Find all functions $f:mathbb{Q} rightarrow mathbb{Q}$ such that $f(x+1)=f(x)+1$ [closed]
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The function $f(x)=x+c$ seems trivial but I thought whether it is the only function.
I tried to used induction but I'm not sure how to do it for $mathbb{Q}$.
functions
asked Dec 14 '18 at 14:43
att eplatt epl
12910
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closed as unclear what you're asking by Nosrati, Did, Paul Frost, Rebellos, Leucippus Dec 15 '18 at 0:56
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
The function $f(x)=x+c$ seems trivial but I thought whether it is the only function.
I tried to used induction but I'm not sure how to do it for $mathbb{Q}$.
functions
asked Dec 14 '18 at 14:43
att eplatt epl
12910
$endgroup$
closed as unclear what you're asking by Nosrati, Did, Paul Frost, Rebellos, Leucippus Dec 15 '18 at 0:56
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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$f(x)=x+a$ also works (for all $ainmathbb{Q}$). Unless I'm wrong these should be the only solutions
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– Yanko
Dec 14 '18 at 14:44
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In a second thought. It may have more solutions, for instance you can take $f(x)=x$ for all $xinmathbb{Z}$ and $f(x)=x+1$ otherwise. It also satisfies this equation.
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– Yanko
Dec 14 '18 at 14:46
3
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Any function defined on $[0,1]capmathbb{Q}$ can be extended to whole $mathbb{Q}$ in a way that satisfies your equality.
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– freakish
Dec 14 '18 at 14:49
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Are there any non linear polynomial solutions ?
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– att epl
Dec 14 '18 at 14:50
1
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@freakish: that should be $[0,1)capBbb Q$.
$endgroup$
– TonyK
Dec 14 '18 at 14:53
|
show 3 more comments
$begingroup$
The function $f(x)=x+c$ seems trivial but I thought whether it is the only function.
I tried to used induction but I'm not sure how to do it for $mathbb{Q}$.
functions
asked Dec 14 '18 at 14:43
att eplatt epl
12910
$endgroup$
The function $f(x)=x+c$ seems trivial but I thought whether it is the only function.
I tried to used induction but I'm not sure how to do it for $mathbb{Q}$.
functions
functions
asked Dec 14 '18 at 14:43
att eplatt epl
12910
asked Dec 14 '18 at 14:43
att eplatt epl
12910
edited Dec 25 '18 at 2:12
att epl
asked Dec 14 '18 at 14:43
att eplatt epl
12910
asked Dec 14 '18 at 14:43
att eplatt epl
12910
asked Dec 14 '18 at 14:43
att eplatt epl
12910
12910
closed as unclear what you're asking by Nosrati, Did, Paul Frost, Rebellos, Leucippus Dec 15 '18 at 0:56
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Nosrati, Did, Paul Frost, Rebellos, Leucippus Dec 15 '18 at 0:56
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
$f(x)=x+a$ also works (for all $ainmathbb{Q}$). Unless I'm wrong these should be the only solutions
$endgroup$
– Yanko
Dec 14 '18 at 14:44
$begingroup$
In a second thought. It may have more solutions, for instance you can take $f(x)=x$ for all $xinmathbb{Z}$ and $f(x)=x+1$ otherwise. It also satisfies this equation.
$endgroup$
– Yanko
Dec 14 '18 at 14:46
3
$begingroup$
Any function defined on $[0,1]capmathbb{Q}$ can be extended to whole $mathbb{Q}$ in a way that satisfies your equality.
$endgroup$
– freakish
Dec 14 '18 at 14:49
$begingroup$
Are there any non linear polynomial solutions ?
$endgroup$
– att epl
Dec 14 '18 at 14:50
1
$begingroup$
@freakish: that should be $[0,1)capBbb Q$.
$endgroup$
– TonyK
Dec 14 '18 at 14:53
|
show 3 more comments
$begingroup$
$f(x)=x+a$ also works (for all $ainmathbb{Q}$). Unless I'm wrong these should be the only solutions
$endgroup$
– Yanko
Dec 14 '18 at 14:44
$begingroup$
In a second thought. It may have more solutions, for instance you can take $f(x)=x$ for all $xinmathbb{Z}$ and $f(x)=x+1$ otherwise. It also satisfies this equation.
$endgroup$
– Yanko
Dec 14 '18 at 14:46
3
$begingroup$
Any function defined on $[0,1]capmathbb{Q}$ can be extended to whole $mathbb{Q}$ in a way that satisfies your equality.
$endgroup$
– freakish
Dec 14 '18 at 14:49
$begingroup$
Are there any non linear polynomial solutions ?
$endgroup$
– att epl
Dec 14 '18 at 14:50
1
$begingroup$
@freakish: that should be $[0,1)capBbb Q$.
$endgroup$
– TonyK
Dec 14 '18 at 14:53
$begingroup$
$f(x)=x+a$ also works (for all $ainmathbb{Q}$). Unless I'm wrong these should be the only solutions
$endgroup$
– Yanko
Dec 14 '18 at 14:44
$begingroup$
$f(x)=x+a$ also works (for all $ainmathbb{Q}$). Unless I'm wrong these should be the only solutions
$endgroup$
– Yanko
Dec 14 '18 at 14:44
$begingroup$
In a second thought. It may have more solutions, for instance you can take $f(x)=x$ for all $xinmathbb{Z}$ and $f(x)=x+1$ otherwise. It also satisfies this equation.
$endgroup$
– Yanko
Dec 14 '18 at 14:46
$begingroup$
In a second thought. It may have more solutions, for instance you can take $f(x)=x$ for all $xinmathbb{Z}$ and $f(x)=x+1$ otherwise. It also satisfies this equation.
$endgroup$
– Yanko
Dec 14 '18 at 14:46
3
3
$begingroup$
Any function defined on $[0,1]capmathbb{Q}$ can be extended to whole $mathbb{Q}$ in a way that satisfies your equality.
$endgroup$
– freakish
Dec 14 '18 at 14:49
$begingroup$
Any function defined on $[0,1]capmathbb{Q}$ can be extended to whole $mathbb{Q}$ in a way that satisfies your equality.
$endgroup$
– freakish
Dec 14 '18 at 14:49
$begingroup$
Are there any non linear polynomial solutions ?
$endgroup$
– att epl
Dec 14 '18 at 14:50
$begingroup$
Are there any non linear polynomial solutions ?
$endgroup$
– att epl
Dec 14 '18 at 14:50
1
1
$begingroup$
@freakish: that should be $[0,1)capBbb Q$.
$endgroup$
– TonyK
Dec 14 '18 at 14:53
$begingroup$
@freakish: that should be $[0,1)capBbb Q$.
$endgroup$
– TonyK
Dec 14 '18 at 14:53
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Given any function $g:Bbb Qcap[0,1)toBbb Q$, the function
$$f(x)=lfloor xrfloor + g({x})$$
satisfies your condition. (Here $lfloor xrfloor$ is the integral part of $x$ and ${x}$ is the fractional part of $x$.)
And indeed any $f$ that satisfies your condition must be of this form.
answered Dec 14 '18 at 14:49
TonyKTonyK
42.9k356135
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given any function $g:Bbb Qcap[0,1)toBbb Q$, the function
$$f(x)=lfloor xrfloor + g({x})$$
satisfies your condition. (Here $lfloor xrfloor$ is the integral part of $x$ and ${x}$ is the fractional part of $x$.)
And indeed any $f$ that satisfies your condition must be of this form.
answered Dec 14 '18 at 14:49
TonyKTonyK
42.9k356135
$endgroup$
add a comment |
$begingroup$
Given any function $g:Bbb Qcap[0,1)toBbb Q$, the function
$$f(x)=lfloor xrfloor + g({x})$$
satisfies your condition. (Here $lfloor xrfloor$ is the integral part of $x$ and ${x}$ is the fractional part of $x$.)
And indeed any $f$ that satisfies your condition must be of this form.
answered Dec 14 '18 at 14:49
TonyKTonyK
42.9k356135
$endgroup$
add a comment |
$begingroup$
Given any function $g:Bbb Qcap[0,1)toBbb Q$, the function
$$f(x)=lfloor xrfloor + g({x})$$
satisfies your condition. (Here $lfloor xrfloor$ is the integral part of $x$ and ${x}$ is the fractional part of $x$.)
And indeed any $f$ that satisfies your condition must be of this form.
answered Dec 14 '18 at 14:49
TonyKTonyK
42.9k356135
$endgroup$
Given any function $g:Bbb Qcap[0,1)toBbb Q$, the function
$$f(x)=lfloor xrfloor + g({x})$$
satisfies your condition. (Here $lfloor xrfloor$ is the integral part of $x$ and ${x}$ is the fractional part of $x$.)
And indeed any $f$ that satisfies your condition must be of this form.
answered Dec 14 '18 at 14:49
TonyKTonyK
42.9k356135
answered Dec 14 '18 at 14:49
TonyKTonyK
42.9k356135
answered Dec 14 '18 at 14:49
TonyKTonyK
42.9k356135
answered Dec 14 '18 at 14:49
TonyKTonyK
42.9k356135
42.9k356135
add a comment |
add a comment |
$begingroup$
$f(x)=x+a$ also works (for all $ainmathbb{Q}$). Unless I'm wrong these should be the only solutions
$endgroup$
– Yanko
Dec 14 '18 at 14:44
$begingroup$
In a second thought. It may have more solutions, for instance you can take $f(x)=x$ for all $xinmathbb{Z}$ and $f(x)=x+1$ otherwise. It also satisfies this equation.
$endgroup$
– Yanko
Dec 14 '18 at 14:46
3
$begingroup$
Any function defined on $[0,1]capmathbb{Q}$ can be extended to whole $mathbb{Q}$ in a way that satisfies your equality.
$endgroup$
– freakish
Dec 14 '18 at 14:49
$begingroup$
Are there any non linear polynomial solutions ?
$endgroup$
– att epl
Dec 14 '18 at 14:50
1
$begingroup$
@freakish: that should be $[0,1)capBbb Q$.
$endgroup$
– TonyK
Dec 14 '18 at 14:53