Prove that $f(n) = [sqrt{n}] - [sqrt{n-1}]$ is multiplicative.
It is about a day i am working on this question with no success. $f(n) = [sqrt{n}] - [sqrt{n-1}]$ is not completely multiplicative example is n=20, m=80. [x] is integer part of x. Thanks in advance for reading and helping out.
for f to be multiplicative, gcd(m,n)=1 is must.
number-theory
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It is about a day i am working on this question with no success. $f(n) = [sqrt{n}] - [sqrt{n-1}]$ is not completely multiplicative example is n=20, m=80. [x] is integer part of x. Thanks in advance for reading and helping out.
for f to be multiplicative, gcd(m,n)=1 is must.
number-theory
asked 2 days ago
A Schizotypal AngelA Schizotypal Angel
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It is about a day i am working on this question with no success. $f(n) = [sqrt{n}] - [sqrt{n-1}]$ is not completely multiplicative example is n=20, m=80. [x] is integer part of x. Thanks in advance for reading and helping out.
for f to be multiplicative, gcd(m,n)=1 is must.
number-theory
asked 2 days ago
A Schizotypal AngelA Schizotypal Angel
143
New contributor
A Schizotypal Angel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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It is about a day i am working on this question with no success. $f(n) = [sqrt{n}] - [sqrt{n-1}]$ is not completely multiplicative example is n=20, m=80. [x] is integer part of x. Thanks in advance for reading and helping out.
for f to be multiplicative, gcd(m,n)=1 is must.
number-theory
number-theory
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A Schizotypal AngelA Schizotypal Angel
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edited 2 days ago
A Schizotypal Angel
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Note that
$$f(n)=begin{cases}1&text{if $n$ is a perfect square}\0&text{otherwise}end{cases} $$
This is so because if $n=k^2$, then $lfloorsqrt nrfloor =k$ and $lfloor sqrt{n-1}rfloor=k-1$, wheras for $n$ with $k^2<n<(k+1)^2$, we have $lfloorsqrt nrfloor =k=lfloor sqrt{n-1}rfloor$.
Now assume $n,m$ are coprime. Then $nm$ is a perfect square iff both $n$ and $m$ are. Hence $f$ is multiplicative.
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
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1 Answer
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1 Answer
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Note that
$$f(n)=begin{cases}1&text{if $n$ is a perfect square}\0&text{otherwise}end{cases} $$
This is so because if $n=k^2$, then $lfloorsqrt nrfloor =k$ and $lfloor sqrt{n-1}rfloor=k-1$, wheras for $n$ with $k^2<n<(k+1)^2$, we have $lfloorsqrt nrfloor =k=lfloor sqrt{n-1}rfloor$.
Now assume $n,m$ are coprime. Then $nm$ is a perfect square iff both $n$ and $m$ are. Hence $f$ is multiplicative.
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
add a comment |
Note that
$$f(n)=begin{cases}1&text{if $n$ is a perfect square}\0&text{otherwise}end{cases} $$
This is so because if $n=k^2$, then $lfloorsqrt nrfloor =k$ and $lfloor sqrt{n-1}rfloor=k-1$, wheras for $n$ with $k^2<n<(k+1)^2$, we have $lfloorsqrt nrfloor =k=lfloor sqrt{n-1}rfloor$.
Now assume $n,m$ are coprime. Then $nm$ is a perfect square iff both $n$ and $m$ are. Hence $f$ is multiplicative.
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
add a comment |
Note that
$$f(n)=begin{cases}1&text{if $n$ is a perfect square}\0&text{otherwise}end{cases} $$
This is so because if $n=k^2$, then $lfloorsqrt nrfloor =k$ and $lfloor sqrt{n-1}rfloor=k-1$, wheras for $n$ with $k^2<n<(k+1)^2$, we have $lfloorsqrt nrfloor =k=lfloor sqrt{n-1}rfloor$.
Now assume $n,m$ are coprime. Then $nm$ is a perfect square iff both $n$ and $m$ are. Hence $f$ is multiplicative.
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
Note that
$$f(n)=begin{cases}1&text{if $n$ is a perfect square}\0&text{otherwise}end{cases} $$
This is so because if $n=k^2$, then $lfloorsqrt nrfloor =k$ and $lfloor sqrt{n-1}rfloor=k-1$, wheras for $n$ with $k^2<n<(k+1)^2$, we have $lfloorsqrt nrfloor =k=lfloor sqrt{n-1}rfloor$.
Now assume $n,m$ are coprime. Then $nm$ is a perfect square iff both $n$ and $m$ are. Hence $f$ is multiplicative.
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
edited 2 days ago
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
answered 2 days ago
Hagen von EitzenHagen von Eitzen
276k21269496
276k21269496
add a comment |
add a comment |
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