick such $a_1,a_2,...,a_n$ and $x$ such that the sum $a_1+a_2+...+a_n$ is minimum. Then we can show that $a_{i}$'s are pairwise coprime.(Suppose that $(a_i,a_j)=dneq 1$ Then replace (a_i,a_j) by (a_i/d,a_j/d) and consider the minimality of the sum $a_1+a_2+...+a_n$.) Then each $a_i$ has a prime divisor(as $a_ineq 1$) which does not divide any other $a_j$'s. Therefore each $a_i$ has a distinct prime divisor. Hence there exists at least $n$ primes that are less than or equal to $x$. This is a contradiction. Hence the conclusion follows.
answered Jan 25 at 11:30
NotoriousJuanGNotoriousJuanG
843
$endgroup$
$begingroup$
Is it possible after we make the sum to be minimum, then some of $a_i$ becomes $1$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:48
1
$begingroup$
In that case, 1 divides the product of the others.
$endgroup$
– NotoriousJuanG
Jan 25 at 17:06
add a comment |
$begingroup$
Assume to the contrary that there exists $x$ and $a_1<a_2<...<a_nleq x$ and no $a_i$ divides the product of the others and $n>pi(x)$. We pick such $a_1,a_2,...,a_n$ and $x$ such that the sum $a_1+a_2+...+a_n$ is minimum. Then we can show that $a_{i}$'s are pairwise coprime.(Suppose that $(a_i,a_j)=dneq 1$ Then replace (a_i,a_j) by (a_i/d,a_j/d) and consider the minimality of the sum $a_1+a_2+...+a_n$.) Then each $a_i$ has a prime divisor(as $a_ineq 1$) which does not divide any other $a_j$'s. Therefore each $a_i$ has a distinct prime divisor. Hence there exists at least $n$ primes that are less than or equal to $x$. This is a contradiction. Hence the conclusion follows.
answered Jan 25 at 11:30
NotoriousJuanGNotoriousJuanG
843
$endgroup$
$begingroup$
Is it possible after we make the sum to be minimum, then some of $a_i$ becomes $1$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:48
1
$begingroup$
In that case, 1 divides the product of the others.
$endgroup$
– NotoriousJuanG
Jan 25 at 17:06
add a comment |
$begingroup$
Assume to the contrary that there exists $x$ and $a_1<a_2<...<a_nleq x$ and no $a_i$ divides the product of the others and $n>pi(x)$. We pick such $a_1,a_2,...,a_n$ and $x$ such that the sum $a_1+a_2+...+a_n$ is minimum. Then we can show that $a_{i}$'s are pairwise coprime.(Suppose that $(a_i,a_j)=dneq 1$ Then replace (a_i,a_j) by (a_i/d,a_j/d) and consider the minimality of the sum $a_1+a_2+...+a_n$.) Then each $a_i$ has a prime divisor(as $a_ineq 1$) which does not divide any other $a_j$'s. Therefore each $a_i$ has a distinct prime divisor. Hence there exists at least $n$ primes that are less than or equal to $x$. This is a contradiction. Hence the conclusion follows.
answered Jan 25 at 11:30
NotoriousJuanGNotoriousJuanG
843
$endgroup$
Assume to the contrary that there exists $x$ and $a_1<a_2<...<a_nleq x$ and no $a_i$ divides the product of the others and $n>pi(x)$. We pick such $a_1,a_2,...,a_n$ and $x$ such that the sum $a_1+a_2+...+a_n$ is minimum. Then we can show that $a_{i}$'s are pairwise coprime.(Suppose that $(a_i,a_j)=dneq 1$ Then replace (a_i,a_j) by (a_i/d,a_j/d) and consider the minimality of the sum $a_1+a_2+...+a_n$.) Then each $a_i$ has a prime divisor(as $a_ineq 1$) which does not divide any other $a_j$'s. Therefore each $a_i$ has a distinct prime divisor. Hence there exists at least $n$ primes that are less than or equal to $x$. This is a contradiction. Hence the conclusion follows.
answered Jan 25 at 11:30
NotoriousJuanGNotoriousJuanG
843
answered Jan 25 at 11:30
NotoriousJuanGNotoriousJuanG
843
answered Jan 25 at 11:30
NotoriousJuanGNotoriousJuanG
843
answered Jan 25 at 11:30
NotoriousJuanGNotoriousJuanG
843
843
$begingroup$
Is it possible after we make the sum to be minimum, then some of $a_i$ becomes $1$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:48
1
$begingroup$
In that case, 1 divides the product of the others.
$endgroup$
– NotoriousJuanG
Jan 25 at 17:06
add a comment |
$begingroup$
Is it possible after we make the sum to be minimum, then some of $a_i$ becomes $1$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:48
1
$begingroup$
In that case, 1 divides the product of the others.
$endgroup$
– NotoriousJuanG
Jan 25 at 17:06
$begingroup$
Is it possible after we make the sum to be minimum, then some of $a_i$ becomes $1$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:48
$begingroup$
Is it possible after we make the sum to be minimum, then some of $a_i$ becomes $1$?
$endgroup$
– kelvin hong 方
Jan 25 at 11:48
1
1
$begingroup$
In that case, 1 divides the product of the others.
$endgroup$
– NotoriousJuanG
Jan 25 at 17:06
$begingroup$
In that case, 1 divides the product of the others.
$endgroup$
– NotoriousJuanG
Jan 25 at 17:06
add a comment |
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