Compare fractions a/b and b/a
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I'm trying to implement a mathematical method for calculation of injustice. The formula depends on two variables $a>=1$ and $b>=1$ and returns a fraction $a/b$.
$a/b=1$ indicates justice. If $a/b<1$ the injustice is to your advantage. If $a/b>1$ the injustice is to your disadvantage.
Now I would like to set aside advantage/disadvantage and only calculate the size of the injustice. I'm having trouble figuring out an easy way to get the same result for $a/b$ as $b/a$.
Anyone have any idea of how I can do this?
fractions
asked Jan 15 at 17:28
DoughDough
31
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add a comment |
$begingroup$
I'm trying to implement a mathematical method for calculation of injustice. The formula depends on two variables $a>=1$ and $b>=1$ and returns a fraction $a/b$.
$a/b=1$ indicates justice. If $a/b<1$ the injustice is to your advantage. If $a/b>1$ the injustice is to your disadvantage.
Now I would like to set aside advantage/disadvantage and only calculate the size of the injustice. I'm having trouble figuring out an easy way to get the same result for $a/b$ as $b/a$.
Anyone have any idea of how I can do this?
fractions
asked Jan 15 at 17:28
DoughDough
31
$endgroup$
2
$begingroup$
Not entirely sure what you are trying to do, but $Fleft(frac abright)=|log frac ab|=|log a - log b|$ might be what you are after. Easy to see that $Fleft(frac abright)=Fleft(frac baright)$ and $Fleft(frac abright)=1implies a=b$.
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– lulu
Jan 15 at 17:31
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Log seems to be a perfect solution. Thank you @lulu!
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– Dough
Jan 15 at 17:34
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No problem. Note: There's a typo in my comment. I meant to write that $Fleft(frac abright)=0 implies a=b$
$endgroup$
– lulu
Jan 15 at 17:38
add a comment |
$begingroup$
I'm trying to implement a mathematical method for calculation of injustice. The formula depends on two variables $a>=1$ and $b>=1$ and returns a fraction $a/b$.
$a/b=1$ indicates justice. If $a/b<1$ the injustice is to your advantage. If $a/b>1$ the injustice is to your disadvantage.
Now I would like to set aside advantage/disadvantage and only calculate the size of the injustice. I'm having trouble figuring out an easy way to get the same result for $a/b$ as $b/a$.
Anyone have any idea of how I can do this?
fractions
asked Jan 15 at 17:28
DoughDough
31
$endgroup$
I'm trying to implement a mathematical method for calculation of injustice. The formula depends on two variables $a>=1$ and $b>=1$ and returns a fraction $a/b$.
$a/b=1$ indicates justice. If $a/b<1$ the injustice is to your advantage. If $a/b>1$ the injustice is to your disadvantage.
Now I would like to set aside advantage/disadvantage and only calculate the size of the injustice. I'm having trouble figuring out an easy way to get the same result for $a/b$ as $b/a$.
Anyone have any idea of how I can do this?
fractions
fractions
asked Jan 15 at 17:28
DoughDough
31
asked Jan 15 at 17:28
DoughDough
31
asked Jan 15 at 17:28
DoughDough
31
asked Jan 15 at 17:28
DoughDough
31
asked Jan 15 at 17:28
DoughDough
31
31
2
$begingroup$
Not entirely sure what you are trying to do, but $Fleft(frac abright)=|log frac ab|=|log a - log b|$ might be what you are after. Easy to see that $Fleft(frac abright)=Fleft(frac baright)$ and $Fleft(frac abright)=1implies a=b$.
$endgroup$
– lulu
Jan 15 at 17:31
$begingroup$
Log seems to be a perfect solution. Thank you @lulu!
$endgroup$
– Dough
Jan 15 at 17:34
$begingroup$
No problem. Note: There's a typo in my comment. I meant to write that $Fleft(frac abright)=0 implies a=b$
$endgroup$
– lulu
Jan 15 at 17:38
add a comment |
2
$begingroup$
Not entirely sure what you are trying to do, but $Fleft(frac abright)=|log frac ab|=|log a - log b|$ might be what you are after. Easy to see that $Fleft(frac abright)=Fleft(frac baright)$ and $Fleft(frac abright)=1implies a=b$.
$endgroup$
– lulu
Jan 15 at 17:31
$begingroup$
Log seems to be a perfect solution. Thank you @lulu!
$endgroup$
– Dough
Jan 15 at 17:34
$begingroup$
No problem. Note: There's a typo in my comment. I meant to write that $Fleft(frac abright)=0 implies a=b$
$endgroup$
– lulu
Jan 15 at 17:38
2
2
$begingroup$
Not entirely sure what you are trying to do, but $Fleft(frac abright)=|log frac ab|=|log a - log b|$ might be what you are after. Easy to see that $Fleft(frac abright)=Fleft(frac baright)$ and $Fleft(frac abright)=1implies a=b$.
$endgroup$
– lulu
Jan 15 at 17:31
$begingroup$
Not entirely sure what you are trying to do, but $Fleft(frac abright)=|log frac ab|=|log a - log b|$ might be what you are after. Easy to see that $Fleft(frac abright)=Fleft(frac baright)$ and $Fleft(frac abright)=1implies a=b$.
$endgroup$
– lulu
Jan 15 at 17:31
$begingroup$
Log seems to be a perfect solution. Thank you @lulu!
$endgroup$
– Dough
Jan 15 at 17:34
$begingroup$
Log seems to be a perfect solution. Thank you @lulu!
$endgroup$
– Dough
Jan 15 at 17:34
$begingroup$
No problem. Note: There's a typo in my comment. I meant to write that $Fleft(frac abright)=0 implies a=b$
$endgroup$
– lulu
Jan 15 at 17:38
$begingroup$
No problem. Note: There's a typo in my comment. I meant to write that $Fleft(frac abright)=0 implies a=b$
$endgroup$
– lulu
Jan 15 at 17:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As noted in the comments:
The function $$Fleft(frac abright)=|log frac ab|=|log a - log b|$$
gets the job done.
It is easy to see that $$Fleft(frac abright)=Fleft(frac baright)$$ and that $$Fleft(frac abright)=0iff a= b$$ which appear to be the two desired properties.
answered Jan 15 at 17:37
lulululu
43.3k25080
$endgroup$
2
$begingroup$
... and $(log a-log b)^2$ is perhaps smoother (if that is desirable)
$endgroup$
– Hagen von Eitzen
Jan 15 at 17:38
add a comment |
$begingroup$
Perhaps simpler than logarithms would be to define the injustice to be the larger of $a/b$ and $b/a$.
answered Jan 15 at 20:35
Andreas BlassAndreas Blass
50.4k452109
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
As noted in the comments:
The function $$Fleft(frac abright)=|log frac ab|=|log a - log b|$$
gets the job done.
It is easy to see that $$Fleft(frac abright)=Fleft(frac baright)$$ and that $$Fleft(frac abright)=0iff a= b$$ which appear to be the two desired properties.
answered Jan 15 at 17:37
lulululu
43.3k25080
$endgroup$
2
$begingroup$
... and $(log a-log b)^2$ is perhaps smoother (if that is desirable)
$endgroup$
– Hagen von Eitzen
Jan 15 at 17:38
add a comment |
$begingroup$
As noted in the comments:
The function $$Fleft(frac abright)=|log frac ab|=|log a - log b|$$
gets the job done.
It is easy to see that $$Fleft(frac abright)=Fleft(frac baright)$$ and that $$Fleft(frac abright)=0iff a= b$$ which appear to be the two desired properties.
answered Jan 15 at 17:37
lulululu
43.3k25080
$endgroup$
2
$begingroup$
... and $(log a-log b)^2$ is perhaps smoother (if that is desirable)
$endgroup$
– Hagen von Eitzen
Jan 15 at 17:38
add a comment |
$begingroup$
As noted in the comments:
The function $$Fleft(frac abright)=|log frac ab|=|log a - log b|$$
gets the job done.
It is easy to see that $$Fleft(frac abright)=Fleft(frac baright)$$ and that $$Fleft(frac abright)=0iff a= b$$ which appear to be the two desired properties.
answered Jan 15 at 17:37
lulululu
43.3k25080
$endgroup$
As noted in the comments:
The function $$Fleft(frac abright)=|log frac ab|=|log a - log b|$$
gets the job done.
It is easy to see that $$Fleft(frac abright)=Fleft(frac baright)$$ and that $$Fleft(frac abright)=0iff a= b$$ which appear to be the two desired properties.
answered Jan 15 at 17:37
lulululu
43.3k25080
answered Jan 15 at 17:37
lulululu
43.3k25080
answered Jan 15 at 17:37
lulululu
43.3k25080
answered Jan 15 at 17:37
lulululu
43.3k25080
43.3k25080
2
$begingroup$
... and $(log a-log b)^2$ is perhaps smoother (if that is desirable)
$endgroup$
– Hagen von Eitzen
Jan 15 at 17:38
add a comment |
2
$begingroup$
... and $(log a-log b)^2$ is perhaps smoother (if that is desirable)
$endgroup$
– Hagen von Eitzen
Jan 15 at 17:38
2
2
$begingroup$
... and $(log a-log b)^2$ is perhaps smoother (if that is desirable)
$endgroup$
– Hagen von Eitzen
Jan 15 at 17:38
$begingroup$
... and $(log a-log b)^2$ is perhaps smoother (if that is desirable)
$endgroup$
– Hagen von Eitzen
Jan 15 at 17:38
add a comment |
$begingroup$
Perhaps simpler than logarithms would be to define the injustice to be the larger of $a/b$ and $b/a$.
answered Jan 15 at 20:35
Andreas BlassAndreas Blass
50.4k452109
$endgroup$
add a comment |
$begingroup$
Perhaps simpler than logarithms would be to define the injustice to be the larger of $a/b$ and $b/a$.
answered Jan 15 at 20:35
Andreas BlassAndreas Blass
50.4k452109
$endgroup$
add a comment |
$begingroup$
Perhaps simpler than logarithms would be to define the injustice to be the larger of $a/b$ and $b/a$.
answered Jan 15 at 20:35
Andreas BlassAndreas Blass
50.4k452109
$endgroup$
Perhaps simpler than logarithms would be to define the injustice to be the larger of $a/b$ and $b/a$.
answered Jan 15 at 20:35
Andreas BlassAndreas Blass
50.4k452109
answered Jan 15 at 20:35
Andreas BlassAndreas Blass
50.4k452109
answered Jan 15 at 20:35
Andreas BlassAndreas Blass
50.4k452109
answered Jan 15 at 20:35
Andreas BlassAndreas Blass
50.4k452109
50.4k452109
add a comment |
add a comment |
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2
$begingroup$
Not entirely sure what you are trying to do, but $Fleft(frac abright)=|log frac ab|=|log a - log b|$ might be what you are after. Easy to see that $Fleft(frac abright)=Fleft(frac baright)$ and $Fleft(frac abright)=1implies a=b$.
$endgroup$
– lulu
Jan 15 at 17:31
$begingroup$
Log seems to be a perfect solution. Thank you @lulu!
$endgroup$
– Dough
Jan 15 at 17:34
$begingroup$
No problem. Note: There's a typo in my comment. I meant to write that $Fleft(frac abright)=0 implies a=b$
$endgroup$
– lulu
Jan 15 at 17:38