Mixing
Finding bounds of the set: $ A = { frac{mcdot n}{m+n}: m,n in mathbb N } $
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I'm having some problems with this.
I know that lower bound will be $dfrac{1}{2}$. Should I just find $m,n$ for which $dfrac{mcdot n}{m+n} lt dfrac{1}{2} + epsilon $? Also I'm not sure how to prove that it's the best possible lower bound.
Also I presume that there isn't any upper bound, thus it goes to infinity, but I have problems coming up with the proof.
Thank you for any hints.
upper-lower-bounds
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amWhy
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Bartosz
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up vote
1
down vote
favorite
I'm having some problems with this.
I know that lower bound will be $dfrac{1}{2}$. Should I just find $m,n$ for which $dfrac{mcdot n}{m+n} lt dfrac{1}{2} + epsilon $? Also I'm not sure how to prove that it's the best possible lower bound.
Also I presume that there isn't any upper bound, thus it goes to infinity, but I have problems coming up with the proof.
Thank you for any hints.
upper-lower-bounds
edited yesterday
amWhy
191k27223437
asked yesterday
Bartosz
84
New contributor
Bartosz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm having some problems with this.
I know that lower bound will be $dfrac{1}{2}$. Should I just find $m,n$ for which $dfrac{mcdot n}{m+n} lt dfrac{1}{2} + epsilon $? Also I'm not sure how to prove that it's the best possible lower bound.
Also I presume that there isn't any upper bound, thus it goes to infinity, but I have problems coming up with the proof.
Thank you for any hints.
upper-lower-bounds
edited yesterday
amWhy
191k27223437
asked yesterday
Bartosz
84
New contributor
Bartosz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm having some problems with this.
I know that lower bound will be $dfrac{1}{2}$. Should I just find $m,n$ for which $dfrac{mcdot n}{m+n} lt dfrac{1}{2} + epsilon $? Also I'm not sure how to prove that it's the best possible lower bound.
Also I presume that there isn't any upper bound, thus it goes to infinity, but I have problems coming up with the proof.
Thank you for any hints.
upper-lower-bounds
upper-lower-bounds
edited yesterday
amWhy
191k27223437
asked yesterday
Bartosz
84
New contributor
Bartosz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
amWhy
191k27223437
asked yesterday
Bartosz
84
New contributor
Bartosz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
amWhy
191k27223437
edited yesterday
amWhy
191k27223437
edited yesterday
amWhy
191k27223437
191k27223437
asked yesterday
Bartosz
84
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Bartosz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked yesterday
Bartosz
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asked yesterday
Bartosz
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84
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New contributor
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3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
The number $dfrac12$ is a lower bound because, for each $m,ninmathbb N$, you have$$frac{mn}{m+n}geqslantfrac12.$$That's so becausebegin{align}frac{mn}{m+n}geqslantfrac12&iff2mngeqslant m+n\&iff2mn-m-ngeqslant0\&iff mn+mn-m-n+1geqslant1\&iff mn+(m-1)(n-1)geqslant1,end{align}which is true. On the other hand $dfrac{1times1}{1+1}=dfrac12$ and therefore $dfrac12$ is the greatest lower bound.
On the other hand, $lim_{ntoinfty}dfrac{n^2}{2n}=infty$, and therefore your set has no upper bound.
answered yesterday
José Carlos Santos
139k18111203
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up vote
1
down vote
Consider $(m-frac{1}{2})(n-frac{1}{2}) ge frac{1}{4}$ if $m,nge1$.
answered yesterday
Richard Martin
1,3238
add a comment |
up vote
0
down vote
Intuitively, you have multiplication on the top and addition on the bottom, so the top should, at least after a point, grow faster than the bottom. Therefore the minimum should come pretty early on.
Let's fix one of the variables and see if we can find the minimum. Suppose $m=1$. Then it becomes $dfrac {n}{n+1}$. Now, how does $dfrac {n}{n+1}$ behave? Well it only gets bigger and bigger: $dfrac 12, dfrac 23, dfrac 34, ...$ Can you prove it using induction?
Next, can you prove that for any fixed $m$, not only $1$, the resulting sequence is increasing as $n$ increases?
If you can do this, then it will follow that the minimum is achieved at $(1, 1)$, so it is $dfrac 12$.
answered yesterday
Ovi
12k938107
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The number $dfrac12$ is a lower bound because, for each $m,ninmathbb N$, you have$$frac{mn}{m+n}geqslantfrac12.$$That's so becausebegin{align}frac{mn}{m+n}geqslantfrac12&iff2mngeqslant m+n\&iff2mn-m-ngeqslant0\&iff mn+mn-m-n+1geqslant1\&iff mn+(m-1)(n-1)geqslant1,end{align}which is true. On the other hand $dfrac{1times1}{1+1}=dfrac12$ and therefore $dfrac12$ is the greatest lower bound.
On the other hand, $lim_{ntoinfty}dfrac{n^2}{2n}=infty$, and therefore your set has no upper bound.
answered yesterday
José Carlos Santos
139k18111203
add a comment |
up vote
2
down vote
accepted
The number $dfrac12$ is a lower bound because, for each $m,ninmathbb N$, you have$$frac{mn}{m+n}geqslantfrac12.$$That's so becausebegin{align}frac{mn}{m+n}geqslantfrac12&iff2mngeqslant m+n\&iff2mn-m-ngeqslant0\&iff mn+mn-m-n+1geqslant1\&iff mn+(m-1)(n-1)geqslant1,end{align}which is true. On the other hand $dfrac{1times1}{1+1}=dfrac12$ and therefore $dfrac12$ is the greatest lower bound.
On the other hand, $lim_{ntoinfty}dfrac{n^2}{2n}=infty$, and therefore your set has no upper bound.
answered yesterday
José Carlos Santos
139k18111203
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The number $dfrac12$ is a lower bound because, for each $m,ninmathbb N$, you have$$frac{mn}{m+n}geqslantfrac12.$$That's so becausebegin{align}frac{mn}{m+n}geqslantfrac12&iff2mngeqslant m+n\&iff2mn-m-ngeqslant0\&iff mn+mn-m-n+1geqslant1\&iff mn+(m-1)(n-1)geqslant1,end{align}which is true. On the other hand $dfrac{1times1}{1+1}=dfrac12$ and therefore $dfrac12$ is the greatest lower bound.
On the other hand, $lim_{ntoinfty}dfrac{n^2}{2n}=infty$, and therefore your set has no upper bound.
answered yesterday
José Carlos Santos
139k18111203
The number $dfrac12$ is a lower bound because, for each $m,ninmathbb N$, you have$$frac{mn}{m+n}geqslantfrac12.$$That's so becausebegin{align}frac{mn}{m+n}geqslantfrac12&iff2mngeqslant m+n\&iff2mn-m-ngeqslant0\&iff mn+mn-m-n+1geqslant1\&iff mn+(m-1)(n-1)geqslant1,end{align}which is true. On the other hand $dfrac{1times1}{1+1}=dfrac12$ and therefore $dfrac12$ is the greatest lower bound.
On the other hand, $lim_{ntoinfty}dfrac{n^2}{2n}=infty$, and therefore your set has no upper bound.
answered yesterday
José Carlos Santos
139k18111203
answered yesterday
José Carlos Santos
139k18111203
answered yesterday
José Carlos Santos
139k18111203
answered yesterday
José Carlos Santos
139k18111203
139k18111203
add a comment |
add a comment |
up vote
1
down vote
Consider $(m-frac{1}{2})(n-frac{1}{2}) ge frac{1}{4}$ if $m,nge1$.
answered yesterday
Richard Martin
1,3238
add a comment |
up vote
1
down vote
Consider $(m-frac{1}{2})(n-frac{1}{2}) ge frac{1}{4}$ if $m,nge1$.
answered yesterday
Richard Martin
1,3238
add a comment |
up vote
1
down vote
up vote
1
down vote
Consider $(m-frac{1}{2})(n-frac{1}{2}) ge frac{1}{4}$ if $m,nge1$.
answered yesterday
Richard Martin
1,3238
Consider $(m-frac{1}{2})(n-frac{1}{2}) ge frac{1}{4}$ if $m,nge1$.
answered yesterday
Richard Martin
1,3238
answered yesterday
Richard Martin
1,3238
answered yesterday
Richard Martin
1,3238
answered yesterday
Richard Martin
1,3238
1,3238
add a comment |
add a comment |
up vote
0
down vote
Intuitively, you have multiplication on the top and addition on the bottom, so the top should, at least after a point, grow faster than the bottom. Therefore the minimum should come pretty early on.
Let's fix one of the variables and see if we can find the minimum. Suppose $m=1$. Then it becomes $dfrac {n}{n+1}$. Now, how does $dfrac {n}{n+1}$ behave? Well it only gets bigger and bigger: $dfrac 12, dfrac 23, dfrac 34, ...$ Can you prove it using induction?
Next, can you prove that for any fixed $m$, not only $1$, the resulting sequence is increasing as $n$ increases?
If you can do this, then it will follow that the minimum is achieved at $(1, 1)$, so it is $dfrac 12$.
answered yesterday
Ovi
12k938107
add a comment |
up vote
0
down vote
Intuitively, you have multiplication on the top and addition on the bottom, so the top should, at least after a point, grow faster than the bottom. Therefore the minimum should come pretty early on.
Let's fix one of the variables and see if we can find the minimum. Suppose $m=1$. Then it becomes $dfrac {n}{n+1}$. Now, how does $dfrac {n}{n+1}$ behave? Well it only gets bigger and bigger: $dfrac 12, dfrac 23, dfrac 34, ...$ Can you prove it using induction?
Next, can you prove that for any fixed $m$, not only $1$, the resulting sequence is increasing as $n$ increases?
If you can do this, then it will follow that the minimum is achieved at $(1, 1)$, so it is $dfrac 12$.
answered yesterday
Ovi
12k938107
add a comment |
up vote
0
down vote
up vote
0
down vote
Intuitively, you have multiplication on the top and addition on the bottom, so the top should, at least after a point, grow faster than the bottom. Therefore the minimum should come pretty early on.
Let's fix one of the variables and see if we can find the minimum. Suppose $m=1$. Then it becomes $dfrac {n}{n+1}$. Now, how does $dfrac {n}{n+1}$ behave? Well it only gets bigger and bigger: $dfrac 12, dfrac 23, dfrac 34, ...$ Can you prove it using induction?
Next, can you prove that for any fixed $m$, not only $1$, the resulting sequence is increasing as $n$ increases?
If you can do this, then it will follow that the minimum is achieved at $(1, 1)$, so it is $dfrac 12$.
answered yesterday
Ovi
12k938107
Intuitively, you have multiplication on the top and addition on the bottom, so the top should, at least after a point, grow faster than the bottom. Therefore the minimum should come pretty early on.
Let's fix one of the variables and see if we can find the minimum. Suppose $m=1$. Then it becomes $dfrac {n}{n+1}$. Now, how does $dfrac {n}{n+1}$ behave? Well it only gets bigger and bigger: $dfrac 12, dfrac 23, dfrac 34, ...$ Can you prove it using induction?
Next, can you prove that for any fixed $m$, not only $1$, the resulting sequence is increasing as $n$ increases?
If you can do this, then it will follow that the minimum is achieved at $(1, 1)$, so it is $dfrac 12$.
answered yesterday
Ovi
12k938107
answered yesterday
Ovi
12k938107
answered yesterday
Ovi
12k938107
answered yesterday
Ovi
12k938107
12k938107
add a comment |
add a comment |
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