Proving an inequality
$begingroup$
Let $x$ and $y$ be positive real numbers. Prove that there exists an integer $n$ such that $x^{-1/2}leq x^{n}yleq x^{1/2}$.
I have no idea how to do this.
inequality
asked Mar 18 '17 at 22:25
pilgrimpilgrim
41828
$endgroup$
add a comment |
$begingroup$
Let $x$ and $y$ be positive real numbers. Prove that there exists an integer $n$ such that $x^{-1/2}leq x^{n}yleq x^{1/2}$.
I have no idea how to do this.
inequality
asked Mar 18 '17 at 22:25
pilgrimpilgrim
41828
$endgroup$
$begingroup$
I suppose $x ge 1$, and $y$ arbitrarily given.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:28
$begingroup$
@GNUSupporter Minor point: I hope we don't have $x=1$. That might make picking a "good" $n$ somewhat challenging :)
$endgroup$
– A. Howells
Mar 18 '17 at 22:45
$begingroup$
@A.Howells Yes, you're right.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:45
add a comment |
$begingroup$
Let $x$ and $y$ be positive real numbers. Prove that there exists an integer $n$ such that $x^{-1/2}leq x^{n}yleq x^{1/2}$.
I have no idea how to do this.
inequality
asked Mar 18 '17 at 22:25
pilgrimpilgrim
41828
$endgroup$
Let $x$ and $y$ be positive real numbers. Prove that there exists an integer $n$ such that $x^{-1/2}leq x^{n}yleq x^{1/2}$.
I have no idea how to do this.
inequality
inequality
asked Mar 18 '17 at 22:25
pilgrimpilgrim
41828
asked Mar 18 '17 at 22:25
pilgrimpilgrim
41828
asked Mar 18 '17 at 22:25
pilgrimpilgrim
41828
asked Mar 18 '17 at 22:25
pilgrimpilgrim
41828
asked Mar 18 '17 at 22:25
pilgrimpilgrim
41828
41828
$begingroup$
I suppose $x ge 1$, and $y$ arbitrarily given.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:28
$begingroup$
@GNUSupporter Minor point: I hope we don't have $x=1$. That might make picking a "good" $n$ somewhat challenging :)
$endgroup$
– A. Howells
Mar 18 '17 at 22:45
$begingroup$
@A.Howells Yes, you're right.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:45
add a comment |
$begingroup$
I suppose $x ge 1$, and $y$ arbitrarily given.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:28
$begingroup$
@GNUSupporter Minor point: I hope we don't have $x=1$. That might make picking a "good" $n$ somewhat challenging :)
$endgroup$
– A. Howells
Mar 18 '17 at 22:45
$begingroup$
@A.Howells Yes, you're right.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:45
$begingroup$
I suppose $x ge 1$, and $y$ arbitrarily given.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:28
$begingroup$
I suppose $x ge 1$, and $y$ arbitrarily given.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:28
$begingroup$
@GNUSupporter Minor point: I hope we don't have $x=1$. That might make picking a "good" $n$ somewhat challenging :)
$endgroup$
– A. Howells
Mar 18 '17 at 22:45
$begingroup$
@GNUSupporter Minor point: I hope we don't have $x=1$. That might make picking a "good" $n$ somewhat challenging :)
$endgroup$
– A. Howells
Mar 18 '17 at 22:45
$begingroup$
@A.Howells Yes, you're right.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:45
$begingroup$
@A.Howells Yes, you're right.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:45
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There is an integer in the interval: $$left[
log_{x}left(frac{1}{y}right)-frac{1}{2} , log_{x}left(frac{1}{y}right)+frac{1}{2}right],$$as its diameter is bigger than $1$.
This number is the answer.
$endgroup$
add a comment |
$begingroup$
Hint. We assume $x>1,y>0$. Then, by dividing by $y$ and by applying $ln$ to both sides, we have
$$
x^{n}yleq x^{1/2}implies n le frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}
$$ similarly,
$$
x^{-1/2}leq x^{n}y implies frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}-1le n
$$ giving the existence of such an integer $n$.
answered Mar 18 '17 at 22:37
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
$begingroup$
But $x,y$ are positive?! how about $0<x<1$?
$endgroup$
– mwomath
Mar 18 '17 at 23:10
$begingroup$
If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:13
$begingroup$
Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
$endgroup$
– mwomath
Mar 18 '17 at 23:16
add a comment |
$begingroup$
For $x,y>0$ the required inequality is equivalent to write
$$1le x^{n+frac{1}{2}}yle x.$$
The right hand side:
Let $f(x,y)=x-x^{n+frac{1}{2}}y$ we want to show that $f(x,y)ge 0$. So that
begin{align}
x-x^{n+frac{1}{2}}y = x left(1-x^{n-frac{1}{2}}yright) ge 0 qquad(?)
end{align}
But since $x>0$ so we need $1-x^{n-frac{1}{2}}y ge 0$ which means
begin{align}
x^{n-frac{1}{2}}y le 1
end{align}
Taking $ln(cdot)$ for both sides (an increasing function), we get
begin{align}
ln x^{n-frac{1}{2}}+ ln y le 0
end{align}
i.e.,
begin{align}
left{ begin{array}{l}
nle frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
\
ngefrac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
end{array} right.
qquad(*)
end{align}
So that the value of such $n$ must chosen taking into account the above condition. The left hand side goes similarly so we can obtain
begin{align}
left{ begin{array}{l}
nge -frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
\
nle-frac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
end{array} right.
qquad(**)
end{align}
Combining the inequalities (*) and (**) we can say that such integer $n$ satisfies
begin{align}
left{ begin{array}{l}
left| {n + frac{{ln y}}{{ln x}}} right| le frac{1}{2},qquad x>1,y>0 \
\
left| {n- frac{{ln y}}{{ln x}}} right| le frac{1}{2}, qquad 1>x>0,y>0
end{array} right.
end{align}
answered Mar 18 '17 at 23:09
mwomathmwomath
988614
$endgroup$
$begingroup$
I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:18
$begingroup$
Yes, I updated my answer. Thank you very much
$endgroup$
– mwomath
Mar 18 '17 at 23:29
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
$begingroup$
There is an integer in the interval: $$left[
log_{x}left(frac{1}{y}right)-frac{1}{2} , log_{x}left(frac{1}{y}right)+frac{1}{2}right],$$as its diameter is bigger than $1$.
This number is the answer.
$endgroup$
add a comment |
$begingroup$
There is an integer in the interval: $$left[
log_{x}left(frac{1}{y}right)-frac{1}{2} , log_{x}left(frac{1}{y}right)+frac{1}{2}right],$$as its diameter is bigger than $1$.
This number is the answer.
$endgroup$
add a comment |
$begingroup$
There is an integer in the interval: $$left[
log_{x}left(frac{1}{y}right)-frac{1}{2} , log_{x}left(frac{1}{y}right)+frac{1}{2}right],$$as its diameter is bigger than $1$.
This number is the answer.
$endgroup$
There is an integer in the interval: $$left[
log_{x}left(frac{1}{y}right)-frac{1}{2} , log_{x}left(frac{1}{y}right)+frac{1}{2}right],$$as its diameter is bigger than $1$.
This number is the answer.
edited Jan 12 at 6:11
community wiki
2 revs, 2 users 67%
Hassan
add a comment |
add a comment |
$begingroup$
Hint. We assume $x>1,y>0$. Then, by dividing by $y$ and by applying $ln$ to both sides, we have
$$
x^{n}yleq x^{1/2}implies n le frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}
$$ similarly,
$$
x^{-1/2}leq x^{n}y implies frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}-1le n
$$ giving the existence of such an integer $n$.
answered Mar 18 '17 at 22:37
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
$begingroup$
But $x,y$ are positive?! how about $0<x<1$?
$endgroup$
– mwomath
Mar 18 '17 at 23:10
$begingroup$
If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:13
$begingroup$
Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
$endgroup$
– mwomath
Mar 18 '17 at 23:16
add a comment |
$begingroup$
Hint. We assume $x>1,y>0$. Then, by dividing by $y$ and by applying $ln$ to both sides, we have
$$
x^{n}yleq x^{1/2}implies n le frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}
$$ similarly,
$$
x^{-1/2}leq x^{n}y implies frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}-1le n
$$ giving the existence of such an integer $n$.
answered Mar 18 '17 at 22:37
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
$begingroup$
But $x,y$ are positive?! how about $0<x<1$?
$endgroup$
– mwomath
Mar 18 '17 at 23:10
$begingroup$
If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:13
$begingroup$
Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
$endgroup$
– mwomath
Mar 18 '17 at 23:16
add a comment |
$begingroup$
Hint. We assume $x>1,y>0$. Then, by dividing by $y$ and by applying $ln$ to both sides, we have
$$
x^{n}yleq x^{1/2}implies n le frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}
$$ similarly,
$$
x^{-1/2}leq x^{n}y implies frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}-1le n
$$ giving the existence of such an integer $n$.
answered Mar 18 '17 at 22:37
Olivier OloaOlivier Oloa
109k17178294
$endgroup$
Hint. We assume $x>1,y>0$. Then, by dividing by $y$ and by applying $ln$ to both sides, we have
$$
x^{n}yleq x^{1/2}implies n le frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}
$$ similarly,
$$
x^{-1/2}leq x^{n}y implies frac{ln left(frac{sqrt{x}}{ y}right)}{ln (x)}-1le n
$$ giving the existence of such an integer $n$.
answered Mar 18 '17 at 22:37
Olivier OloaOlivier Oloa
109k17178294
answered Mar 18 '17 at 22:37
Olivier OloaOlivier Oloa
109k17178294
answered Mar 18 '17 at 22:37
Olivier OloaOlivier Oloa
109k17178294
answered Mar 18 '17 at 22:37
Olivier OloaOlivier Oloa
109k17178294
109k17178294
$begingroup$
But $x,y$ are positive?! how about $0<x<1$?
$endgroup$
– mwomath
Mar 18 '17 at 23:10
$begingroup$
If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:13
$begingroup$
Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
$endgroup$
– mwomath
Mar 18 '17 at 23:16
add a comment |
$begingroup$
But $x,y$ are positive?! how about $0<x<1$?
$endgroup$
– mwomath
Mar 18 '17 at 23:10
$begingroup$
If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:13
$begingroup$
Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
$endgroup$
– mwomath
Mar 18 '17 at 23:16
$begingroup$
But $x,y$ are positive?! how about $0<x<1$?
$endgroup$
– mwomath
Mar 18 '17 at 23:10
$begingroup$
But $x,y$ are positive?! how about $0<x<1$?
$endgroup$
– mwomath
Mar 18 '17 at 23:10
$begingroup$
If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:13
$begingroup$
If $0<x<1$, how can $x^{-1/2}leq x^{1/2}$ holds?
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:13
$begingroup$
Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
$endgroup$
– mwomath
Mar 18 '17 at 23:16
$begingroup$
Right, thanks. But there is another way to demonstrate the question. Would you read my solution below.
$endgroup$
– mwomath
Mar 18 '17 at 23:16
add a comment |
$begingroup$
For $x,y>0$ the required inequality is equivalent to write
$$1le x^{n+frac{1}{2}}yle x.$$
The right hand side:
Let $f(x,y)=x-x^{n+frac{1}{2}}y$ we want to show that $f(x,y)ge 0$. So that
begin{align}
x-x^{n+frac{1}{2}}y = x left(1-x^{n-frac{1}{2}}yright) ge 0 qquad(?)
end{align}
But since $x>0$ so we need $1-x^{n-frac{1}{2}}y ge 0$ which means
begin{align}
x^{n-frac{1}{2}}y le 1
end{align}
Taking $ln(cdot)$ for both sides (an increasing function), we get
begin{align}
ln x^{n-frac{1}{2}}+ ln y le 0
end{align}
i.e.,
begin{align}
left{ begin{array}{l}
nle frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
\
ngefrac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
end{array} right.
qquad(*)
end{align}
So that the value of such $n$ must chosen taking into account the above condition. The left hand side goes similarly so we can obtain
begin{align}
left{ begin{array}{l}
nge -frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
\
nle-frac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
end{array} right.
qquad(**)
end{align}
Combining the inequalities (*) and (**) we can say that such integer $n$ satisfies
begin{align}
left{ begin{array}{l}
left| {n + frac{{ln y}}{{ln x}}} right| le frac{1}{2},qquad x>1,y>0 \
\
left| {n- frac{{ln y}}{{ln x}}} right| le frac{1}{2}, qquad 1>x>0,y>0
end{array} right.
end{align}
answered Mar 18 '17 at 23:09
mwomathmwomath
988614
$endgroup$
$begingroup$
I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:18
$begingroup$
Yes, I updated my answer. Thank you very much
$endgroup$
– mwomath
Mar 18 '17 at 23:29
add a comment |
$begingroup$
For $x,y>0$ the required inequality is equivalent to write
$$1le x^{n+frac{1}{2}}yle x.$$
The right hand side:
Let $f(x,y)=x-x^{n+frac{1}{2}}y$ we want to show that $f(x,y)ge 0$. So that
begin{align}
x-x^{n+frac{1}{2}}y = x left(1-x^{n-frac{1}{2}}yright) ge 0 qquad(?)
end{align}
But since $x>0$ so we need $1-x^{n-frac{1}{2}}y ge 0$ which means
begin{align}
x^{n-frac{1}{2}}y le 1
end{align}
Taking $ln(cdot)$ for both sides (an increasing function), we get
begin{align}
ln x^{n-frac{1}{2}}+ ln y le 0
end{align}
i.e.,
begin{align}
left{ begin{array}{l}
nle frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
\
ngefrac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
end{array} right.
qquad(*)
end{align}
So that the value of such $n$ must chosen taking into account the above condition. The left hand side goes similarly so we can obtain
begin{align}
left{ begin{array}{l}
nge -frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
\
nle-frac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
end{array} right.
qquad(**)
end{align}
Combining the inequalities (*) and (**) we can say that such integer $n$ satisfies
begin{align}
left{ begin{array}{l}
left| {n + frac{{ln y}}{{ln x}}} right| le frac{1}{2},qquad x>1,y>0 \
\
left| {n- frac{{ln y}}{{ln x}}} right| le frac{1}{2}, qquad 1>x>0,y>0
end{array} right.
end{align}
answered Mar 18 '17 at 23:09
mwomathmwomath
988614
$endgroup$
$begingroup$
I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:18
$begingroup$
Yes, I updated my answer. Thank you very much
$endgroup$
– mwomath
Mar 18 '17 at 23:29
add a comment |
$begingroup$
For $x,y>0$ the required inequality is equivalent to write
$$1le x^{n+frac{1}{2}}yle x.$$
The right hand side:
Let $f(x,y)=x-x^{n+frac{1}{2}}y$ we want to show that $f(x,y)ge 0$. So that
begin{align}
x-x^{n+frac{1}{2}}y = x left(1-x^{n-frac{1}{2}}yright) ge 0 qquad(?)
end{align}
But since $x>0$ so we need $1-x^{n-frac{1}{2}}y ge 0$ which means
begin{align}
x^{n-frac{1}{2}}y le 1
end{align}
Taking $ln(cdot)$ for both sides (an increasing function), we get
begin{align}
ln x^{n-frac{1}{2}}+ ln y le 0
end{align}
i.e.,
begin{align}
left{ begin{array}{l}
nle frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
\
ngefrac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
end{array} right.
qquad(*)
end{align}
So that the value of such $n$ must chosen taking into account the above condition. The left hand side goes similarly so we can obtain
begin{align}
left{ begin{array}{l}
nge -frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
\
nle-frac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
end{array} right.
qquad(**)
end{align}
Combining the inequalities (*) and (**) we can say that such integer $n$ satisfies
begin{align}
left{ begin{array}{l}
left| {n + frac{{ln y}}{{ln x}}} right| le frac{1}{2},qquad x>1,y>0 \
\
left| {n- frac{{ln y}}{{ln x}}} right| le frac{1}{2}, qquad 1>x>0,y>0
end{array} right.
end{align}
answered Mar 18 '17 at 23:09
mwomathmwomath
988614
$endgroup$
For $x,y>0$ the required inequality is equivalent to write
$$1le x^{n+frac{1}{2}}yle x.$$
The right hand side:
Let $f(x,y)=x-x^{n+frac{1}{2}}y$ we want to show that $f(x,y)ge 0$. So that
begin{align}
x-x^{n+frac{1}{2}}y = x left(1-x^{n-frac{1}{2}}yright) ge 0 qquad(?)
end{align}
But since $x>0$ so we need $1-x^{n-frac{1}{2}}y ge 0$ which means
begin{align}
x^{n-frac{1}{2}}y le 1
end{align}
Taking $ln(cdot)$ for both sides (an increasing function), we get
begin{align}
ln x^{n-frac{1}{2}}+ ln y le 0
end{align}
i.e.,
begin{align}
left{ begin{array}{l}
nle frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
\
ngefrac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
end{array} right.
qquad(*)
end{align}
So that the value of such $n$ must chosen taking into account the above condition. The left hand side goes similarly so we can obtain
begin{align}
left{ begin{array}{l}
nge -frac{1}{2}- frac{ln y}{ln x}, qquad x>1,y>0 \
\
nle-frac{1}{2}+ frac{ln y}{ln x}, qquad 1>x>0,y>0 \
end{array} right.
qquad(**)
end{align}
Combining the inequalities (*) and (**) we can say that such integer $n$ satisfies
begin{align}
left{ begin{array}{l}
left| {n + frac{{ln y}}{{ln x}}} right| le frac{1}{2},qquad x>1,y>0 \
\
left| {n- frac{{ln y}}{{ln x}}} right| le frac{1}{2}, qquad 1>x>0,y>0
end{array} right.
end{align}
answered Mar 18 '17 at 23:09
mwomathmwomath
988614
edited Mar 18 '17 at 23:28
answered Mar 18 '17 at 23:09
mwomathmwomath
988614
answered Mar 18 '17 at 23:09
mwomathmwomath
988614
answered Mar 18 '17 at 23:09
mwomathmwomath
988614
988614
$begingroup$
I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:18
$begingroup$
Yes, I updated my answer. Thank you very much
$endgroup$
– mwomath
Mar 18 '17 at 23:29
add a comment |
$begingroup$
I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:18
$begingroup$
Yes, I updated my answer. Thank you very much
$endgroup$
– mwomath
Mar 18 '17 at 23:29
$begingroup$
I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:18
$begingroup$
I think there is a mistake when obtaining $(*)$ from the previous line, you have divided by $ln x$ and you have implicitely assumed $ln x>0$ which is not true if $0<x<1$.
$endgroup$
– Olivier Oloa
Mar 18 '17 at 23:18
$begingroup$
Yes, I updated my answer. Thank you very much
$endgroup$
– mwomath
Mar 18 '17 at 23:29
$begingroup$
Yes, I updated my answer. Thank you very much
$endgroup$
– mwomath
Mar 18 '17 at 23:29
add a comment |
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$begingroup$
I suppose $x ge 1$, and $y$ arbitrarily given.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:28
$begingroup$
@GNUSupporter Minor point: I hope we don't have $x=1$. That might make picking a "good" $n$ somewhat challenging :)
$endgroup$
– A. Howells
Mar 18 '17 at 22:45
$begingroup$
@A.Howells Yes, you're right.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Mar 18 '17 at 22:45