Mixing
How to show that $lfloor n/1rfloor+lfloor n/2 rfloor+…+lfloor n/nrfloor+lfloor{sqrt{n}}rfloor$ is even?
$begingroup$
Let $n$ is a natural number. Prove that $$leftlfloorfrac{n}{1}rightrfloor+leftlfloorfrac{n}{2}rightrfloor+....+leftlfloorfrac{n}{n}rightrfloor+leftlfloor{sqrt{n}}rightrfloor$$
is even.
I used induction and the inequality $$x-1<lfloor{x}rfloorle{x}$$ to prove it.
Is there any other, nicer way to do it?
elementary-number-theory alternative-proof
edited Jun 19 '14 at 11:51
Martin Sleziak
44.9k10121274
asked Feb 2 '14 at 18:07
MathGodMathGod
4,31912549
$endgroup$
add a comment |
$begingroup$
Let $n$ is a natural number. Prove that $$leftlfloorfrac{n}{1}rightrfloor+leftlfloorfrac{n}{2}rightrfloor+....+leftlfloorfrac{n}{n}rightrfloor+leftlfloor{sqrt{n}}rightrfloor$$
is even.
I used induction and the inequality $$x-1<lfloor{x}rfloorle{x}$$ to prove it.
Is there any other, nicer way to do it?
elementary-number-theory alternative-proof
edited Jun 19 '14 at 11:51
Martin Sleziak
44.9k10121274
asked Feb 2 '14 at 18:07
MathGodMathGod
4,31912549
$endgroup$
$begingroup$
Induction is a good way to go and the inequality you used is all we have about the floor function, so what could be a better way?
$endgroup$
– Hagen von Eitzen
Feb 2 '14 at 18:27
1
$begingroup$
A duplicate of this question was asked later. You might find my answer to that question useful.
$endgroup$
– David
Feb 18 '15 at 4:58
$begingroup$
@David +1. Truly remarkable answer by you in that post.
$endgroup$
– MathGod
Feb 18 '15 at 7:36
add a comment |
$begingroup$
Let $n$ is a natural number. Prove that $$leftlfloorfrac{n}{1}rightrfloor+leftlfloorfrac{n}{2}rightrfloor+....+leftlfloorfrac{n}{n}rightrfloor+leftlfloor{sqrt{n}}rightrfloor$$
is even.
I used induction and the inequality $$x-1<lfloor{x}rfloorle{x}$$ to prove it.
Is there any other, nicer way to do it?
elementary-number-theory alternative-proof
edited Jun 19 '14 at 11:51
Martin Sleziak
44.9k10121274
asked Feb 2 '14 at 18:07
MathGodMathGod
4,31912549
$endgroup$
Let $n$ is a natural number. Prove that $$leftlfloorfrac{n}{1}rightrfloor+leftlfloorfrac{n}{2}rightrfloor+....+leftlfloorfrac{n}{n}rightrfloor+leftlfloor{sqrt{n}}rightrfloor$$
is even.
I used induction and the inequality $$x-1<lfloor{x}rfloorle{x}$$ to prove it.
Is there any other, nicer way to do it?
elementary-number-theory alternative-proof
elementary-number-theory alternative-proof
edited Jun 19 '14 at 11:51
Martin Sleziak
44.9k10121274
asked Feb 2 '14 at 18:07
MathGodMathGod
4,31912549
edited Jun 19 '14 at 11:51
Martin Sleziak
44.9k10121274
asked Feb 2 '14 at 18:07
MathGodMathGod
4,31912549
edited Jun 19 '14 at 11:51
Martin Sleziak
44.9k10121274
edited Jun 19 '14 at 11:51
Martin Sleziak
44.9k10121274
edited Jun 19 '14 at 11:51
Martin Sleziak
44.9k10121274
44.9k10121274
asked Feb 2 '14 at 18:07
MathGodMathGod
4,31912549
asked Feb 2 '14 at 18:07
MathGodMathGod
4,31912549
asked Feb 2 '14 at 18:07
MathGodMathGod
4,31912549
4,31912549
$begingroup$
Induction is a good way to go and the inequality you used is all we have about the floor function, so what could be a better way?
$endgroup$
– Hagen von Eitzen
Feb 2 '14 at 18:27
1
$begingroup$
A duplicate of this question was asked later. You might find my answer to that question useful.
$endgroup$
– David
Feb 18 '15 at 4:58
$begingroup$
@David +1. Truly remarkable answer by you in that post.
$endgroup$
– MathGod
Feb 18 '15 at 7:36
add a comment |
$begingroup$
Induction is a good way to go and the inequality you used is all we have about the floor function, so what could be a better way?
$endgroup$
– Hagen von Eitzen
Feb 2 '14 at 18:27
1
$begingroup$
A duplicate of this question was asked later. You might find my answer to that question useful.
$endgroup$
– David
Feb 18 '15 at 4:58
$begingroup$
@David +1. Truly remarkable answer by you in that post.
$endgroup$
– MathGod
Feb 18 '15 at 7:36
$begingroup$
Induction is a good way to go and the inequality you used is all we have about the floor function, so what could be a better way?
$endgroup$
– Hagen von Eitzen
Feb 2 '14 at 18:27
$begingroup$
Induction is a good way to go and the inequality you used is all we have about the floor function, so what could be a better way?
$endgroup$
– Hagen von Eitzen
Feb 2 '14 at 18:27
1
1
$begingroup$
A duplicate of this question was asked later. You might find my answer to that question useful.
$endgroup$
– David
Feb 18 '15 at 4:58
$begingroup$
A duplicate of this question was asked later. You might find my answer to that question useful.
$endgroup$
– David
Feb 18 '15 at 4:58
$begingroup$
@David +1. Truly remarkable answer by you in that post.
$endgroup$
– MathGod
Feb 18 '15 at 7:36
$begingroup$
@David +1. Truly remarkable answer by you in that post.
$endgroup$
– MathGod
Feb 18 '15 at 7:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let
$$S = {,(a,b)inmathbb N^2mid able n,} $$
and $$ T={,(a,a)inmathbb N^2mid a^2le n,}.$$
Then
$$ |S|=sum_{bge 1}|{,ainmathbb Nmid aletfrac nb,}|=sum_{bge 1}lfloor tfrac nbrfloor =leftlfloor frac n1rightrfloor+leftlfloor frac n2rightrfloor+ldots +leftlfloor frac nnright rfloor$$
and $$ |T|=lfloor sqrt nrfloor.$$
The map $(a,b)mapsto (b,a)$ is a fixedpoint-free involution of $Ssetminus T$, hence
$|Ssetminus T|$ is even. Since
$$ leftlfloor frac n1rightrfloor+leftlfloor frac n2rightrfloor+ldots +leftlfloor frac nnright rfloor+lfloorsqrt nrfloor = |Ssetminus T|+2|T|$$
we are done.
edited Jan 26 at 1:27
darij grinberg
11.2k33167
answered Feb 2 '14 at 18:40
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let
$$S = {,(a,b)inmathbb N^2mid able n,} $$
and $$ T={,(a,a)inmathbb N^2mid a^2le n,}.$$
Then
$$ |S|=sum_{bge 1}|{,ainmathbb Nmid aletfrac nb,}|=sum_{bge 1}lfloor tfrac nbrfloor =leftlfloor frac n1rightrfloor+leftlfloor frac n2rightrfloor+ldots +leftlfloor frac nnright rfloor$$
and $$ |T|=lfloor sqrt nrfloor.$$
The map $(a,b)mapsto (b,a)$ is a fixedpoint-free involution of $Ssetminus T$, hence
$|Ssetminus T|$ is even. Since
$$ leftlfloor frac n1rightrfloor+leftlfloor frac n2rightrfloor+ldots +leftlfloor frac nnright rfloor+lfloorsqrt nrfloor = |Ssetminus T|+2|T|$$
we are done.
edited Jan 26 at 1:27
darij grinberg
11.2k33167
answered Feb 2 '14 at 18:40
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
add a comment |
$begingroup$
Let
$$S = {,(a,b)inmathbb N^2mid able n,} $$
and $$ T={,(a,a)inmathbb N^2mid a^2le n,}.$$
Then
$$ |S|=sum_{bge 1}|{,ainmathbb Nmid aletfrac nb,}|=sum_{bge 1}lfloor tfrac nbrfloor =leftlfloor frac n1rightrfloor+leftlfloor frac n2rightrfloor+ldots +leftlfloor frac nnright rfloor$$
and $$ |T|=lfloor sqrt nrfloor.$$
The map $(a,b)mapsto (b,a)$ is a fixedpoint-free involution of $Ssetminus T$, hence
$|Ssetminus T|$ is even. Since
$$ leftlfloor frac n1rightrfloor+leftlfloor frac n2rightrfloor+ldots +leftlfloor frac nnright rfloor+lfloorsqrt nrfloor = |Ssetminus T|+2|T|$$
we are done.
edited Jan 26 at 1:27
darij grinberg
11.2k33167
answered Feb 2 '14 at 18:40
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
add a comment |
$begingroup$
Let
$$S = {,(a,b)inmathbb N^2mid able n,} $$
and $$ T={,(a,a)inmathbb N^2mid a^2le n,}.$$
Then
$$ |S|=sum_{bge 1}|{,ainmathbb Nmid aletfrac nb,}|=sum_{bge 1}lfloor tfrac nbrfloor =leftlfloor frac n1rightrfloor+leftlfloor frac n2rightrfloor+ldots +leftlfloor frac nnright rfloor$$
and $$ |T|=lfloor sqrt nrfloor.$$
The map $(a,b)mapsto (b,a)$ is a fixedpoint-free involution of $Ssetminus T$, hence
$|Ssetminus T|$ is even. Since
$$ leftlfloor frac n1rightrfloor+leftlfloor frac n2rightrfloor+ldots +leftlfloor frac nnright rfloor+lfloorsqrt nrfloor = |Ssetminus T|+2|T|$$
we are done.
edited Jan 26 at 1:27
darij grinberg
11.2k33167
answered Feb 2 '14 at 18:40
Hagen von EitzenHagen von Eitzen
283k23272507
$endgroup$
Let
$$S = {,(a,b)inmathbb N^2mid able n,} $$
and $$ T={,(a,a)inmathbb N^2mid a^2le n,}.$$
Then
$$ |S|=sum_{bge 1}|{,ainmathbb Nmid aletfrac nb,}|=sum_{bge 1}lfloor tfrac nbrfloor =leftlfloor frac n1rightrfloor+leftlfloor frac n2rightrfloor+ldots +leftlfloor frac nnright rfloor$$
and $$ |T|=lfloor sqrt nrfloor.$$
The map $(a,b)mapsto (b,a)$ is a fixedpoint-free involution of $Ssetminus T$, hence
$|Ssetminus T|$ is even. Since
$$ leftlfloor frac n1rightrfloor+leftlfloor frac n2rightrfloor+ldots +leftlfloor frac nnright rfloor+lfloorsqrt nrfloor = |Ssetminus T|+2|T|$$
we are done.
edited Jan 26 at 1:27
darij grinberg
11.2k33167
answered Feb 2 '14 at 18:40
Hagen von EitzenHagen von Eitzen
283k23272507
edited Jan 26 at 1:27
darij grinberg
11.2k33167
edited Jan 26 at 1:27
darij grinberg
11.2k33167
edited Jan 26 at 1:27
darij grinberg
11.2k33167
11.2k33167
answered Feb 2 '14 at 18:40
Hagen von EitzenHagen von Eitzen
283k23272507
answered Feb 2 '14 at 18:40
Hagen von EitzenHagen von Eitzen
283k23272507
answered Feb 2 '14 at 18:40
Hagen von EitzenHagen von Eitzen
283k23272507
283k23272507
add a comment |
add a comment |
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$begingroup$
Induction is a good way to go and the inequality you used is all we have about the floor function, so what could be a better way?
$endgroup$
– Hagen von Eitzen
Feb 2 '14 at 18:27
1
$begingroup$
A duplicate of this question was asked later. You might find my answer to that question useful.
$endgroup$
– David
Feb 18 '15 at 4:58
$begingroup$
@David +1. Truly remarkable answer by you in that post.
$endgroup$
– MathGod
Feb 18 '15 at 7:36