Mixing
Does $leftlfloordfrac{x^2+x}{i}rightrfloor - leftlfloordfrac{x^2}{i}rightrfloor =...
Does $leftlfloordfrac{x^2+x}{i}rightrfloor - leftlfloordfrac{x^2}{i}rightrfloor = leftlfloordfrac{x}{i}rightrfloor$?
where $i le x^2$ and $x$ are any positive integer.
Intuitively, this doesn't seem correct to me but here's my argument which appears valid:
(1) There exists an integer $a$ such that: $x equiv a pmod i$ where $0 le a < i$
(2) $leftlfloordfrac{x^2+x}{i}rightrfloor - leftlfloordfrac{x^2}{i}rightrfloor = dfrac{x^2 + x - a^2 - a}{i} - dfrac{x^2 - a^2}{i} = dfrac{x-a}{i} = leftlfloordfrac{x}{i}rightrfloor$
Is my argument wrong? Is my intuition wrong?
proof-verification floor-function
asked Nov 20 '18 at 16:45
Larry Freeman
3,23921239
add a comment |
Does $leftlfloordfrac{x^2+x}{i}rightrfloor - leftlfloordfrac{x^2}{i}rightrfloor = leftlfloordfrac{x}{i}rightrfloor$?
where $i le x^2$ and $x$ are any positive integer.
Intuitively, this doesn't seem correct to me but here's my argument which appears valid:
(1) There exists an integer $a$ such that: $x equiv a pmod i$ where $0 le a < i$
(2) $leftlfloordfrac{x^2+x}{i}rightrfloor - leftlfloordfrac{x^2}{i}rightrfloor = dfrac{x^2 + x - a^2 - a}{i} - dfrac{x^2 - a^2}{i} = dfrac{x-a}{i} = leftlfloordfrac{x}{i}rightrfloor$
Is my argument wrong? Is my intuition wrong?
proof-verification floor-function
asked Nov 20 '18 at 16:45
Larry Freeman
3,23921239
2
$leftlfloor frac{4 + 2}{3} rightrfloor - leftlfloor frac{4}{3} rightrfloor neq leftlfloor frac{2}{3} rightrfloor$
– Connor Harris
Nov 20 '18 at 16:53
If $0 leq a < i$ then you can't conclude that $0 leq a^2 < i$.
– Connor Harris
Nov 20 '18 at 16:55
1
Nice. So the mistake is the subtraction by $a^2$. Thanks. I suspected it was wrong.
– Larry Freeman
Nov 20 '18 at 16:56
So, it is only true when $leftlfloordfrac{a^2+a}{i}rightrfloor = leftlfloordfrac{a^2}{i}rightrfloor$?
– Larry Freeman
Nov 20 '18 at 17:02
add a comment |
Does $leftlfloordfrac{x^2+x}{i}rightrfloor - leftlfloordfrac{x^2}{i}rightrfloor = leftlfloordfrac{x}{i}rightrfloor$?
where $i le x^2$ and $x$ are any positive integer.
Intuitively, this doesn't seem correct to me but here's my argument which appears valid:
(1) There exists an integer $a$ such that: $x equiv a pmod i$ where $0 le a < i$
(2) $leftlfloordfrac{x^2+x}{i}rightrfloor - leftlfloordfrac{x^2}{i}rightrfloor = dfrac{x^2 + x - a^2 - a}{i} - dfrac{x^2 - a^2}{i} = dfrac{x-a}{i} = leftlfloordfrac{x}{i}rightrfloor$
Is my argument wrong? Is my intuition wrong?
proof-verification floor-function
asked Nov 20 '18 at 16:45
Larry Freeman
3,23921239
Does $leftlfloordfrac{x^2+x}{i}rightrfloor - leftlfloordfrac{x^2}{i}rightrfloor = leftlfloordfrac{x}{i}rightrfloor$?
where $i le x^2$ and $x$ are any positive integer.
Intuitively, this doesn't seem correct to me but here's my argument which appears valid:
(1) There exists an integer $a$ such that: $x equiv a pmod i$ where $0 le a < i$
(2) $leftlfloordfrac{x^2+x}{i}rightrfloor - leftlfloordfrac{x^2}{i}rightrfloor = dfrac{x^2 + x - a^2 - a}{i} - dfrac{x^2 - a^2}{i} = dfrac{x-a}{i} = leftlfloordfrac{x}{i}rightrfloor$
Is my argument wrong? Is my intuition wrong?
proof-verification floor-function
proof-verification floor-function
asked Nov 20 '18 at 16:45
Larry Freeman
3,23921239
asked Nov 20 '18 at 16:45
Larry Freeman
3,23921239
edited Nov 20 '18 at 16:50
asked Nov 20 '18 at 16:45
Larry Freeman
3,23921239
asked Nov 20 '18 at 16:45
Larry Freeman
3,23921239
asked Nov 20 '18 at 16:45
Larry Freeman
3,23921239
3,23921239
2
$leftlfloor frac{4 + 2}{3} rightrfloor - leftlfloor frac{4}{3} rightrfloor neq leftlfloor frac{2}{3} rightrfloor$
– Connor Harris
Nov 20 '18 at 16:53
If $0 leq a < i$ then you can't conclude that $0 leq a^2 < i$.
– Connor Harris
Nov 20 '18 at 16:55
1
Nice. So the mistake is the subtraction by $a^2$. Thanks. I suspected it was wrong.
– Larry Freeman
Nov 20 '18 at 16:56
So, it is only true when $leftlfloordfrac{a^2+a}{i}rightrfloor = leftlfloordfrac{a^2}{i}rightrfloor$?
– Larry Freeman
Nov 20 '18 at 17:02
add a comment |
2
$leftlfloor frac{4 + 2}{3} rightrfloor - leftlfloor frac{4}{3} rightrfloor neq leftlfloor frac{2}{3} rightrfloor$
– Connor Harris
Nov 20 '18 at 16:53
If $0 leq a < i$ then you can't conclude that $0 leq a^2 < i$.
– Connor Harris
Nov 20 '18 at 16:55
1
Nice. So the mistake is the subtraction by $a^2$. Thanks. I suspected it was wrong.
– Larry Freeman
Nov 20 '18 at 16:56
So, it is only true when $leftlfloordfrac{a^2+a}{i}rightrfloor = leftlfloordfrac{a^2}{i}rightrfloor$?
– Larry Freeman
Nov 20 '18 at 17:02
2
2
$leftlfloor frac{4 + 2}{3} rightrfloor - leftlfloor frac{4}{3} rightrfloor neq leftlfloor frac{2}{3} rightrfloor$
– Connor Harris
Nov 20 '18 at 16:53
$leftlfloor frac{4 + 2}{3} rightrfloor - leftlfloor frac{4}{3} rightrfloor neq leftlfloor frac{2}{3} rightrfloor$
– Connor Harris
Nov 20 '18 at 16:53
If $0 leq a < i$ then you can't conclude that $0 leq a^2 < i$.
– Connor Harris
Nov 20 '18 at 16:55
If $0 leq a < i$ then you can't conclude that $0 leq a^2 < i$.
– Connor Harris
Nov 20 '18 at 16:55
1
1
Nice. So the mistake is the subtraction by $a^2$. Thanks. I suspected it was wrong.
– Larry Freeman
Nov 20 '18 at 16:56
Nice. So the mistake is the subtraction by $a^2$. Thanks. I suspected it was wrong.
– Larry Freeman
Nov 20 '18 at 16:56
So, it is only true when $leftlfloordfrac{a^2+a}{i}rightrfloor = leftlfloordfrac{a^2}{i}rightrfloor$?
– Larry Freeman
Nov 20 '18 at 17:02
So, it is only true when $leftlfloordfrac{a^2+a}{i}rightrfloor = leftlfloordfrac{a^2}{i}rightrfloor$?
– Larry Freeman
Nov 20 '18 at 17:02
add a comment |
1 Answer
1
active
oldest
votes
Consider x = 7 and i = 5.
you get
$leftlfloordfrac{49+7}{5}rightrfloor = 11$
$leftlfloordfrac{49}{5}rightrfloor = 9$
$leftlfloordfrac{7}{5}rightrfloor = 1$
as you see the equation doesn't hold.
Your intuition is right and the argument is wrong. The same argument should follow the following logic
$dfrac{x}{i} = k + dfrac{a}{i}$ where $k$ is highest possible positive integer without making term a negative
$dfrac{x^2}{i} = l + dfrac{a^2}{i}$
$leftlfloordfrac{x^2+x}{ i}rightrfloor = k + l + leftlfloordfrac{a+a^2}{i}rightrfloor$
$leftlfloordfrac{x^2}{i}rightrfloor = l + leftlfloordfrac{a^2}{i}rightrfloor$
$leftlfloordfrac{x}{i}rightrfloor = k + leftlfloordfrac{a}{i}rightrfloor$
as you see the floor terms are not necessarily equal. There could be cases when $a+a^2$
is higher than $i$ when both $a$ and $a^2$ are less than $i$.
Hope it helps
edited Nov 20 '18 at 17:37
Larry Freeman
3,23921239
answered Nov 20 '18 at 17:03
Ofya
5048
1
Thanks for your answer. I hope it's ok that I converted your answer to latex. I think that it makes it more readable. Cheers.
– Larry Freeman
Nov 20 '18 at 17:25
Thank you very much, I wanted to make it more readable but didn't have much time and didn't want to leave the question unanswered.
– Ofya
Nov 20 '18 at 20:24
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
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Consider x = 7 and i = 5.
you get
$leftlfloordfrac{49+7}{5}rightrfloor = 11$
$leftlfloordfrac{49}{5}rightrfloor = 9$
$leftlfloordfrac{7}{5}rightrfloor = 1$
as you see the equation doesn't hold.
Your intuition is right and the argument is wrong. The same argument should follow the following logic
$dfrac{x}{i} = k + dfrac{a}{i}$ where $k$ is highest possible positive integer without making term a negative
$dfrac{x^2}{i} = l + dfrac{a^2}{i}$
$leftlfloordfrac{x^2+x}{ i}rightrfloor = k + l + leftlfloordfrac{a+a^2}{i}rightrfloor$
$leftlfloordfrac{x^2}{i}rightrfloor = l + leftlfloordfrac{a^2}{i}rightrfloor$
$leftlfloordfrac{x}{i}rightrfloor = k + leftlfloordfrac{a}{i}rightrfloor$
as you see the floor terms are not necessarily equal. There could be cases when $a+a^2$
is higher than $i$ when both $a$ and $a^2$ are less than $i$.
Hope it helps
edited Nov 20 '18 at 17:37
Larry Freeman
3,23921239
answered Nov 20 '18 at 17:03
Ofya
5048
1
Thanks for your answer. I hope it's ok that I converted your answer to latex. I think that it makes it more readable. Cheers.
– Larry Freeman
Nov 20 '18 at 17:25
Thank you very much, I wanted to make it more readable but didn't have much time and didn't want to leave the question unanswered.
– Ofya
Nov 20 '18 at 20:24
add a comment |
Consider x = 7 and i = 5.
you get
$leftlfloordfrac{49+7}{5}rightrfloor = 11$
$leftlfloordfrac{49}{5}rightrfloor = 9$
$leftlfloordfrac{7}{5}rightrfloor = 1$
as you see the equation doesn't hold.
Your intuition is right and the argument is wrong. The same argument should follow the following logic
$dfrac{x}{i} = k + dfrac{a}{i}$ where $k$ is highest possible positive integer without making term a negative
$dfrac{x^2}{i} = l + dfrac{a^2}{i}$
$leftlfloordfrac{x^2+x}{ i}rightrfloor = k + l + leftlfloordfrac{a+a^2}{i}rightrfloor$
$leftlfloordfrac{x^2}{i}rightrfloor = l + leftlfloordfrac{a^2}{i}rightrfloor$
$leftlfloordfrac{x}{i}rightrfloor = k + leftlfloordfrac{a}{i}rightrfloor$
as you see the floor terms are not necessarily equal. There could be cases when $a+a^2$
is higher than $i$ when both $a$ and $a^2$ are less than $i$.
Hope it helps
edited Nov 20 '18 at 17:37
Larry Freeman
3,23921239
answered Nov 20 '18 at 17:03
Ofya
5048
1
Thanks for your answer. I hope it's ok that I converted your answer to latex. I think that it makes it more readable. Cheers.
– Larry Freeman
Nov 20 '18 at 17:25
Thank you very much, I wanted to make it more readable but didn't have much time and didn't want to leave the question unanswered.
– Ofya
Nov 20 '18 at 20:24
add a comment |
Consider x = 7 and i = 5.
you get
$leftlfloordfrac{49+7}{5}rightrfloor = 11$
$leftlfloordfrac{49}{5}rightrfloor = 9$
$leftlfloordfrac{7}{5}rightrfloor = 1$
as you see the equation doesn't hold.
Your intuition is right and the argument is wrong. The same argument should follow the following logic
$dfrac{x}{i} = k + dfrac{a}{i}$ where $k$ is highest possible positive integer without making term a negative
$dfrac{x^2}{i} = l + dfrac{a^2}{i}$
$leftlfloordfrac{x^2+x}{ i}rightrfloor = k + l + leftlfloordfrac{a+a^2}{i}rightrfloor$
$leftlfloordfrac{x^2}{i}rightrfloor = l + leftlfloordfrac{a^2}{i}rightrfloor$
$leftlfloordfrac{x}{i}rightrfloor = k + leftlfloordfrac{a}{i}rightrfloor$
as you see the floor terms are not necessarily equal. There could be cases when $a+a^2$
is higher than $i$ when both $a$ and $a^2$ are less than $i$.
Hope it helps
edited Nov 20 '18 at 17:37
Larry Freeman
3,23921239
answered Nov 20 '18 at 17:03
Ofya
5048
Consider x = 7 and i = 5.
you get
$leftlfloordfrac{49+7}{5}rightrfloor = 11$
$leftlfloordfrac{49}{5}rightrfloor = 9$
$leftlfloordfrac{7}{5}rightrfloor = 1$
as you see the equation doesn't hold.
Your intuition is right and the argument is wrong. The same argument should follow the following logic
$dfrac{x}{i} = k + dfrac{a}{i}$ where $k$ is highest possible positive integer without making term a negative
$dfrac{x^2}{i} = l + dfrac{a^2}{i}$
$leftlfloordfrac{x^2+x}{ i}rightrfloor = k + l + leftlfloordfrac{a+a^2}{i}rightrfloor$
$leftlfloordfrac{x^2}{i}rightrfloor = l + leftlfloordfrac{a^2}{i}rightrfloor$
$leftlfloordfrac{x}{i}rightrfloor = k + leftlfloordfrac{a}{i}rightrfloor$
as you see the floor terms are not necessarily equal. There could be cases when $a+a^2$
is higher than $i$ when both $a$ and $a^2$ are less than $i$.
Hope it helps
edited Nov 20 '18 at 17:37
Larry Freeman
3,23921239
answered Nov 20 '18 at 17:03
Ofya
5048
edited Nov 20 '18 at 17:37
Larry Freeman
3,23921239
edited Nov 20 '18 at 17:37
Larry Freeman
3,23921239
edited Nov 20 '18 at 17:37
Larry Freeman
3,23921239
3,23921239
answered Nov 20 '18 at 17:03
Ofya
5048
answered Nov 20 '18 at 17:03
Ofya
5048
answered Nov 20 '18 at 17:03
Ofya
5048
5048
1
Thanks for your answer. I hope it's ok that I converted your answer to latex. I think that it makes it more readable. Cheers.
– Larry Freeman
Nov 20 '18 at 17:25
Thank you very much, I wanted to make it more readable but didn't have much time and didn't want to leave the question unanswered.
– Ofya
Nov 20 '18 at 20:24
add a comment |
1
Thanks for your answer. I hope it's ok that I converted your answer to latex. I think that it makes it more readable. Cheers.
– Larry Freeman
Nov 20 '18 at 17:25
Thank you very much, I wanted to make it more readable but didn't have much time and didn't want to leave the question unanswered.
– Ofya
Nov 20 '18 at 20:24
1
1
Thanks for your answer. I hope it's ok that I converted your answer to latex. I think that it makes it more readable. Cheers.
– Larry Freeman
Nov 20 '18 at 17:25
Thanks for your answer. I hope it's ok that I converted your answer to latex. I think that it makes it more readable. Cheers.
– Larry Freeman
Nov 20 '18 at 17:25
Thank you very much, I wanted to make it more readable but didn't have much time and didn't want to leave the question unanswered.
– Ofya
Nov 20 '18 at 20:24
Thank you very much, I wanted to make it more readable but didn't have much time and didn't want to leave the question unanswered.
– Ofya
Nov 20 '18 at 20:24
add a comment |
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2
$leftlfloor frac{4 + 2}{3} rightrfloor - leftlfloor frac{4}{3} rightrfloor neq leftlfloor frac{2}{3} rightrfloor$
– Connor Harris
Nov 20 '18 at 16:53
If $0 leq a < i$ then you can't conclude that $0 leq a^2 < i$.
– Connor Harris
Nov 20 '18 at 16:55
1
Nice. So the mistake is the subtraction by $a^2$. Thanks. I suspected it was wrong.
– Larry Freeman
Nov 20 '18 at 16:56
So, it is only true when $leftlfloordfrac{a^2+a}{i}rightrfloor = leftlfloordfrac{a^2}{i}rightrfloor$?
– Larry Freeman
Nov 20 '18 at 17:02