Is the any math function that change the power of denominator of an input fraction?












0












$begingroup$


I'm seeking for a function that gets a fraction as input and change its denominator power in the log function like follow:



$f(frac{a}{b}) = log(frac{a}{b^k})$



Is it possible to find a function like f?



p.s: The input of function is not a and b, it's $frac{a}{b}$ and I need the definition of function, like $f(r) = r^2$.






functions fractions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Such a function does not appear to be well-defined (as a function of a single variable). The problem is that two different representations of the same fraction can result in different answers. You could define it as a function of two variables, however.
    $endgroup$
    – Michael Burr
    Dec 13 '18 at 14:05
















0












$begingroup$


I'm seeking for a function that gets a fraction as input and change its denominator power in the log function like follow:



$f(frac{a}{b}) = log(frac{a}{b^k})$



Is it possible to find a function like f?



p.s: The input of function is not a and b, it's $frac{a}{b}$ and I need the definition of function, like $f(r) = r^2$.






functions fractions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Such a function does not appear to be well-defined (as a function of a single variable). The problem is that two different representations of the same fraction can result in different answers. You could define it as a function of two variables, however.
    $endgroup$
    – Michael Burr
    Dec 13 '18 at 14:05














0












0








0





$begingroup$


I'm seeking for a function that gets a fraction as input and change its denominator power in the log function like follow:



$f(frac{a}{b}) = log(frac{a}{b^k})$



Is it possible to find a function like f?



p.s: The input of function is not a and b, it's $frac{a}{b}$ and I need the definition of function, like $f(r) = r^2$.






functions fractions







share|cite|improve this question











$endgroup$




I'm seeking for a function that gets a fraction as input and change its denominator power in the log function like follow:



$f(frac{a}{b}) = log(frac{a}{b^k})$



Is it possible to find a function like f?



p.s: The input of function is not a and b, it's $frac{a}{b}$ and I need the definition of function, like $f(r) = r^2$.






functions fractions




functions fractions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 14:30







user137927

















asked Dec 13 '18 at 14:01









user137927user137927

14318




14318












  • $begingroup$
    Such a function does not appear to be well-defined (as a function of a single variable). The problem is that two different representations of the same fraction can result in different answers. You could define it as a function of two variables, however.
    $endgroup$
    – Michael Burr
    Dec 13 '18 at 14:05


















  • $begingroup$
    Such a function does not appear to be well-defined (as a function of a single variable). The problem is that two different representations of the same fraction can result in different answers. You could define it as a function of two variables, however.
    $endgroup$
    – Michael Burr
    Dec 13 '18 at 14:05
















$begingroup$
Such a function does not appear to be well-defined (as a function of a single variable). The problem is that two different representations of the same fraction can result in different answers. You could define it as a function of two variables, however.
$endgroup$
– Michael Burr
Dec 13 '18 at 14:05




$begingroup$
Such a function does not appear to be well-defined (as a function of a single variable). The problem is that two different representations of the same fraction can result in different answers. You could define it as a function of two variables, however.
$endgroup$
– Michael Burr
Dec 13 '18 at 14:05










3 Answers
3






active

oldest

votes


















1












$begingroup$

You can define a function as follows. Let $f(a/b) := log(a/b^2)$ where $a$ and $b$ are relatively prime positive integers. The function $f$ "maps" positive rational numbers (i.e. fractions) to real numbers. Note that $f$ is not smooth or continuous.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    When $a$ and $b$ are not relatively prime, you can divide them by their lcd prior to calculating $a/b^2$
    $endgroup$
    – Damien
    Dec 13 '18 at 14:31



















0












$begingroup$

If the input is not $a$ and $b$ but a fraction $r$, you can use continued fractions to find the corresponding $a$ and $b$ such that $r = a/b$, where $a$ and $b$ are relatively prime, and then apply your formula. But as stated by @irchans the obtained function is not continuous






share|cite|improve this answer









$endgroup$













  • $begingroup$
    When they are not relatively prime and we don't know the value of a and b, it's not possible to make them relatively prime by dividing them by their lcd.
    $endgroup$
    – user137927
    Dec 13 '18 at 14:48










  • $begingroup$
    @user137927 The continued fractions algorithm provides a pair $(a, b)$ where $a$ and $b$ are relatively prime. If it is not the case, for example because you use another algorithm to get $a$ and $b$, we can then use the obtained values to calculate their lcd, and then obtain new values which are relatively prime
    $endgroup$
    – Damien
    Dec 13 '18 at 15:00



















-1












$begingroup$

Many functions (taking $a$ and $b$ as inputs) can be manufactured. How about $$f(a,b) = log(frac{a}{b})+ sum_{n=1}^{k-1}log(frac{1}{b})$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The input is not a and b, it's their fraction, and that's the problem.
    $endgroup$
    – user137927
    Dec 13 '18 at 14:31











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You can define a function as follows. Let $f(a/b) := log(a/b^2)$ where $a$ and $b$ are relatively prime positive integers. The function $f$ "maps" positive rational numbers (i.e. fractions) to real numbers. Note that $f$ is not smooth or continuous.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    When $a$ and $b$ are not relatively prime, you can divide them by their lcd prior to calculating $a/b^2$
    $endgroup$
    – Damien
    Dec 13 '18 at 14:31
















1












$begingroup$

You can define a function as follows. Let $f(a/b) := log(a/b^2)$ where $a$ and $b$ are relatively prime positive integers. The function $f$ "maps" positive rational numbers (i.e. fractions) to real numbers. Note that $f$ is not smooth or continuous.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    When $a$ and $b$ are not relatively prime, you can divide them by their lcd prior to calculating $a/b^2$
    $endgroup$
    – Damien
    Dec 13 '18 at 14:31














1












1








1





$begingroup$

You can define a function as follows. Let $f(a/b) := log(a/b^2)$ where $a$ and $b$ are relatively prime positive integers. The function $f$ "maps" positive rational numbers (i.e. fractions) to real numbers. Note that $f$ is not smooth or continuous.






share|cite|improve this answer









$endgroup$



You can define a function as follows. Let $f(a/b) := log(a/b^2)$ where $a$ and $b$ are relatively prime positive integers. The function $f$ "maps" positive rational numbers (i.e. fractions) to real numbers. Note that $f$ is not smooth or continuous.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 14:18









irchansirchans

1,11239




1,11239








  • 1




    $begingroup$
    When $a$ and $b$ are not relatively prime, you can divide them by their lcd prior to calculating $a/b^2$
    $endgroup$
    – Damien
    Dec 13 '18 at 14:31














  • 1




    $begingroup$
    When $a$ and $b$ are not relatively prime, you can divide them by their lcd prior to calculating $a/b^2$
    $endgroup$
    – Damien
    Dec 13 '18 at 14:31








1




1




$begingroup$
When $a$ and $b$ are not relatively prime, you can divide them by their lcd prior to calculating $a/b^2$
$endgroup$
– Damien
Dec 13 '18 at 14:31




$begingroup$
When $a$ and $b$ are not relatively prime, you can divide them by their lcd prior to calculating $a/b^2$
$endgroup$
– Damien
Dec 13 '18 at 14:31











0












$begingroup$

If the input is not $a$ and $b$ but a fraction $r$, you can use continued fractions to find the corresponding $a$ and $b$ such that $r = a/b$, where $a$ and $b$ are relatively prime, and then apply your formula. But as stated by @irchans the obtained function is not continuous






share|cite|improve this answer









$endgroup$













  • $begingroup$
    When they are not relatively prime and we don't know the value of a and b, it's not possible to make them relatively prime by dividing them by their lcd.
    $endgroup$
    – user137927
    Dec 13 '18 at 14:48










  • $begingroup$
    @user137927 The continued fractions algorithm provides a pair $(a, b)$ where $a$ and $b$ are relatively prime. If it is not the case, for example because you use another algorithm to get $a$ and $b$, we can then use the obtained values to calculate their lcd, and then obtain new values which are relatively prime
    $endgroup$
    – Damien
    Dec 13 '18 at 15:00
















0












$begingroup$

If the input is not $a$ and $b$ but a fraction $r$, you can use continued fractions to find the corresponding $a$ and $b$ such that $r = a/b$, where $a$ and $b$ are relatively prime, and then apply your formula. But as stated by @irchans the obtained function is not continuous






share|cite|improve this answer









$endgroup$













  • $begingroup$
    When they are not relatively prime and we don't know the value of a and b, it's not possible to make them relatively prime by dividing them by their lcd.
    $endgroup$
    – user137927
    Dec 13 '18 at 14:48










  • $begingroup$
    @user137927 The continued fractions algorithm provides a pair $(a, b)$ where $a$ and $b$ are relatively prime. If it is not the case, for example because you use another algorithm to get $a$ and $b$, we can then use the obtained values to calculate their lcd, and then obtain new values which are relatively prime
    $endgroup$
    – Damien
    Dec 13 '18 at 15:00














0












0








0





$begingroup$

If the input is not $a$ and $b$ but a fraction $r$, you can use continued fractions to find the corresponding $a$ and $b$ such that $r = a/b$, where $a$ and $b$ are relatively prime, and then apply your formula. But as stated by @irchans the obtained function is not continuous






share|cite|improve this answer









$endgroup$



If the input is not $a$ and $b$ but a fraction $r$, you can use continued fractions to find the corresponding $a$ and $b$ such that $r = a/b$, where $a$ and $b$ are relatively prime, and then apply your formula. But as stated by @irchans the obtained function is not continuous







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 14:36









DamienDamien

59714




59714












  • $begingroup$
    When they are not relatively prime and we don't know the value of a and b, it's not possible to make them relatively prime by dividing them by their lcd.
    $endgroup$
    – user137927
    Dec 13 '18 at 14:48










  • $begingroup$
    @user137927 The continued fractions algorithm provides a pair $(a, b)$ where $a$ and $b$ are relatively prime. If it is not the case, for example because you use another algorithm to get $a$ and $b$, we can then use the obtained values to calculate their lcd, and then obtain new values which are relatively prime
    $endgroup$
    – Damien
    Dec 13 '18 at 15:00


















  • $begingroup$
    When they are not relatively prime and we don't know the value of a and b, it's not possible to make them relatively prime by dividing them by their lcd.
    $endgroup$
    – user137927
    Dec 13 '18 at 14:48










  • $begingroup$
    @user137927 The continued fractions algorithm provides a pair $(a, b)$ where $a$ and $b$ are relatively prime. If it is not the case, for example because you use another algorithm to get $a$ and $b$, we can then use the obtained values to calculate their lcd, and then obtain new values which are relatively prime
    $endgroup$
    – Damien
    Dec 13 '18 at 15:00
















$begingroup$
When they are not relatively prime and we don't know the value of a and b, it's not possible to make them relatively prime by dividing them by their lcd.
$endgroup$
– user137927
Dec 13 '18 at 14:48




$begingroup$
When they are not relatively prime and we don't know the value of a and b, it's not possible to make them relatively prime by dividing them by their lcd.
$endgroup$
– user137927
Dec 13 '18 at 14:48












$begingroup$
@user137927 The continued fractions algorithm provides a pair $(a, b)$ where $a$ and $b$ are relatively prime. If it is not the case, for example because you use another algorithm to get $a$ and $b$, we can then use the obtained values to calculate their lcd, and then obtain new values which are relatively prime
$endgroup$
– Damien
Dec 13 '18 at 15:00




$begingroup$
@user137927 The continued fractions algorithm provides a pair $(a, b)$ where $a$ and $b$ are relatively prime. If it is not the case, for example because you use another algorithm to get $a$ and $b$, we can then use the obtained values to calculate their lcd, and then obtain new values which are relatively prime
$endgroup$
– Damien
Dec 13 '18 at 15:00











-1












$begingroup$

Many functions (taking $a$ and $b$ as inputs) can be manufactured. How about $$f(a,b) = log(frac{a}{b})+ sum_{n=1}^{k-1}log(frac{1}{b})$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The input is not a and b, it's their fraction, and that's the problem.
    $endgroup$
    – user137927
    Dec 13 '18 at 14:31
















-1












$begingroup$

Many functions (taking $a$ and $b$ as inputs) can be manufactured. How about $$f(a,b) = log(frac{a}{b})+ sum_{n=1}^{k-1}log(frac{1}{b})$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The input is not a and b, it's their fraction, and that's the problem.
    $endgroup$
    – user137927
    Dec 13 '18 at 14:31














-1












-1








-1





$begingroup$

Many functions (taking $a$ and $b$ as inputs) can be manufactured. How about $$f(a,b) = log(frac{a}{b})+ sum_{n=1}^{k-1}log(frac{1}{b})$$






share|cite|improve this answer









$endgroup$



Many functions (taking $a$ and $b$ as inputs) can be manufactured. How about $$f(a,b) = log(frac{a}{b})+ sum_{n=1}^{k-1}log(frac{1}{b})$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 14:07









NazimJNazimJ

3067




3067












  • $begingroup$
    The input is not a and b, it's their fraction, and that's the problem.
    $endgroup$
    – user137927
    Dec 13 '18 at 14:31


















  • $begingroup$
    The input is not a and b, it's their fraction, and that's the problem.
    $endgroup$
    – user137927
    Dec 13 '18 at 14:31
















$begingroup$
The input is not a and b, it's their fraction, and that's the problem.
$endgroup$
– user137927
Dec 13 '18 at 14:31




$begingroup$
The input is not a and b, it's their fraction, and that's the problem.
$endgroup$
– user137927
Dec 13 '18 at 14:31


















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