Mixing
Show that for any $q$ in the interval $(1,2)$ exists $n$ to let the inequality of $m$ does not have positive...
$begingroup$
Show that $forall q in (1,2)$, $exists n in N^+$ let there is no $m in N^+$ satisfy the inequality$$q^n-q^{n-1}+1 le q^mlt q^{n+1}-q^n+1$$
The inequality can be written as $$q^{n-1}lefrac{q^m-1}{q-1}lt q^n$$
The sum of the geometric sequence is bigger than the $n-1$th term and the $n$th
term of a geometric sequence with a common ratio of $q$.
But I don't know what I can do next.
I am not sure whether the statement is correct.
inequality
asked Jan 6 at 3:40
yuanming luoyuanming luo
777
$endgroup$
add a comment |
$begingroup$
Show that $forall q in (1,2)$, $exists n in N^+$ let there is no $m in N^+$ satisfy the inequality$$q^n-q^{n-1}+1 le q^mlt q^{n+1}-q^n+1$$
The inequality can be written as $$q^{n-1}lefrac{q^m-1}{q-1}lt q^n$$
The sum of the geometric sequence is bigger than the $n-1$th term and the $n$th
term of a geometric sequence with a common ratio of $q$.
But I don't know what I can do next.
I am not sure whether the statement is correct.
inequality
asked Jan 6 at 3:40
yuanming luoyuanming luo
777
$endgroup$
add a comment |
$begingroup$
Show that $forall q in (1,2)$, $exists n in N^+$ let there is no $m in N^+$ satisfy the inequality$$q^n-q^{n-1}+1 le q^mlt q^{n+1}-q^n+1$$
The inequality can be written as $$q^{n-1}lefrac{q^m-1}{q-1}lt q^n$$
The sum of the geometric sequence is bigger than the $n-1$th term and the $n$th
term of a geometric sequence with a common ratio of $q$.
But I don't know what I can do next.
I am not sure whether the statement is correct.
inequality
asked Jan 6 at 3:40
yuanming luoyuanming luo
777
$endgroup$
Show that $forall q in (1,2)$, $exists n in N^+$ let there is no $m in N^+$ satisfy the inequality$$q^n-q^{n-1}+1 le q^mlt q^{n+1}-q^n+1$$
The inequality can be written as $$q^{n-1}lefrac{q^m-1}{q-1}lt q^n$$
The sum of the geometric sequence is bigger than the $n-1$th term and the $n$th
term of a geometric sequence with a common ratio of $q$.
But I don't know what I can do next.
I am not sure whether the statement is correct.
inequality
inequality
asked Jan 6 at 3:40
yuanming luoyuanming luo
777
asked Jan 6 at 3:40
yuanming luoyuanming luo
777
edited Jan 7 at 13:31
yuanming luo
asked Jan 6 at 3:40
yuanming luoyuanming luo
777
asked Jan 6 at 3:40
yuanming luoyuanming luo
777
asked Jan 6 at 3:40
yuanming luoyuanming luo
777
777
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Consider fixed arbitrary $q in (1,2)$. Consider $q^i(2-q)$ as $i$ varies. This diverges to infinity, so in particular is eventually $geq 1$. Also $q^0(2-q)=2-q<1$. Thus there exists a smallest integer $k$ with $q^k(2-q) geq 1$, and we have $k geq 1$. By minimality of $k$ we get $q^{k-1}(2-q)<1$.
Take $n=k in mathbb{N}^{+}$. We get
$$q^{k-1}=q^{k-1}(q-1)+q^{k-1}(2-q)<q^{k-1}(q-1)+1=q^k-q^{k-1}+1$$
$$q^k=q^k(q-1)+q^k(2-q) geq q^k(q-1)+1=q^{k+1}-q^k+1$$
We are thus done, as $q^k-q^{k-1}+1 leq q^m<q^{k+1}-q^k+1$ then implies $k-1<m<k$, a contradiction, so no such $m$ exists.
Remark: In fact it is possible to show that $q(2-q)<1$ for $qin (1,2)$, thus $k geq 2$, which reconciles with the fact that $m=1$ always works when $n=1$.
answered Jan 7 at 14:03
user632469user632469
161
$endgroup$
add a comment |
$begingroup$
Let $epsilon_n in [0,1] forall n$ with $epsilon_{t+1}>epsilon_t$.
We have that $q$ is expressible as $1+epsilon_1$. This leads to: $$1+epsilon_2-epsilon_3leq q^m < 1+epsilon_3 - epsilon_2$$
This draws that the only possible integer solution is $0$, which isn't positive.
answered Jan 6 at 4:51
Rhys HughesRhys Hughes
6,0171530
$endgroup$
$begingroup$
I do not understand why the unique integer solution is 0.
$endgroup$
– yuanming luo
Jan 6 at 5:16
$begingroup$
Two reasons. $$|epsilon_3-epsilon_2|<1$$ and $q^0=1$
$endgroup$
– Rhys Hughes
Jan 6 at 6:05
$begingroup$
I still cannot catch your point and your reason. Could you explain your method with more details?
$endgroup$
– yuanming luo
Jan 7 at 13:27
$begingroup$
I implied that $$(1+epsilon_1)^n=1+epsilon_2$$, and $$(1+epsilon_1)^{n+1}=1+epsilon_3$$ and then applied that to the original statement
$endgroup$
– Rhys Hughes
Jan 7 at 15:25
add a comment |
$begingroup$
$q^n-q^{n-1}+1
le q^m
lt q^{n+1}-q^n+1
$
The right inequality
implies $m le n$ because
$q^{n+1}-q^n+1
lt q^{n+1}
$
so $m < n+1$
so $m le n$.
The left inequality,
in contrast,
has many solutions.
Write it as
$q^{n-1}(q-1)+1
le q^m
$
and let
$q = 1+x$
where
$0 < x < 1$.
It becomes
$x(1+x)^{n-1}+1
le (1+x)^m
$.
If $frac1{n} le x le frac1{n-1}$,
then
$(1+x)^{n-1}
le (1+frac1{n-1})^{n-1}
lt e
$
so the left side is less than
$1+frac{e}{n-1}
$.
For the right side,
$q^m
=(1+x)^m
ge 1+mx
ge 1+frac{m}{n}
$
so the left inequality is true if
$1+frac{e}{n-1}
lt 1+frac{m}{n}
$
or
$m > frac{en}{n-1}
gt 5
$
if
$n ge 2
$.
Therefore,
for any $m > 5$ and $n > 2$
we can find a $q$
so that the left inequality holds.
answered Jan 7 at 1:41
marty cohenmarty cohen
73.3k549128
$endgroup$
$begingroup$
actually, the question I ask is to show for all q in the interval there is always existing n to let no positive integer m satisfy the inequality. I will edit again the question to let it less confusing.
$endgroup$
– yuanming luo
Jan 7 at 12:00
$begingroup$
I have already tried to close this question. The question posted on this web is clearer than this. If you still have the interest in the question I asked. math.stackexchange.com/questions/3064930/…
$endgroup$
– yuanming luo
Jan 7 at 12:06
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider fixed arbitrary $q in (1,2)$. Consider $q^i(2-q)$ as $i$ varies. This diverges to infinity, so in particular is eventually $geq 1$. Also $q^0(2-q)=2-q<1$. Thus there exists a smallest integer $k$ with $q^k(2-q) geq 1$, and we have $k geq 1$. By minimality of $k$ we get $q^{k-1}(2-q)<1$.
Take $n=k in mathbb{N}^{+}$. We get
$$q^{k-1}=q^{k-1}(q-1)+q^{k-1}(2-q)<q^{k-1}(q-1)+1=q^k-q^{k-1}+1$$
$$q^k=q^k(q-1)+q^k(2-q) geq q^k(q-1)+1=q^{k+1}-q^k+1$$
We are thus done, as $q^k-q^{k-1}+1 leq q^m<q^{k+1}-q^k+1$ then implies $k-1<m<k$, a contradiction, so no such $m$ exists.
Remark: In fact it is possible to show that $q(2-q)<1$ for $qin (1,2)$, thus $k geq 2$, which reconciles with the fact that $m=1$ always works when $n=1$.
answered Jan 7 at 14:03
user632469user632469
161
$endgroup$
add a comment |
$begingroup$
Consider fixed arbitrary $q in (1,2)$. Consider $q^i(2-q)$ as $i$ varies. This diverges to infinity, so in particular is eventually $geq 1$. Also $q^0(2-q)=2-q<1$. Thus there exists a smallest integer $k$ with $q^k(2-q) geq 1$, and we have $k geq 1$. By minimality of $k$ we get $q^{k-1}(2-q)<1$.
Take $n=k in mathbb{N}^{+}$. We get
$$q^{k-1}=q^{k-1}(q-1)+q^{k-1}(2-q)<q^{k-1}(q-1)+1=q^k-q^{k-1}+1$$
$$q^k=q^k(q-1)+q^k(2-q) geq q^k(q-1)+1=q^{k+1}-q^k+1$$
We are thus done, as $q^k-q^{k-1}+1 leq q^m<q^{k+1}-q^k+1$ then implies $k-1<m<k$, a contradiction, so no such $m$ exists.
Remark: In fact it is possible to show that $q(2-q)<1$ for $qin (1,2)$, thus $k geq 2$, which reconciles with the fact that $m=1$ always works when $n=1$.
answered Jan 7 at 14:03
user632469user632469
161
$endgroup$
add a comment |
$begingroup$
Consider fixed arbitrary $q in (1,2)$. Consider $q^i(2-q)$ as $i$ varies. This diverges to infinity, so in particular is eventually $geq 1$. Also $q^0(2-q)=2-q<1$. Thus there exists a smallest integer $k$ with $q^k(2-q) geq 1$, and we have $k geq 1$. By minimality of $k$ we get $q^{k-1}(2-q)<1$.
Take $n=k in mathbb{N}^{+}$. We get
$$q^{k-1}=q^{k-1}(q-1)+q^{k-1}(2-q)<q^{k-1}(q-1)+1=q^k-q^{k-1}+1$$
$$q^k=q^k(q-1)+q^k(2-q) geq q^k(q-1)+1=q^{k+1}-q^k+1$$
We are thus done, as $q^k-q^{k-1}+1 leq q^m<q^{k+1}-q^k+1$ then implies $k-1<m<k$, a contradiction, so no such $m$ exists.
Remark: In fact it is possible to show that $q(2-q)<1$ for $qin (1,2)$, thus $k geq 2$, which reconciles with the fact that $m=1$ always works when $n=1$.
answered Jan 7 at 14:03
user632469user632469
161
$endgroup$
Consider fixed arbitrary $q in (1,2)$. Consider $q^i(2-q)$ as $i$ varies. This diverges to infinity, so in particular is eventually $geq 1$. Also $q^0(2-q)=2-q<1$. Thus there exists a smallest integer $k$ with $q^k(2-q) geq 1$, and we have $k geq 1$. By minimality of $k$ we get $q^{k-1}(2-q)<1$.
Take $n=k in mathbb{N}^{+}$. We get
$$q^{k-1}=q^{k-1}(q-1)+q^{k-1}(2-q)<q^{k-1}(q-1)+1=q^k-q^{k-1}+1$$
$$q^k=q^k(q-1)+q^k(2-q) geq q^k(q-1)+1=q^{k+1}-q^k+1$$
We are thus done, as $q^k-q^{k-1}+1 leq q^m<q^{k+1}-q^k+1$ then implies $k-1<m<k$, a contradiction, so no such $m$ exists.
Remark: In fact it is possible to show that $q(2-q)<1$ for $qin (1,2)$, thus $k geq 2$, which reconciles with the fact that $m=1$ always works when $n=1$.
answered Jan 7 at 14:03
user632469user632469
161
edited Jan 7 at 14:16
answered Jan 7 at 14:03
user632469user632469
161
answered Jan 7 at 14:03
user632469user632469
161
answered Jan 7 at 14:03
user632469user632469
161
161
add a comment |
add a comment |
$begingroup$
Let $epsilon_n in [0,1] forall n$ with $epsilon_{t+1}>epsilon_t$.
We have that $q$ is expressible as $1+epsilon_1$. This leads to: $$1+epsilon_2-epsilon_3leq q^m < 1+epsilon_3 - epsilon_2$$
This draws that the only possible integer solution is $0$, which isn't positive.
answered Jan 6 at 4:51
Rhys HughesRhys Hughes
6,0171530
$endgroup$
$begingroup$
I do not understand why the unique integer solution is 0.
$endgroup$
– yuanming luo
Jan 6 at 5:16
$begingroup$
Two reasons. $$|epsilon_3-epsilon_2|<1$$ and $q^0=1$
$endgroup$
– Rhys Hughes
Jan 6 at 6:05
$begingroup$
I still cannot catch your point and your reason. Could you explain your method with more details?
$endgroup$
– yuanming luo
Jan 7 at 13:27
$begingroup$
I implied that $$(1+epsilon_1)^n=1+epsilon_2$$, and $$(1+epsilon_1)^{n+1}=1+epsilon_3$$ and then applied that to the original statement
$endgroup$
– Rhys Hughes
Jan 7 at 15:25
add a comment |
$begingroup$
Let $epsilon_n in [0,1] forall n$ with $epsilon_{t+1}>epsilon_t$.
We have that $q$ is expressible as $1+epsilon_1$. This leads to: $$1+epsilon_2-epsilon_3leq q^m < 1+epsilon_3 - epsilon_2$$
This draws that the only possible integer solution is $0$, which isn't positive.
answered Jan 6 at 4:51
Rhys HughesRhys Hughes
6,0171530
$endgroup$
$begingroup$
I do not understand why the unique integer solution is 0.
$endgroup$
– yuanming luo
Jan 6 at 5:16
$begingroup$
Two reasons. $$|epsilon_3-epsilon_2|<1$$ and $q^0=1$
$endgroup$
– Rhys Hughes
Jan 6 at 6:05
$begingroup$
I still cannot catch your point and your reason. Could you explain your method with more details?
$endgroup$
– yuanming luo
Jan 7 at 13:27
$begingroup$
I implied that $$(1+epsilon_1)^n=1+epsilon_2$$, and $$(1+epsilon_1)^{n+1}=1+epsilon_3$$ and then applied that to the original statement
$endgroup$
– Rhys Hughes
Jan 7 at 15:25
add a comment |
$begingroup$
Let $epsilon_n in [0,1] forall n$ with $epsilon_{t+1}>epsilon_t$.
We have that $q$ is expressible as $1+epsilon_1$. This leads to: $$1+epsilon_2-epsilon_3leq q^m < 1+epsilon_3 - epsilon_2$$
This draws that the only possible integer solution is $0$, which isn't positive.
answered Jan 6 at 4:51
Rhys HughesRhys Hughes
6,0171530
$endgroup$
Let $epsilon_n in [0,1] forall n$ with $epsilon_{t+1}>epsilon_t$.
We have that $q$ is expressible as $1+epsilon_1$. This leads to: $$1+epsilon_2-epsilon_3leq q^m < 1+epsilon_3 - epsilon_2$$
This draws that the only possible integer solution is $0$, which isn't positive.
answered Jan 6 at 4:51
Rhys HughesRhys Hughes
6,0171530
answered Jan 6 at 4:51
Rhys HughesRhys Hughes
6,0171530
answered Jan 6 at 4:51
Rhys HughesRhys Hughes
6,0171530
answered Jan 6 at 4:51
Rhys HughesRhys Hughes
6,0171530
6,0171530
$begingroup$
I do not understand why the unique integer solution is 0.
$endgroup$
– yuanming luo
Jan 6 at 5:16
$begingroup$
Two reasons. $$|epsilon_3-epsilon_2|<1$$ and $q^0=1$
$endgroup$
– Rhys Hughes
Jan 6 at 6:05
$begingroup$
I still cannot catch your point and your reason. Could you explain your method with more details?
$endgroup$
– yuanming luo
Jan 7 at 13:27
$begingroup$
I implied that $$(1+epsilon_1)^n=1+epsilon_2$$, and $$(1+epsilon_1)^{n+1}=1+epsilon_3$$ and then applied that to the original statement
$endgroup$
– Rhys Hughes
Jan 7 at 15:25
add a comment |
$begingroup$
I do not understand why the unique integer solution is 0.
$endgroup$
– yuanming luo
Jan 6 at 5:16
$begingroup$
Two reasons. $$|epsilon_3-epsilon_2|<1$$ and $q^0=1$
$endgroup$
– Rhys Hughes
Jan 6 at 6:05
$begingroup$
I still cannot catch your point and your reason. Could you explain your method with more details?
$endgroup$
– yuanming luo
Jan 7 at 13:27
$begingroup$
I implied that $$(1+epsilon_1)^n=1+epsilon_2$$, and $$(1+epsilon_1)^{n+1}=1+epsilon_3$$ and then applied that to the original statement
$endgroup$
– Rhys Hughes
Jan 7 at 15:25
$begingroup$
I do not understand why the unique integer solution is 0.
$endgroup$
– yuanming luo
Jan 6 at 5:16
$begingroup$
I do not understand why the unique integer solution is 0.
$endgroup$
– yuanming luo
Jan 6 at 5:16
$begingroup$
Two reasons. $$|epsilon_3-epsilon_2|<1$$ and $q^0=1$
$endgroup$
– Rhys Hughes
Jan 6 at 6:05
$begingroup$
Two reasons. $$|epsilon_3-epsilon_2|<1$$ and $q^0=1$
$endgroup$
– Rhys Hughes
Jan 6 at 6:05
$begingroup$
I still cannot catch your point and your reason. Could you explain your method with more details?
$endgroup$
– yuanming luo
Jan 7 at 13:27
$begingroup$
I still cannot catch your point and your reason. Could you explain your method with more details?
$endgroup$
– yuanming luo
Jan 7 at 13:27
$begingroup$
I implied that $$(1+epsilon_1)^n=1+epsilon_2$$, and $$(1+epsilon_1)^{n+1}=1+epsilon_3$$ and then applied that to the original statement
$endgroup$
– Rhys Hughes
Jan 7 at 15:25
$begingroup$
I implied that $$(1+epsilon_1)^n=1+epsilon_2$$, and $$(1+epsilon_1)^{n+1}=1+epsilon_3$$ and then applied that to the original statement
$endgroup$
– Rhys Hughes
Jan 7 at 15:25
add a comment |
$begingroup$
$q^n-q^{n-1}+1
le q^m
lt q^{n+1}-q^n+1
$
The right inequality
implies $m le n$ because
$q^{n+1}-q^n+1
lt q^{n+1}
$
so $m < n+1$
so $m le n$.
The left inequality,
in contrast,
has many solutions.
Write it as
$q^{n-1}(q-1)+1
le q^m
$
and let
$q = 1+x$
where
$0 < x < 1$.
It becomes
$x(1+x)^{n-1}+1
le (1+x)^m
$.
If $frac1{n} le x le frac1{n-1}$,
then
$(1+x)^{n-1}
le (1+frac1{n-1})^{n-1}
lt e
$
so the left side is less than
$1+frac{e}{n-1}
$.
For the right side,
$q^m
=(1+x)^m
ge 1+mx
ge 1+frac{m}{n}
$
so the left inequality is true if
$1+frac{e}{n-1}
lt 1+frac{m}{n}
$
or
$m > frac{en}{n-1}
gt 5
$
if
$n ge 2
$.
Therefore,
for any $m > 5$ and $n > 2$
we can find a $q$
so that the left inequality holds.
answered Jan 7 at 1:41
marty cohenmarty cohen
73.3k549128
$endgroup$
$begingroup$
actually, the question I ask is to show for all q in the interval there is always existing n to let no positive integer m satisfy the inequality. I will edit again the question to let it less confusing.
$endgroup$
– yuanming luo
Jan 7 at 12:00
$begingroup$
I have already tried to close this question. The question posted on this web is clearer than this. If you still have the interest in the question I asked. math.stackexchange.com/questions/3064930/…
$endgroup$
– yuanming luo
Jan 7 at 12:06
add a comment |
$begingroup$
$q^n-q^{n-1}+1
le q^m
lt q^{n+1}-q^n+1
$
The right inequality
implies $m le n$ because
$q^{n+1}-q^n+1
lt q^{n+1}
$
so $m < n+1$
so $m le n$.
The left inequality,
in contrast,
has many solutions.
Write it as
$q^{n-1}(q-1)+1
le q^m
$
and let
$q = 1+x$
where
$0 < x < 1$.
It becomes
$x(1+x)^{n-1}+1
le (1+x)^m
$.
If $frac1{n} le x le frac1{n-1}$,
then
$(1+x)^{n-1}
le (1+frac1{n-1})^{n-1}
lt e
$
so the left side is less than
$1+frac{e}{n-1}
$.
For the right side,
$q^m
=(1+x)^m
ge 1+mx
ge 1+frac{m}{n}
$
so the left inequality is true if
$1+frac{e}{n-1}
lt 1+frac{m}{n}
$
or
$m > frac{en}{n-1}
gt 5
$
if
$n ge 2
$.
Therefore,
for any $m > 5$ and $n > 2$
we can find a $q$
so that the left inequality holds.
answered Jan 7 at 1:41
marty cohenmarty cohen
73.3k549128
$endgroup$
$begingroup$
actually, the question I ask is to show for all q in the interval there is always existing n to let no positive integer m satisfy the inequality. I will edit again the question to let it less confusing.
$endgroup$
– yuanming luo
Jan 7 at 12:00
$begingroup$
I have already tried to close this question. The question posted on this web is clearer than this. If you still have the interest in the question I asked. math.stackexchange.com/questions/3064930/…
$endgroup$
– yuanming luo
Jan 7 at 12:06
add a comment |
$begingroup$
$q^n-q^{n-1}+1
le q^m
lt q^{n+1}-q^n+1
$
The right inequality
implies $m le n$ because
$q^{n+1}-q^n+1
lt q^{n+1}
$
so $m < n+1$
so $m le n$.
The left inequality,
in contrast,
has many solutions.
Write it as
$q^{n-1}(q-1)+1
le q^m
$
and let
$q = 1+x$
where
$0 < x < 1$.
It becomes
$x(1+x)^{n-1}+1
le (1+x)^m
$.
If $frac1{n} le x le frac1{n-1}$,
then
$(1+x)^{n-1}
le (1+frac1{n-1})^{n-1}
lt e
$
so the left side is less than
$1+frac{e}{n-1}
$.
For the right side,
$q^m
=(1+x)^m
ge 1+mx
ge 1+frac{m}{n}
$
so the left inequality is true if
$1+frac{e}{n-1}
lt 1+frac{m}{n}
$
or
$m > frac{en}{n-1}
gt 5
$
if
$n ge 2
$.
Therefore,
for any $m > 5$ and $n > 2$
we can find a $q$
so that the left inequality holds.
answered Jan 7 at 1:41
marty cohenmarty cohen
73.3k549128
$endgroup$
$q^n-q^{n-1}+1
le q^m
lt q^{n+1}-q^n+1
$
The right inequality
implies $m le n$ because
$q^{n+1}-q^n+1
lt q^{n+1}
$
so $m < n+1$
so $m le n$.
The left inequality,
in contrast,
has many solutions.
Write it as
$q^{n-1}(q-1)+1
le q^m
$
and let
$q = 1+x$
where
$0 < x < 1$.
It becomes
$x(1+x)^{n-1}+1
le (1+x)^m
$.
If $frac1{n} le x le frac1{n-1}$,
then
$(1+x)^{n-1}
le (1+frac1{n-1})^{n-1}
lt e
$
so the left side is less than
$1+frac{e}{n-1}
$.
For the right side,
$q^m
=(1+x)^m
ge 1+mx
ge 1+frac{m}{n}
$
so the left inequality is true if
$1+frac{e}{n-1}
lt 1+frac{m}{n}
$
or
$m > frac{en}{n-1}
gt 5
$
if
$n ge 2
$.
Therefore,
for any $m > 5$ and $n > 2$
we can find a $q$
so that the left inequality holds.
answered Jan 7 at 1:41
marty cohenmarty cohen
73.3k549128
answered Jan 7 at 1:41
marty cohenmarty cohen
73.3k549128
answered Jan 7 at 1:41
marty cohenmarty cohen
73.3k549128
answered Jan 7 at 1:41
marty cohenmarty cohen
73.3k549128
73.3k549128
$begingroup$
actually, the question I ask is to show for all q in the interval there is always existing n to let no positive integer m satisfy the inequality. I will edit again the question to let it less confusing.
$endgroup$
– yuanming luo
Jan 7 at 12:00
$begingroup$
I have already tried to close this question. The question posted on this web is clearer than this. If you still have the interest in the question I asked. math.stackexchange.com/questions/3064930/…
$endgroup$
– yuanming luo
Jan 7 at 12:06
add a comment |
$begingroup$
actually, the question I ask is to show for all q in the interval there is always existing n to let no positive integer m satisfy the inequality. I will edit again the question to let it less confusing.
$endgroup$
– yuanming luo
Jan 7 at 12:00
$begingroup$
I have already tried to close this question. The question posted on this web is clearer than this. If you still have the interest in the question I asked. math.stackexchange.com/questions/3064930/…
$endgroup$
– yuanming luo
Jan 7 at 12:06
$begingroup$
actually, the question I ask is to show for all q in the interval there is always existing n to let no positive integer m satisfy the inequality. I will edit again the question to let it less confusing.
$endgroup$
– yuanming luo
Jan 7 at 12:00
$begingroup$
actually, the question I ask is to show for all q in the interval there is always existing n to let no positive integer m satisfy the inequality. I will edit again the question to let it less confusing.
$endgroup$
– yuanming luo
Jan 7 at 12:00
$begingroup$
I have already tried to close this question. The question posted on this web is clearer than this. If you still have the interest in the question I asked. math.stackexchange.com/questions/3064930/…
$endgroup$
– yuanming luo
Jan 7 at 12:06
$begingroup$
I have already tried to close this question. The question posted on this web is clearer than this. If you still have the interest in the question I asked. math.stackexchange.com/questions/3064930/…
$endgroup$
– yuanming luo
Jan 7 at 12:06
add a comment |
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