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Prove that $ leftlfloorfrac xnrightrfloor= leftlfloorlfloorxrfloorover nrightrfloor$ where $n ge 1, n in mathbbN$ [duplicate]
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This question already has an answer here:
How to prove or disprove $forall xinBbbR, forall ninBbbN,ngt 0implies lfloorfraclfloor xrfloornrfloor=lfloorfracxnrfloor$.
5 answers
Prove that $ leftlfloorfrac xnrightrfloor= leftlfloorlfloorxrfloorover nrightrfloor$ where $n ge 1, n in mathbbN$ and $lfloor.rfloor$ represents Greatest Integer $mathbfle x$ or floor function
I tried to prove it by writing $x = lfloorxrfloor + x $ where $ .$ represents Fractional Part function and $ 0 le x < 1$
So we get,
$ lfloorfrac xnrfloor= lfloorlfloor xrfloorover n+ xover nrfloor tag1$
Then I tried to use the property,
$lfloorx+yrfloor =begincases
lfloor xrfloor + lfloor yrfloor& textif $0le x + y$ < 1 tag2\
1+ lfloor xrfloor + lfloor yrfloor & textif $1le x + y$ < 2 \
endcases $
So if I can prove $(1)$ = first case of $(2) $
I’ll have ,
$ lfloorfrac xnrfloor= lfloorlfloor xrfloorover nrfloor+ lfloorxover nrfloor = lfloorlfloorxrfloorover nrfloor$ as the second term will come out to be zero. However, I am unable to prove this.
Can someone help me out with this proof by showing me how $mathbf(1)$= first case of $mathbf (2)$ and proving the question using this method and also giving a clear proof using a simpler method
functions floor-function fractional-part
asked Jul 18 at 14:17
Hola
304
marked as duplicate by Xander Henderson, Michael Lugo, Ken, Strants, José Carlos Santos Jul 18 at 21:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
4
down vote
favorite
This question already has an answer here:
How to prove or disprove $forall xinBbbR, forall ninBbbN,ngt 0implies lfloorfraclfloor xrfloornrfloor=lfloorfracxnrfloor$.
5 answers
Prove that $ leftlfloorfrac xnrightrfloor= leftlfloorlfloorxrfloorover nrightrfloor$ where $n ge 1, n in mathbbN$ and $lfloor.rfloor$ represents Greatest Integer $mathbfle x$ or floor function
I tried to prove it by writing $x = lfloorxrfloor + x $ where $ .$ represents Fractional Part function and $ 0 le x < 1$
So we get,
$ lfloorfrac xnrfloor= lfloorlfloor xrfloorover n+ xover nrfloor tag1$
Then I tried to use the property,
$lfloorx+yrfloor =begincases
lfloor xrfloor + lfloor yrfloor& textif $0le x + y$ < 1 tag2\
1+ lfloor xrfloor + lfloor yrfloor & textif $1le x + y$ < 2 \
endcases $
So if I can prove $(1)$ = first case of $(2) $
I’ll have ,
$ lfloorfrac xnrfloor= lfloorlfloor xrfloorover nrfloor+ lfloorxover nrfloor = lfloorlfloorxrfloorover nrfloor$ as the second term will come out to be zero. However, I am unable to prove this.
Can someone help me out with this proof by showing me how $mathbf(1)$= first case of $mathbf (2)$ and proving the question using this method and also giving a clear proof using a simpler method
functions floor-function fractional-part
asked Jul 18 at 14:17
Hola
304
marked as duplicate by Xander Henderson, Michael Lugo, Ken, Strants, José Carlos Santos Jul 18 at 21:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Also Prove $bigglfloorfraclfloor x rfloor mbiggrfloor =bigglfloorfracx mbiggrfloor $.
– Xander Henderson
Jul 18 at 14:48
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
This question already has an answer here:
How to prove or disprove $forall xinBbbR, forall ninBbbN,ngt 0implies lfloorfraclfloor xrfloornrfloor=lfloorfracxnrfloor$.
5 answers
Prove that $ leftlfloorfrac xnrightrfloor= leftlfloorlfloorxrfloorover nrightrfloor$ where $n ge 1, n in mathbbN$ and $lfloor.rfloor$ represents Greatest Integer $mathbfle x$ or floor function
I tried to prove it by writing $x = lfloorxrfloor + x $ where $ .$ represents Fractional Part function and $ 0 le x < 1$
So we get,
$ lfloorfrac xnrfloor= lfloorlfloor xrfloorover n+ xover nrfloor tag1$
Then I tried to use the property,
$lfloorx+yrfloor =begincases
lfloor xrfloor + lfloor yrfloor& textif $0le x + y$ < 1 tag2\
1+ lfloor xrfloor + lfloor yrfloor & textif $1le x + y$ < 2 \
endcases $
So if I can prove $(1)$ = first case of $(2) $
I’ll have ,
$ lfloorfrac xnrfloor= lfloorlfloor xrfloorover nrfloor+ lfloorxover nrfloor = lfloorlfloorxrfloorover nrfloor$ as the second term will come out to be zero. However, I am unable to prove this.
Can someone help me out with this proof by showing me how $mathbf(1)$= first case of $mathbf (2)$ and proving the question using this method and also giving a clear proof using a simpler method
functions floor-function fractional-part
asked Jul 18 at 14:17
Hola
304
This question already has an answer here:
How to prove or disprove $forall xinBbbR, forall ninBbbN,ngt 0implies lfloorfraclfloor xrfloornrfloor=lfloorfracxnrfloor$.
5 answers
Prove that $ leftlfloorfrac xnrightrfloor= leftlfloorlfloorxrfloorover nrightrfloor$ where $n ge 1, n in mathbbN$ and $lfloor.rfloor$ represents Greatest Integer $mathbfle x$ or floor function
I tried to prove it by writing $x = lfloorxrfloor + x $ where $ .$ represents Fractional Part function and $ 0 le x < 1$
So we get,
$ lfloorfrac xnrfloor= lfloorlfloor xrfloorover n+ xover nrfloor tag1$
Then I tried to use the property,
$lfloorx+yrfloor =begincases
lfloor xrfloor + lfloor yrfloor& textif $0le x + y$ < 1 tag2\
1+ lfloor xrfloor + lfloor yrfloor & textif $1le x + y$ < 2 \
endcases $
So if I can prove $(1)$ = first case of $(2) $
I’ll have ,
$ lfloorfrac xnrfloor= lfloorlfloor xrfloorover nrfloor+ lfloorxover nrfloor = lfloorlfloorxrfloorover nrfloor$ as the second term will come out to be zero. However, I am unable to prove this.
Can someone help me out with this proof by showing me how $mathbf(1)$= first case of $mathbf (2)$ and proving the question using this method and also giving a clear proof using a simpler method
This question already has an answer here:
How to prove or disprove $forall xinBbbR, forall ninBbbN,ngt 0implies lfloorfraclfloor xrfloornrfloor=lfloorfracxnrfloor$.
5 answers
functions floor-function fractional-part
asked Jul 18 at 14:17
Hola
304
edited Jul 18 at 14:32
asked Jul 18 at 14:17
Hola
304
asked Jul 18 at 14:17
Hola
304
asked Jul 18 at 14:17
Hola
304
304
marked as duplicate by Xander Henderson, Michael Lugo, Ken, Strants, José Carlos Santos Jul 18 at 21:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Xander Henderson, Michael Lugo, Ken, Strants, José Carlos Santos Jul 18 at 21:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Also Prove $bigglfloorfraclfloor x rfloor mbiggrfloor =bigglfloorfracx mbiggrfloor $.
– Xander Henderson
Jul 18 at 14:48
add a comment |Â
Also Prove $bigglfloorfraclfloor x rfloor mbiggrfloor =bigglfloorfracx mbiggrfloor $.
– Xander Henderson
Jul 18 at 14:48
Also Prove $bigglfloorfraclfloor x rfloor mbiggrfloor =bigglfloorfracx mbiggrfloor $.
– Xander Henderson
Jul 18 at 14:48
Also Prove $bigglfloorfraclfloor x rfloor mbiggrfloor =bigglfloorfracx mbiggrfloor $.
– Xander Henderson
Jul 18 at 14:48
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
The OP's proof method is a good one. It remains to calculate $$leftfraclfloor x rfloornright$$
and
$$leftfracx nright$$
Note that the first one is at most $fracn-1n$, and the second is strictly less than $frac1n$, hence their sum is strictly less than $1$.
More details, as requested:
Since $lfloor xrfloor$ is an integer, write it as $qn+r$, where $q,r$ are integers and $0le rle n-1$. This is the division algorithm. Now $fraclfloor x rfloorn=fracqn+rn=q+fracrn$. This has fractional part $fracrn$, which is at most $fracn-1n$.
Now, $0le x<1$, so $0le fracxn<frac1n$. It is its own fractional part (since it's between $0$ and $1$), which is strictly less than $frac1n$.
answered Jul 18 at 14:31
vadim123
73.8k895184
Thanks for answering. I’m sorry I didn’t understand. Can you please explain how first one is at most (n-1)/n and second is strictly less than 1/n?
– Hola
Jul 18 at 14:52
1
x/n is strictly less than 1/n. What can I say about x/n ? And I can’t understand how first one is at most (n-1)/n
– Hola
Jul 18 at 14:54
2
If $ 0 le y < y$ then $y = y$ So $frac xn =frac xn < frac 1n$. As for then first $[x]= k$ is an integer so $frac kn = a + frac bn$ for integers $a$ and $b$ and $b$ is at most $n - 1$.
– fleablood
Jul 18 at 15:27
add a comment |Â
up vote
4
down vote
Simpler method, using only that :
if $m$ is an integer and $y$ is a real then $y ge m Longleftrightarrow lfloor y rfloor ge m$.
beginalign*
biglfloor fracxnbigrfloor &=maxbig fracxn ge kbig=max big x ge nkbig\
&=max bigkinmathbbZ =maxbig fraclfloor xrfloornge kbig = biglfloorfraclfloor xrfloornbigrfloor.
endalign*
answered Jul 18 at 14:31
Charles Madeline
3,1681836
1
Nice. Might be easier to read without all the $max$ calls. For $kin mathbb Z: $ $$kleq fracxniff nkleq xiff nkleqlfloor xrfloor iff kleqfraclfloor xrfloorn$$ Your way of writing it, while correct, hides the simplicity of the argument.
– Thomas Andrews
Jul 18 at 15:37
@Thomas Andrews the way you wrote it is indeed better, thanks
– Charles Madeline
Jul 18 at 15:48
In the original answer can you explain how $k$ went from $k in mathbbZ $ to $k in mathbbN $ ? I know it doesn’t make a difference to the proof but I was curious to know. It should remain $ k in mathbbZ $ right ?
– Hola
Jul 19 at 6:54
@Hola indeed, will correct that
– Charles Madeline
Jul 19 at 9:06
add a comment |Â
up vote
1
down vote
Consider by the archimedian principal there are a unique integers $k,m$ so that $kn le kn + mle x< kn + m + 1 le (k+1)n$.
So $frac [x]n = k + frac mn$
So $frac [x]n = frac mnle frac n-1n$ and $frac xn < frac 1n$ so $frac xn = frac xn < frac 1n$.
So $frac [x]n + frac xn = frac mn + frac xn < frac n-1n + frac 1n = 1$.
answered Jul 18 at 15:19
fleablood
60.5k22575
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The OP's proof method is a good one. It remains to calculate $$leftfraclfloor x rfloornright$$
and
$$leftfracx nright$$
Note that the first one is at most $fracn-1n$, and the second is strictly less than $frac1n$, hence their sum is strictly less than $1$.
More details, as requested:
Since $lfloor xrfloor$ is an integer, write it as $qn+r$, where $q,r$ are integers and $0le rle n-1$. This is the division algorithm. Now $fraclfloor x rfloorn=fracqn+rn=q+fracrn$. This has fractional part $fracrn$, which is at most $fracn-1n$.
Now, $0le x<1$, so $0le fracxn<frac1n$. It is its own fractional part (since it's between $0$ and $1$), which is strictly less than $frac1n$.
answered Jul 18 at 14:31
vadim123
73.8k895184
Thanks for answering. I’m sorry I didn’t understand. Can you please explain how first one is at most (n-1)/n and second is strictly less than 1/n?
– Hola
Jul 18 at 14:52
1
x/n is strictly less than 1/n. What can I say about x/n ? And I can’t understand how first one is at most (n-1)/n
– Hola
Jul 18 at 14:54
2
If $ 0 le y < y$ then $y = y$ So $frac xn =frac xn < frac 1n$. As for then first $[x]= k$ is an integer so $frac kn = a + frac bn$ for integers $a$ and $b$ and $b$ is at most $n - 1$.
– fleablood
Jul 18 at 15:27
add a comment |Â
up vote
2
down vote
accepted
The OP's proof method is a good one. It remains to calculate $$leftfraclfloor x rfloornright$$
and
$$leftfracx nright$$
Note that the first one is at most $fracn-1n$, and the second is strictly less than $frac1n$, hence their sum is strictly less than $1$.
More details, as requested:
Since $lfloor xrfloor$ is an integer, write it as $qn+r$, where $q,r$ are integers and $0le rle n-1$. This is the division algorithm. Now $fraclfloor x rfloorn=fracqn+rn=q+fracrn$. This has fractional part $fracrn$, which is at most $fracn-1n$.
Now, $0le x<1$, so $0le fracxn<frac1n$. It is its own fractional part (since it's between $0$ and $1$), which is strictly less than $frac1n$.
answered Jul 18 at 14:31
vadim123
73.8k895184
Thanks for answering. I’m sorry I didn’t understand. Can you please explain how first one is at most (n-1)/n and second is strictly less than 1/n?
– Hola
Jul 18 at 14:52
1
x/n is strictly less than 1/n. What can I say about x/n ? And I can’t understand how first one is at most (n-1)/n
– Hola
Jul 18 at 14:54
2
If $ 0 le y < y$ then $y = y$ So $frac xn =frac xn < frac 1n$. As for then first $[x]= k$ is an integer so $frac kn = a + frac bn$ for integers $a$ and $b$ and $b$ is at most $n - 1$.
– fleablood
Jul 18 at 15:27
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The OP's proof method is a good one. It remains to calculate $$leftfraclfloor x rfloornright$$
and
$$leftfracx nright$$
Note that the first one is at most $fracn-1n$, and the second is strictly less than $frac1n$, hence their sum is strictly less than $1$.
More details, as requested:
Since $lfloor xrfloor$ is an integer, write it as $qn+r$, where $q,r$ are integers and $0le rle n-1$. This is the division algorithm. Now $fraclfloor x rfloorn=fracqn+rn=q+fracrn$. This has fractional part $fracrn$, which is at most $fracn-1n$.
Now, $0le x<1$, so $0le fracxn<frac1n$. It is its own fractional part (since it's between $0$ and $1$), which is strictly less than $frac1n$.
answered Jul 18 at 14:31
vadim123
73.8k895184
The OP's proof method is a good one. It remains to calculate $$leftfraclfloor x rfloornright$$
and
$$leftfracx nright$$
Note that the first one is at most $fracn-1n$, and the second is strictly less than $frac1n$, hence their sum is strictly less than $1$.
More details, as requested:
Since $lfloor xrfloor$ is an integer, write it as $qn+r$, where $q,r$ are integers and $0le rle n-1$. This is the division algorithm. Now $fraclfloor x rfloorn=fracqn+rn=q+fracrn$. This has fractional part $fracrn$, which is at most $fracn-1n$.
Now, $0le x<1$, so $0le fracxn<frac1n$. It is its own fractional part (since it's between $0$ and $1$), which is strictly less than $frac1n$.
answered Jul 18 at 14:31
vadim123
73.8k895184
edited Jul 18 at 19:29
answered Jul 18 at 14:31
vadim123
73.8k895184
answered Jul 18 at 14:31
vadim123
73.8k895184
answered Jul 18 at 14:31
vadim123
73.8k895184
73.8k895184
Thanks for answering. I’m sorry I didn’t understand. Can you please explain how first one is at most (n-1)/n and second is strictly less than 1/n?
– Hola
Jul 18 at 14:52
1
x/n is strictly less than 1/n. What can I say about x/n ? And I can’t understand how first one is at most (n-1)/n
– Hola
Jul 18 at 14:54
2
If $ 0 le y < y$ then $y = y$ So $frac xn =frac xn < frac 1n$. As for then first $[x]= k$ is an integer so $frac kn = a + frac bn$ for integers $a$ and $b$ and $b$ is at most $n - 1$.
– fleablood
Jul 18 at 15:27
add a comment |Â
Thanks for answering. I’m sorry I didn’t understand. Can you please explain how first one is at most (n-1)/n and second is strictly less than 1/n?
– Hola
Jul 18 at 14:52
1
x/n is strictly less than 1/n. What can I say about x/n ? And I can’t understand how first one is at most (n-1)/n
– Hola
Jul 18 at 14:54
2
If $ 0 le y < y$ then $y = y$ So $frac xn =frac xn < frac 1n$. As for then first $[x]= k$ is an integer so $frac kn = a + frac bn$ for integers $a$ and $b$ and $b$ is at most $n - 1$.
– fleablood
Jul 18 at 15:27
Thanks for answering. I’m sorry I didn’t understand. Can you please explain how first one is at most (n-1)/n and second is strictly less than 1/n?
– Hola
Jul 18 at 14:52
Thanks for answering. I’m sorry I didn’t understand. Can you please explain how first one is at most (n-1)/n and second is strictly less than 1/n?
– Hola
Jul 18 at 14:52
1
1
x/n is strictly less than 1/n. What can I say about x/n ? And I can’t understand how first one is at most (n-1)/n
– Hola
Jul 18 at 14:54
x/n is strictly less than 1/n. What can I say about x/n ? And I can’t understand how first one is at most (n-1)/n
– Hola
Jul 18 at 14:54
2
2
If $ 0 le y < y$ then $y = y$ So $frac xn =frac xn < frac 1n$. As for then first $[x]= k$ is an integer so $frac kn = a + frac bn$ for integers $a$ and $b$ and $b$ is at most $n - 1$.
– fleablood
Jul 18 at 15:27
If $ 0 le y < y$ then $y = y$ So $frac xn =frac xn < frac 1n$. As for then first $[x]= k$ is an integer so $frac kn = a + frac bn$ for integers $a$ and $b$ and $b$ is at most $n - 1$.
– fleablood
Jul 18 at 15:27
add a comment |Â
up vote
4
down vote
Simpler method, using only that :
if $m$ is an integer and $y$ is a real then $y ge m Longleftrightarrow lfloor y rfloor ge m$.
beginalign*
biglfloor fracxnbigrfloor &=maxbig fracxn ge kbig=max big x ge nkbig\
&=max bigkinmathbbZ =maxbig fraclfloor xrfloornge kbig = biglfloorfraclfloor xrfloornbigrfloor.
endalign*
answered Jul 18 at 14:31
Charles Madeline
3,1681836
1
Nice. Might be easier to read without all the $max$ calls. For $kin mathbb Z: $ $$kleq fracxniff nkleq xiff nkleqlfloor xrfloor iff kleqfraclfloor xrfloorn$$ Your way of writing it, while correct, hides the simplicity of the argument.
– Thomas Andrews
Jul 18 at 15:37
@Thomas Andrews the way you wrote it is indeed better, thanks
– Charles Madeline
Jul 18 at 15:48
In the original answer can you explain how $k$ went from $k in mathbbZ $ to $k in mathbbN $ ? I know it doesn’t make a difference to the proof but I was curious to know. It should remain $ k in mathbbZ $ right ?
– Hola
Jul 19 at 6:54
@Hola indeed, will correct that
– Charles Madeline
Jul 19 at 9:06
add a comment |Â
up vote
4
down vote
Simpler method, using only that :
if $m$ is an integer and $y$ is a real then $y ge m Longleftrightarrow lfloor y rfloor ge m$.
beginalign*
biglfloor fracxnbigrfloor &=maxbig fracxn ge kbig=max big x ge nkbig\
&=max bigkinmathbbZ =maxbig fraclfloor xrfloornge kbig = biglfloorfraclfloor xrfloornbigrfloor.
endalign*
answered Jul 18 at 14:31
Charles Madeline
3,1681836
1
Nice. Might be easier to read without all the $max$ calls. For $kin mathbb Z: $ $$kleq fracxniff nkleq xiff nkleqlfloor xrfloor iff kleqfraclfloor xrfloorn$$ Your way of writing it, while correct, hides the simplicity of the argument.
– Thomas Andrews
Jul 18 at 15:37
@Thomas Andrews the way you wrote it is indeed better, thanks
– Charles Madeline
Jul 18 at 15:48
In the original answer can you explain how $k$ went from $k in mathbbZ $ to $k in mathbbN $ ? I know it doesn’t make a difference to the proof but I was curious to know. It should remain $ k in mathbbZ $ right ?
– Hola
Jul 19 at 6:54
@Hola indeed, will correct that
– Charles Madeline
Jul 19 at 9:06
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Simpler method, using only that :
if $m$ is an integer and $y$ is a real then $y ge m Longleftrightarrow lfloor y rfloor ge m$.
beginalign*
biglfloor fracxnbigrfloor &=maxbig fracxn ge kbig=max big x ge nkbig\
&=max bigkinmathbbZ =maxbig fraclfloor xrfloornge kbig = biglfloorfraclfloor xrfloornbigrfloor.
endalign*
answered Jul 18 at 14:31
Charles Madeline
3,1681836
Simpler method, using only that :
if $m$ is an integer and $y$ is a real then $y ge m Longleftrightarrow lfloor y rfloor ge m$.
beginalign*
biglfloor fracxnbigrfloor &=maxbig fracxn ge kbig=max big x ge nkbig\
&=max bigkinmathbbZ =maxbig fraclfloor xrfloornge kbig = biglfloorfraclfloor xrfloornbigrfloor.
endalign*
answered Jul 18 at 14:31
Charles Madeline
3,1681836
edited Jul 19 at 9:07
answered Jul 18 at 14:31
Charles Madeline
3,1681836
answered Jul 18 at 14:31
Charles Madeline
3,1681836
answered Jul 18 at 14:31
Charles Madeline
3,1681836
3,1681836
1
Nice. Might be easier to read without all the $max$ calls. For $kin mathbb Z: $ $$kleq fracxniff nkleq xiff nkleqlfloor xrfloor iff kleqfraclfloor xrfloorn$$ Your way of writing it, while correct, hides the simplicity of the argument.
– Thomas Andrews
Jul 18 at 15:37
@Thomas Andrews the way you wrote it is indeed better, thanks
– Charles Madeline
Jul 18 at 15:48
In the original answer can you explain how $k$ went from $k in mathbbZ $ to $k in mathbbN $ ? I know it doesn’t make a difference to the proof but I was curious to know. It should remain $ k in mathbbZ $ right ?
– Hola
Jul 19 at 6:54
@Hola indeed, will correct that
– Charles Madeline
Jul 19 at 9:06
add a comment |Â
1
Nice. Might be easier to read without all the $max$ calls. For $kin mathbb Z: $ $$kleq fracxniff nkleq xiff nkleqlfloor xrfloor iff kleqfraclfloor xrfloorn$$ Your way of writing it, while correct, hides the simplicity of the argument.
– Thomas Andrews
Jul 18 at 15:37
@Thomas Andrews the way you wrote it is indeed better, thanks
– Charles Madeline
Jul 18 at 15:48
In the original answer can you explain how $k$ went from $k in mathbbZ $ to $k in mathbbN $ ? I know it doesn’t make a difference to the proof but I was curious to know. It should remain $ k in mathbbZ $ right ?
– Hola
Jul 19 at 6:54
@Hola indeed, will correct that
– Charles Madeline
Jul 19 at 9:06
1
1
Nice. Might be easier to read without all the $max$ calls. For $kin mathbb Z: $ $$kleq fracxniff nkleq xiff nkleqlfloor xrfloor iff kleqfraclfloor xrfloorn$$ Your way of writing it, while correct, hides the simplicity of the argument.
– Thomas Andrews
Jul 18 at 15:37
Nice. Might be easier to read without all the $max$ calls. For $kin mathbb Z: $ $$kleq fracxniff nkleq xiff nkleqlfloor xrfloor iff kleqfraclfloor xrfloorn$$ Your way of writing it, while correct, hides the simplicity of the argument.
– Thomas Andrews
Jul 18 at 15:37
@Thomas Andrews the way you wrote it is indeed better, thanks
– Charles Madeline
Jul 18 at 15:48
@Thomas Andrews the way you wrote it is indeed better, thanks
– Charles Madeline
Jul 18 at 15:48
In the original answer can you explain how $k$ went from $k in mathbbZ $ to $k in mathbbN $ ? I know it doesn’t make a difference to the proof but I was curious to know. It should remain $ k in mathbbZ $ right ?
– Hola
Jul 19 at 6:54
In the original answer can you explain how $k$ went from $k in mathbbZ $ to $k in mathbbN $ ? I know it doesn’t make a difference to the proof but I was curious to know. It should remain $ k in mathbbZ $ right ?
– Hola
Jul 19 at 6:54
@Hola indeed, will correct that
– Charles Madeline
Jul 19 at 9:06
@Hola indeed, will correct that
– Charles Madeline
Jul 19 at 9:06
add a comment |Â
up vote
1
down vote
Consider by the archimedian principal there are a unique integers $k,m$ so that $kn le kn + mle x< kn + m + 1 le (k+1)n$.
So $frac [x]n = k + frac mn$
So $frac [x]n = frac mnle frac n-1n$ and $frac xn < frac 1n$ so $frac xn = frac xn < frac 1n$.
So $frac [x]n + frac xn = frac mn + frac xn < frac n-1n + frac 1n = 1$.
answered Jul 18 at 15:19
fleablood
60.5k22575
add a comment |Â
up vote
1
down vote
Consider by the archimedian principal there are a unique integers $k,m$ so that $kn le kn + mle x< kn + m + 1 le (k+1)n$.
So $frac [x]n = k + frac mn$
So $frac [x]n = frac mnle frac n-1n$ and $frac xn < frac 1n$ so $frac xn = frac xn < frac 1n$.
So $frac [x]n + frac xn = frac mn + frac xn < frac n-1n + frac 1n = 1$.
answered Jul 18 at 15:19
fleablood
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Consider by the archimedian principal there are a unique integers $k,m$ so that $kn le kn + mle x< kn + m + 1 le (k+1)n$.
So $frac [x]n = k + frac mn$
So $frac [x]n = frac mnle frac n-1n$ and $frac xn < frac 1n$ so $frac xn = frac xn < frac 1n$.
So $frac [x]n + frac xn = frac mn + frac xn < frac n-1n + frac 1n = 1$.
answered Jul 18 at 15:19
fleablood
60.5k22575
Consider by the archimedian principal there are a unique integers $k,m$ so that $kn le kn + mle x< kn + m + 1 le (k+1)n$.
So $frac [x]n = k + frac mn$
So $frac [x]n = frac mnle frac n-1n$ and $frac xn < frac 1n$ so $frac xn = frac xn < frac 1n$.
So $frac [x]n + frac xn = frac mn + frac xn < frac n-1n + frac 1n = 1$.
answered Jul 18 at 15:19
fleablood
60.5k22575
answered Jul 18 at 15:19
fleablood
60.5k22575
answered Jul 18 at 15:19
fleablood
60.5k22575
answered Jul 18 at 15:19
fleablood
60.5k22575
60.5k22575
add a comment |Â
add a comment |Â
Also Prove $bigglfloorfraclfloor x rfloor mbiggrfloor =bigglfloorfracx mbiggrfloor $.
– Xander Henderson
Jul 18 at 14:48