Commuting floor functions
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Positive irrational numbers $a,b>0$ are such that $lfloor alfloor bxrfloorrfloor = lfloor blfloor axrfloorrfloor$ for all $x>0$. Must it be that $a=b$?
If $a$ and $b$ are allowed to be rational, this is not always true. For example, we can take $a=1$ and $b=frac12$. It is true that $lfloor x/2rfloor = lfloorlfloor xrfloor/2rfloor$ for any real number $x>0$, as when $2nleq x<2n+2$ for some nonnegative integer $n$, both sides evaluate to $n$.
real-analysis algebra-precalculus
asked Sep 8 at 13:24
pi66
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up vote
3
down vote
favorite
Positive irrational numbers $a,b>0$ are such that $lfloor alfloor bxrfloorrfloor = lfloor blfloor axrfloorrfloor$ for all $x>0$. Must it be that $a=b$?
If $a$ and $b$ are allowed to be rational, this is not always true. For example, we can take $a=1$ and $b=frac12$. It is true that $lfloor x/2rfloor = lfloorlfloor xrfloor/2rfloor$ for any real number $x>0$, as when $2nleq x<2n+2$ for some nonnegative integer $n$, both sides evaluate to $n$.
real-analysis algebra-precalculus
asked Sep 8 at 13:24
pi66
3,9291037
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Positive irrational numbers $a,b>0$ are such that $lfloor alfloor bxrfloorrfloor = lfloor blfloor axrfloorrfloor$ for all $x>0$. Must it be that $a=b$?
If $a$ and $b$ are allowed to be rational, this is not always true. For example, we can take $a=1$ and $b=frac12$. It is true that $lfloor x/2rfloor = lfloorlfloor xrfloor/2rfloor$ for any real number $x>0$, as when $2nleq x<2n+2$ for some nonnegative integer $n$, both sides evaluate to $n$.
real-analysis algebra-precalculus
asked Sep 8 at 13:24
pi66
3,9291037
Positive irrational numbers $a,b>0$ are such that $lfloor alfloor bxrfloorrfloor = lfloor blfloor axrfloorrfloor$ for all $x>0$. Must it be that $a=b$?
If $a$ and $b$ are allowed to be rational, this is not always true. For example, we can take $a=1$ and $b=frac12$. It is true that $lfloor x/2rfloor = lfloorlfloor xrfloor/2rfloor$ for any real number $x>0$, as when $2nleq x<2n+2$ for some nonnegative integer $n$, both sides evaluate to $n$.
real-analysis algebra-precalculus
real-analysis algebra-precalculus
asked Sep 8 at 13:24
pi66
3,9291037
asked Sep 8 at 13:24
pi66
3,9291037
edited Sep 9 at 1:34
asked Sep 8 at 13:24
pi66
3,9291037
asked Sep 8 at 13:24
pi66
3,9291037
asked Sep 8 at 13:24
pi66
3,9291037
3,9291037
add a comment |Â
add a comment |Â
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