Mixing
Proving a property of Rationals using definition of Rationals (as an ordered field)
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I'm trying to find more behind the theorem:
Given $c,q in mathbbQ$ and $c,q > 0$ and $n in mathbbN$. If $q^n > c$ then $exists epsilon in mathbbQ$ with $0 < epsilon < 1 $ such that $(q-epsilon)^n > c$.
Is this a common theorem? Does it have a name? Is there a proof lying around. I cannot for the life of me find how to prove this on my own, so even if you have any tips on how to approach this, I'd greatly appreciate it. Thanks.
real-analysis
asked Jul 22 at 8:21
Florian Suess
1049
add a comment |Â
up vote
2
down vote
favorite
I'm trying to find more behind the theorem:
Given $c,q in mathbbQ$ and $c,q > 0$ and $n in mathbbN$. If $q^n > c$ then $exists epsilon in mathbbQ$ with $0 < epsilon < 1 $ such that $(q-epsilon)^n > c$.
Is this a common theorem? Does it have a name? Is there a proof lying around. I cannot for the life of me find how to prove this on my own, so even if you have any tips on how to approach this, I'd greatly appreciate it. Thanks.
real-analysis
asked Jul 22 at 8:21
Florian Suess
1049
You can find inspiration from the fact that the map $q mapsto q^n$ is continuous.
– mathcounterexamples.net
Jul 22 at 8:35
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to find more behind the theorem:
Given $c,q in mathbbQ$ and $c,q > 0$ and $n in mathbbN$. If $q^n > c$ then $exists epsilon in mathbbQ$ with $0 < epsilon < 1 $ such that $(q-epsilon)^n > c$.
Is this a common theorem? Does it have a name? Is there a proof lying around. I cannot for the life of me find how to prove this on my own, so even if you have any tips on how to approach this, I'd greatly appreciate it. Thanks.
real-analysis
asked Jul 22 at 8:21
Florian Suess
1049
I'm trying to find more behind the theorem:
Given $c,q in mathbbQ$ and $c,q > 0$ and $n in mathbbN$. If $q^n > c$ then $exists epsilon in mathbbQ$ with $0 < epsilon < 1 $ such that $(q-epsilon)^n > c$.
Is this a common theorem? Does it have a name? Is there a proof lying around. I cannot for the life of me find how to prove this on my own, so even if you have any tips on how to approach this, I'd greatly appreciate it. Thanks.
real-analysis
asked Jul 22 at 8:21
Florian Suess
1049
asked Jul 22 at 8:21
Florian Suess
1049
asked Jul 22 at 8:21
Florian Suess
1049
asked Jul 22 at 8:21
Florian Suess
1049
1049
You can find inspiration from the fact that the map $q mapsto q^n$ is continuous.
– mathcounterexamples.net
Jul 22 at 8:35
add a comment |Â
You can find inspiration from the fact that the map $q mapsto q^n$ is continuous.
– mathcounterexamples.net
Jul 22 at 8:35
You can find inspiration from the fact that the map $q mapsto q^n$ is continuous.
– mathcounterexamples.net
Jul 22 at 8:35
You can find inspiration from the fact that the map $q mapsto q^n$ is continuous.
– mathcounterexamples.net
Jul 22 at 8:35
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
It's not important enough to have a name. It can be proved easily enough
using analysis: it's true for small positive real $epsilon$, and so
by the Archimdean property it's true for some $epsilon = 1/N$, but you want to avoid analysis.
The inequality is equivalent to $(1-epsilon/q)^n>cq^-n=1-d$
where $0<d<1$. By Bernoulli's inequality it will hold provided that
$nepsilon/q <d$, that is if $epsilon <dq/n$.
(One form of) Bernoulli's inequality states that
$$(1-x)^nge 1-nx$$
for $0<x<1$ and $ninBbb N$. It's readily proved by induction
on $n$.
answered Jul 22 at 8:34
Lord Shark the Unknown
85.2k950111
add a comment |Â
up vote
1
down vote
Just note that $$q^n-(q-h) ^n=hq^n-1+q^n-2(q-h)+dots $$ and if $0<h<q$ then the RHS of the above equation is less than $nhq^n-1$ and we just need to ensure this to be less than $q^n-c$ and our job is done. In concrete terms any $h$ with $$0<h<minleft(1,q,fracq^n-cnq^n-1right)$$ works. An example is $c=1000,q=2,n=10$ and $(q^n-c) /nq^n-1=3/640$ and hence $h=1/1000=0.001$ works fine and you can check that $1.999^10>1000$.
The above uses $h$ instead of $epsilon$ to save typing effort.
answered Jul 22 at 19:02
Paramanand Singh
45.2k553142
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It's not important enough to have a name. It can be proved easily enough
using analysis: it's true for small positive real $epsilon$, and so
by the Archimdean property it's true for some $epsilon = 1/N$, but you want to avoid analysis.
The inequality is equivalent to $(1-epsilon/q)^n>cq^-n=1-d$
where $0<d<1$. By Bernoulli's inequality it will hold provided that
$nepsilon/q <d$, that is if $epsilon <dq/n$.
(One form of) Bernoulli's inequality states that
$$(1-x)^nge 1-nx$$
for $0<x<1$ and $ninBbb N$. It's readily proved by induction
on $n$.
answered Jul 22 at 8:34
Lord Shark the Unknown
85.2k950111
add a comment |Â
up vote
1
down vote
accepted
It's not important enough to have a name. It can be proved easily enough
using analysis: it's true for small positive real $epsilon$, and so
by the Archimdean property it's true for some $epsilon = 1/N$, but you want to avoid analysis.
The inequality is equivalent to $(1-epsilon/q)^n>cq^-n=1-d$
where $0<d<1$. By Bernoulli's inequality it will hold provided that
$nepsilon/q <d$, that is if $epsilon <dq/n$.
(One form of) Bernoulli's inequality states that
$$(1-x)^nge 1-nx$$
for $0<x<1$ and $ninBbb N$. It's readily proved by induction
on $n$.
answered Jul 22 at 8:34
Lord Shark the Unknown
85.2k950111
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It's not important enough to have a name. It can be proved easily enough
using analysis: it's true for small positive real $epsilon$, and so
by the Archimdean property it's true for some $epsilon = 1/N$, but you want to avoid analysis.
The inequality is equivalent to $(1-epsilon/q)^n>cq^-n=1-d$
where $0<d<1$. By Bernoulli's inequality it will hold provided that
$nepsilon/q <d$, that is if $epsilon <dq/n$.
(One form of) Bernoulli's inequality states that
$$(1-x)^nge 1-nx$$
for $0<x<1$ and $ninBbb N$. It's readily proved by induction
on $n$.
answered Jul 22 at 8:34
Lord Shark the Unknown
85.2k950111
It's not important enough to have a name. It can be proved easily enough
using analysis: it's true for small positive real $epsilon$, and so
by the Archimdean property it's true for some $epsilon = 1/N$, but you want to avoid analysis.
The inequality is equivalent to $(1-epsilon/q)^n>cq^-n=1-d$
where $0<d<1$. By Bernoulli's inequality it will hold provided that
$nepsilon/q <d$, that is if $epsilon <dq/n$.
(One form of) Bernoulli's inequality states that
$$(1-x)^nge 1-nx$$
for $0<x<1$ and $ninBbb N$. It's readily proved by induction
on $n$.
answered Jul 22 at 8:34
Lord Shark the Unknown
85.2k950111
answered Jul 22 at 8:34
Lord Shark the Unknown
85.2k950111
answered Jul 22 at 8:34
Lord Shark the Unknown
85.2k950111
answered Jul 22 at 8:34
Lord Shark the Unknown
85.2k950111
85.2k950111
add a comment |Â
add a comment |Â
up vote
1
down vote
Just note that $$q^n-(q-h) ^n=hq^n-1+q^n-2(q-h)+dots $$ and if $0<h<q$ then the RHS of the above equation is less than $nhq^n-1$ and we just need to ensure this to be less than $q^n-c$ and our job is done. In concrete terms any $h$ with $$0<h<minleft(1,q,fracq^n-cnq^n-1right)$$ works. An example is $c=1000,q=2,n=10$ and $(q^n-c) /nq^n-1=3/640$ and hence $h=1/1000=0.001$ works fine and you can check that $1.999^10>1000$.
The above uses $h$ instead of $epsilon$ to save typing effort.
answered Jul 22 at 19:02
Paramanand Singh
45.2k553142
add a comment |Â
up vote
1
down vote
Just note that $$q^n-(q-h) ^n=hq^n-1+q^n-2(q-h)+dots $$ and if $0<h<q$ then the RHS of the above equation is less than $nhq^n-1$ and we just need to ensure this to be less than $q^n-c$ and our job is done. In concrete terms any $h$ with $$0<h<minleft(1,q,fracq^n-cnq^n-1right)$$ works. An example is $c=1000,q=2,n=10$ and $(q^n-c) /nq^n-1=3/640$ and hence $h=1/1000=0.001$ works fine and you can check that $1.999^10>1000$.
The above uses $h$ instead of $epsilon$ to save typing effort.
answered Jul 22 at 19:02
Paramanand Singh
45.2k553142
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Just note that $$q^n-(q-h) ^n=hq^n-1+q^n-2(q-h)+dots $$ and if $0<h<q$ then the RHS of the above equation is less than $nhq^n-1$ and we just need to ensure this to be less than $q^n-c$ and our job is done. In concrete terms any $h$ with $$0<h<minleft(1,q,fracq^n-cnq^n-1right)$$ works. An example is $c=1000,q=2,n=10$ and $(q^n-c) /nq^n-1=3/640$ and hence $h=1/1000=0.001$ works fine and you can check that $1.999^10>1000$.
The above uses $h$ instead of $epsilon$ to save typing effort.
answered Jul 22 at 19:02
Paramanand Singh
45.2k553142
Just note that $$q^n-(q-h) ^n=hq^n-1+q^n-2(q-h)+dots $$ and if $0<h<q$ then the RHS of the above equation is less than $nhq^n-1$ and we just need to ensure this to be less than $q^n-c$ and our job is done. In concrete terms any $h$ with $$0<h<minleft(1,q,fracq^n-cnq^n-1right)$$ works. An example is $c=1000,q=2,n=10$ and $(q^n-c) /nq^n-1=3/640$ and hence $h=1/1000=0.001$ works fine and you can check that $1.999^10>1000$.
The above uses $h$ instead of $epsilon$ to save typing effort.
answered Jul 22 at 19:02
Paramanand Singh
45.2k553142
edited Jul 22 at 19:07
answered Jul 22 at 19:02
Paramanand Singh
45.2k553142
answered Jul 22 at 19:02
Paramanand Singh
45.2k553142
answered Jul 22 at 19:02
Paramanand Singh
45.2k553142
45.2k553142
add a comment |Â
add a comment |Â
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You can find inspiration from the fact that the map $q mapsto q^n$ is continuous.
– mathcounterexamples.net
Jul 22 at 8:35