Mixing
If $x geq y>x/2$ then is it true that $x pmod y < x/2$?
$begingroup$
I'm not sure how to prove this statement, which I believe is true:
Given $x,y in Z$ such that $x geq y > frac {x}{2}$ then $x$ (mod $y$) < $frac {x}{2}$
Edit:
Would this be an acceptable sketch of the proof?
Suppose that $x geq y geq frac{x}{2}$ then consider the bounds where $y = x$ or $y = frac{x}{2}$.
In the case in which $x=y$ then $x$ (mod $y$) = $x$ (mod $x) = 0$.
On the other hand, if $y = frac{x}{2}$. then $x$ (mod $y$) = $x$ (mod $frac {x}{2}) = frac {x}{2}$.
Therefore, $0 leq x$ (mod $y$) < $frac {x}{2}$.
discrete-mathematics modular-arithmetic
asked Jan 13 at 5:04
Matteo CiccozziMatteo Ciccozzi
469
$endgroup$
|
show 4 more comments
$begingroup$
I'm not sure how to prove this statement, which I believe is true:
Given $x,y in Z$ such that $x geq y > frac {x}{2}$ then $x$ (mod $y$) < $frac {x}{2}$
Edit:
Would this be an acceptable sketch of the proof?
Suppose that $x geq y geq frac{x}{2}$ then consider the bounds where $y = x$ or $y = frac{x}{2}$.
In the case in which $x=y$ then $x$ (mod $y$) = $x$ (mod $x) = 0$.
On the other hand, if $y = frac{x}{2}$. then $x$ (mod $y$) = $x$ (mod $frac {x}{2}) = frac {x}{2}$.
Therefore, $0 leq x$ (mod $y$) < $frac {x}{2}$.
discrete-mathematics modular-arithmetic
asked Jan 13 at 5:04
Matteo CiccozziMatteo Ciccozzi
469
$endgroup$
$begingroup$
How can $x$ be greater than $x/2$?
$endgroup$
– Randall
Jan 13 at 5:07
$begingroup$
corrected title my apologies
$endgroup$
– Matteo Ciccozzi
Jan 13 at 5:09
$begingroup$
Your title implies that $x > frac{x}{2}$.
$endgroup$
– Randall
Jan 13 at 5:09
$begingroup$
x is greater than x/2 for $x geq 1$
$endgroup$
– Matteo Ciccozzi
Jan 13 at 5:11
$begingroup$
I am a moron... Unbelievable.
$endgroup$
– Randall
Jan 13 at 5:12
|
show 4 more comments
$begingroup$
I'm not sure how to prove this statement, which I believe is true:
Given $x,y in Z$ such that $x geq y > frac {x}{2}$ then $x$ (mod $y$) < $frac {x}{2}$
Edit:
Would this be an acceptable sketch of the proof?
Suppose that $x geq y geq frac{x}{2}$ then consider the bounds where $y = x$ or $y = frac{x}{2}$.
In the case in which $x=y$ then $x$ (mod $y$) = $x$ (mod $x) = 0$.
On the other hand, if $y = frac{x}{2}$. then $x$ (mod $y$) = $x$ (mod $frac {x}{2}) = frac {x}{2}$.
Therefore, $0 leq x$ (mod $y$) < $frac {x}{2}$.
discrete-mathematics modular-arithmetic
asked Jan 13 at 5:04
Matteo CiccozziMatteo Ciccozzi
469
$endgroup$
I'm not sure how to prove this statement, which I believe is true:
Given $x,y in Z$ such that $x geq y > frac {x}{2}$ then $x$ (mod $y$) < $frac {x}{2}$
Edit:
Would this be an acceptable sketch of the proof?
Suppose that $x geq y geq frac{x}{2}$ then consider the bounds where $y = x$ or $y = frac{x}{2}$.
In the case in which $x=y$ then $x$ (mod $y$) = $x$ (mod $x) = 0$.
On the other hand, if $y = frac{x}{2}$. then $x$ (mod $y$) = $x$ (mod $frac {x}{2}) = frac {x}{2}$.
Therefore, $0 leq x$ (mod $y$) < $frac {x}{2}$.
discrete-mathematics modular-arithmetic
discrete-mathematics modular-arithmetic
asked Jan 13 at 5:04
Matteo CiccozziMatteo Ciccozzi
469
asked Jan 13 at 5:04
Matteo CiccozziMatteo Ciccozzi
469
edited Jan 13 at 5:22
Matteo Ciccozzi
asked Jan 13 at 5:04
Matteo CiccozziMatteo Ciccozzi
469
asked Jan 13 at 5:04
Matteo CiccozziMatteo Ciccozzi
469
asked Jan 13 at 5:04
Matteo CiccozziMatteo Ciccozzi
469
469
$begingroup$
How can $x$ be greater than $x/2$?
$endgroup$
– Randall
Jan 13 at 5:07
$begingroup$
corrected title my apologies
$endgroup$
– Matteo Ciccozzi
Jan 13 at 5:09
$begingroup$
Your title implies that $x > frac{x}{2}$.
$endgroup$
– Randall
Jan 13 at 5:09
$begingroup$
x is greater than x/2 for $x geq 1$
$endgroup$
– Matteo Ciccozzi
Jan 13 at 5:11
$begingroup$
I am a moron... Unbelievable.
$endgroup$
– Randall
Jan 13 at 5:12
|
show 4 more comments
$begingroup$
How can $x$ be greater than $x/2$?
$endgroup$
– Randall
Jan 13 at 5:07
$begingroup$
corrected title my apologies
$endgroup$
– Matteo Ciccozzi
Jan 13 at 5:09
$begingroup$
Your title implies that $x > frac{x}{2}$.
$endgroup$
– Randall
Jan 13 at 5:09
$begingroup$
x is greater than x/2 for $x geq 1$
$endgroup$
– Matteo Ciccozzi
Jan 13 at 5:11
$begingroup$
I am a moron... Unbelievable.
$endgroup$
– Randall
Jan 13 at 5:12
$begingroup$
How can $x$ be greater than $x/2$?
$endgroup$
– Randall
Jan 13 at 5:07
$begingroup$
How can $x$ be greater than $x/2$?
$endgroup$
– Randall
Jan 13 at 5:07
$begingroup$
corrected title my apologies
$endgroup$
– Matteo Ciccozzi
Jan 13 at 5:09
$begingroup$
corrected title my apologies
$endgroup$
– Matteo Ciccozzi
Jan 13 at 5:09
$begingroup$
Your title implies that $x > frac{x}{2}$.
$endgroup$
– Randall
Jan 13 at 5:09
$begingroup$
Your title implies that $x > frac{x}{2}$.
$endgroup$
– Randall
Jan 13 at 5:09
$begingroup$
x is greater than x/2 for $x geq 1$
$endgroup$
– Matteo Ciccozzi
Jan 13 at 5:11
$begingroup$
x is greater than x/2 for $x geq 1$
$endgroup$
– Matteo Ciccozzi
Jan 13 at 5:11
$begingroup$
I am a moron... Unbelievable.
$endgroup$
– Randall
Jan 13 at 5:12
$begingroup$
I am a moron... Unbelievable.
$endgroup$
– Randall
Jan 13 at 5:12
|
show 4 more comments
2 Answers
2
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oldest
votes
$begingroup$
This is true. Suppose that $0leq frac{x}{2} < y leq x$. Long divide $x$ by $y$ by the Division Algorithm to write $x=yq+r$ where $0 leq r <y$. Here, $r$ is your $x bmod y$. We claim that $r < frac{x}{2}$.
Suppose not, so that $r geq frac{x}{2}$. Then $x=yq+r geq yq+frac{x}{2}$. Subtracting $frac{x}{2}$ from each side gives $frac{x}{2} geq yq$. But $y >frac{x}{2}$ so this gives $frac{x}{2} > frac{x}{2}q$ so $q < 1$. Hence $q=0$ and $x=r < y leq x$, a contradiction.
answered Jan 13 at 5:40
RandallRandall
9,73111230
$endgroup$
$begingroup$
There may be a faster, less convoluted way to get there, but it's late and I've had 2 beers and 0 coffees.
$endgroup$
– Randall
Jan 13 at 5:48
add a comment |
$begingroup$
The answer to the title is yes.
Assuming $x > 0$, $x ge y > x/2 implies 1≤x/y<2$, so the remainder when $x$ is divided by $y$ is $x−y$ and also $x ge y > x/2 implies 0 le x-y < x/2$.
answered Jan 13 at 5:44
J. W. TannerJ. W. Tanner
1,792114
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
This is true. Suppose that $0leq frac{x}{2} < y leq x$. Long divide $x$ by $y$ by the Division Algorithm to write $x=yq+r$ where $0 leq r <y$. Here, $r$ is your $x bmod y$. We claim that $r < frac{x}{2}$.
Suppose not, so that $r geq frac{x}{2}$. Then $x=yq+r geq yq+frac{x}{2}$. Subtracting $frac{x}{2}$ from each side gives $frac{x}{2} geq yq$. But $y >frac{x}{2}$ so this gives $frac{x}{2} > frac{x}{2}q$ so $q < 1$. Hence $q=0$ and $x=r < y leq x$, a contradiction.
answered Jan 13 at 5:40
RandallRandall
9,73111230
$endgroup$
$begingroup$
There may be a faster, less convoluted way to get there, but it's late and I've had 2 beers and 0 coffees.
$endgroup$
– Randall
Jan 13 at 5:48
add a comment |
$begingroup$
This is true. Suppose that $0leq frac{x}{2} < y leq x$. Long divide $x$ by $y$ by the Division Algorithm to write $x=yq+r$ where $0 leq r <y$. Here, $r$ is your $x bmod y$. We claim that $r < frac{x}{2}$.
Suppose not, so that $r geq frac{x}{2}$. Then $x=yq+r geq yq+frac{x}{2}$. Subtracting $frac{x}{2}$ from each side gives $frac{x}{2} geq yq$. But $y >frac{x}{2}$ so this gives $frac{x}{2} > frac{x}{2}q$ so $q < 1$. Hence $q=0$ and $x=r < y leq x$, a contradiction.
answered Jan 13 at 5:40
RandallRandall
9,73111230
$endgroup$
$begingroup$
There may be a faster, less convoluted way to get there, but it's late and I've had 2 beers and 0 coffees.
$endgroup$
– Randall
Jan 13 at 5:48
add a comment |
$begingroup$
This is true. Suppose that $0leq frac{x}{2} < y leq x$. Long divide $x$ by $y$ by the Division Algorithm to write $x=yq+r$ where $0 leq r <y$. Here, $r$ is your $x bmod y$. We claim that $r < frac{x}{2}$.
Suppose not, so that $r geq frac{x}{2}$. Then $x=yq+r geq yq+frac{x}{2}$. Subtracting $frac{x}{2}$ from each side gives $frac{x}{2} geq yq$. But $y >frac{x}{2}$ so this gives $frac{x}{2} > frac{x}{2}q$ so $q < 1$. Hence $q=0$ and $x=r < y leq x$, a contradiction.
answered Jan 13 at 5:40
RandallRandall
9,73111230
$endgroup$
This is true. Suppose that $0leq frac{x}{2} < y leq x$. Long divide $x$ by $y$ by the Division Algorithm to write $x=yq+r$ where $0 leq r <y$. Here, $r$ is your $x bmod y$. We claim that $r < frac{x}{2}$.
Suppose not, so that $r geq frac{x}{2}$. Then $x=yq+r geq yq+frac{x}{2}$. Subtracting $frac{x}{2}$ from each side gives $frac{x}{2} geq yq$. But $y >frac{x}{2}$ so this gives $frac{x}{2} > frac{x}{2}q$ so $q < 1$. Hence $q=0$ and $x=r < y leq x$, a contradiction.
answered Jan 13 at 5:40
RandallRandall
9,73111230
answered Jan 13 at 5:40
RandallRandall
9,73111230
answered Jan 13 at 5:40
RandallRandall
9,73111230
answered Jan 13 at 5:40
RandallRandall
9,73111230
9,73111230
$begingroup$
There may be a faster, less convoluted way to get there, but it's late and I've had 2 beers and 0 coffees.
$endgroup$
– Randall
Jan 13 at 5:48
add a comment |
$begingroup$
There may be a faster, less convoluted way to get there, but it's late and I've had 2 beers and 0 coffees.
$endgroup$
– Randall
Jan 13 at 5:48
$begingroup$
There may be a faster, less convoluted way to get there, but it's late and I've had 2 beers and 0 coffees.
$endgroup$
– Randall
Jan 13 at 5:48
$begingroup$
There may be a faster, less convoluted way to get there, but it's late and I've had 2 beers and 0 coffees.
$endgroup$
– Randall
Jan 13 at 5:48
add a comment |
$begingroup$
The answer to the title is yes.
Assuming $x > 0$, $x ge y > x/2 implies 1≤x/y<2$, so the remainder when $x$ is divided by $y$ is $x−y$ and also $x ge y > x/2 implies 0 le x-y < x/2$.
answered Jan 13 at 5:44
J. W. TannerJ. W. Tanner
1,792114
$endgroup$
add a comment |
$begingroup$
The answer to the title is yes.
Assuming $x > 0$, $x ge y > x/2 implies 1≤x/y<2$, so the remainder when $x$ is divided by $y$ is $x−y$ and also $x ge y > x/2 implies 0 le x-y < x/2$.
answered Jan 13 at 5:44
J. W. TannerJ. W. Tanner
1,792114
$endgroup$
add a comment |
$begingroup$
The answer to the title is yes.
Assuming $x > 0$, $x ge y > x/2 implies 1≤x/y<2$, so the remainder when $x$ is divided by $y$ is $x−y$ and also $x ge y > x/2 implies 0 le x-y < x/2$.
answered Jan 13 at 5:44
J. W. TannerJ. W. Tanner
1,792114
$endgroup$
The answer to the title is yes.
Assuming $x > 0$, $x ge y > x/2 implies 1≤x/y<2$, so the remainder when $x$ is divided by $y$ is $x−y$ and also $x ge y > x/2 implies 0 le x-y < x/2$.
answered Jan 13 at 5:44
J. W. TannerJ. W. Tanner
1,792114
edited Jan 13 at 19:10
answered Jan 13 at 5:44
J. W. TannerJ. W. Tanner
1,792114
answered Jan 13 at 5:44
J. W. TannerJ. W. Tanner
1,792114
answered Jan 13 at 5:44
J. W. TannerJ. W. Tanner
1,792114
1,792114
add a comment |
add a comment |
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$begingroup$
How can $x$ be greater than $x/2$?
$endgroup$
– Randall
Jan 13 at 5:07
$begingroup$
corrected title my apologies
$endgroup$
– Matteo Ciccozzi
Jan 13 at 5:09
$begingroup$
Your title implies that $x > frac{x}{2}$.
$endgroup$
– Randall
Jan 13 at 5:09
$begingroup$
x is greater than x/2 for $x geq 1$
$endgroup$
– Matteo Ciccozzi
Jan 13 at 5:11
$begingroup$
I am a moron... Unbelievable.
$endgroup$
– Randall
Jan 13 at 5:12