Mixing
Show that if $x,yin [0,1]$ and $|x-y|leq a + b$ then there exists $zin [0,1]$ such that $|x-z|leq a$ and...
$begingroup$
Show that if $x,yin [0,1], a,bgeq 0$ and $|x-y|leq a + b$ then there exists $zin [0,1]$ such that $|x-z|leq a$ and $|z-y|leq b.$
I tried to come up with a general $z$ that works but I am unable to do so. Any hints would be much appreciated.
inequality
asked Jan 7 at 23:29
Hello_WorldHello_World
4,12621731
$endgroup$
add a comment |
$begingroup$
Show that if $x,yin [0,1], a,bgeq 0$ and $|x-y|leq a + b$ then there exists $zin [0,1]$ such that $|x-z|leq a$ and $|z-y|leq b.$
I tried to come up with a general $z$ that works but I am unable to do so. Any hints would be much appreciated.
inequality
asked Jan 7 at 23:29
Hello_WorldHello_World
4,12621731
$endgroup$
$begingroup$
As a hint about how to proceed, try to find the unique $z$ when $|x - y| = a + b$. It will lie somewhere in the interval $[x, y]$ (or $[y, x]$), so there must be some $lambda in [0, 1]$ such that $z = lambda x + (1 - lambda) y$. Try to find that $lambda$ as a function of $a$ and $b$.
$endgroup$
– Theo Bendit
Jan 7 at 23:38
$begingroup$
Also, I assume $a, b ge 0$, otherwise the result is clearly false. :-)
$endgroup$
– Theo Bendit
Jan 7 at 23:40
$begingroup$
@TheoBendit OP stated that $a,bgeq 0$. He just missed it in the title.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:44
add a comment |
$begingroup$
Show that if $x,yin [0,1], a,bgeq 0$ and $|x-y|leq a + b$ then there exists $zin [0,1]$ such that $|x-z|leq a$ and $|z-y|leq b.$
I tried to come up with a general $z$ that works but I am unable to do so. Any hints would be much appreciated.
inequality
asked Jan 7 at 23:29
Hello_WorldHello_World
4,12621731
$endgroup$
Show that if $x,yin [0,1], a,bgeq 0$ and $|x-y|leq a + b$ then there exists $zin [0,1]$ such that $|x-z|leq a$ and $|z-y|leq b.$
I tried to come up with a general $z$ that works but I am unable to do so. Any hints would be much appreciated.
inequality
inequality
asked Jan 7 at 23:29
Hello_WorldHello_World
4,12621731
asked Jan 7 at 23:29
Hello_WorldHello_World
4,12621731
edited Jan 7 at 23:41
Hello_World
asked Jan 7 at 23:29
Hello_WorldHello_World
4,12621731
asked Jan 7 at 23:29
Hello_WorldHello_World
4,12621731
asked Jan 7 at 23:29
Hello_WorldHello_World
4,12621731
4,12621731
$begingroup$
As a hint about how to proceed, try to find the unique $z$ when $|x - y| = a + b$. It will lie somewhere in the interval $[x, y]$ (or $[y, x]$), so there must be some $lambda in [0, 1]$ such that $z = lambda x + (1 - lambda) y$. Try to find that $lambda$ as a function of $a$ and $b$.
$endgroup$
– Theo Bendit
Jan 7 at 23:38
$begingroup$
Also, I assume $a, b ge 0$, otherwise the result is clearly false. :-)
$endgroup$
– Theo Bendit
Jan 7 at 23:40
$begingroup$
@TheoBendit OP stated that $a,bgeq 0$. He just missed it in the title.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:44
add a comment |
$begingroup$
As a hint about how to proceed, try to find the unique $z$ when $|x - y| = a + b$. It will lie somewhere in the interval $[x, y]$ (or $[y, x]$), so there must be some $lambda in [0, 1]$ such that $z = lambda x + (1 - lambda) y$. Try to find that $lambda$ as a function of $a$ and $b$.
$endgroup$
– Theo Bendit
Jan 7 at 23:38
$begingroup$
Also, I assume $a, b ge 0$, otherwise the result is clearly false. :-)
$endgroup$
– Theo Bendit
Jan 7 at 23:40
$begingroup$
@TheoBendit OP stated that $a,bgeq 0$. He just missed it in the title.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:44
$begingroup$
As a hint about how to proceed, try to find the unique $z$ when $|x - y| = a + b$. It will lie somewhere in the interval $[x, y]$ (or $[y, x]$), so there must be some $lambda in [0, 1]$ such that $z = lambda x + (1 - lambda) y$. Try to find that $lambda$ as a function of $a$ and $b$.
$endgroup$
– Theo Bendit
Jan 7 at 23:38
$begingroup$
As a hint about how to proceed, try to find the unique $z$ when $|x - y| = a + b$. It will lie somewhere in the interval $[x, y]$ (or $[y, x]$), so there must be some $lambda in [0, 1]$ such that $z = lambda x + (1 - lambda) y$. Try to find that $lambda$ as a function of $a$ and $b$.
$endgroup$
– Theo Bendit
Jan 7 at 23:38
$begingroup$
Also, I assume $a, b ge 0$, otherwise the result is clearly false. :-)
$endgroup$
– Theo Bendit
Jan 7 at 23:40
$begingroup$
Also, I assume $a, b ge 0$, otherwise the result is clearly false. :-)
$endgroup$
– Theo Bendit
Jan 7 at 23:40
$begingroup$
@TheoBendit OP stated that $a,bgeq 0$. He just missed it in the title.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:44
$begingroup$
@TheoBendit OP stated that $a,bgeq 0$. He just missed it in the title.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $a,binmathbb{R}$ have no restriction is false because you can choose one of them negative, so you can't find $z$ which satisfy the request.
However, if $a,bge 0$ you can reason like this: suppose $x<y$ and take $z=frac{a}{a+b}y+frac{b}{a+b}x$.
By choice we have $zin[x,y]subseteq[0,1]$, and moreover
begin{align}
|x-z|=z-x &=frac{a}{a+b}y+frac{b}{a+b}x-x=\
&=frac{a}{a+b}y+frac{b-a-b}{a+b}x=\
&=frac{a}{a+b}y-frac{a}{a+b}x=\
&=frac{a}{a+b}(y-x)
end{align}
So $|z-x|=frac{a}{a+b}|y-x|<frac{a}{a+b}(a+b)=a$.
In the same way you can prove $|z-y|<b$.
edited Jan 8 at 0:30
Theo Bendit
17.4k12150
answered Jan 8 at 0:03
ecrinecrin
3827
$endgroup$
$begingroup$
Do you really need $x<y$?
$endgroup$
– Hello_World
Jan 8 at 0:19
$begingroup$
No, it works anyway thanks to ||, I made this assumption in order to semplify the notation
$endgroup$
– ecrin
Jan 8 at 9:03
add a comment |
$begingroup$
Consider the intervals $[x-a,x+a]$ and $[y-b,y+b]$. If they have a point in common then a common point $z$ will do the trick. Prove that neither interval lies completely to the left of the other , i.e. we cannot have $y+b <x-a$ or $x+a<y-b$. This proves that the intervals intersect. Consider the case $x-a leq y-bleq x+a leq y+b$. If the interval $[y-b,x+a]$ does not intersect $[0,1]$ then either $y-b >1$ or $x+a<0$ and both these are clearly false. Hence we can choose $z$ to be between $0$ and $1$. Similarly argument holds in the other case.
answered Jan 7 at 23:42
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
$begingroup$
How can I conclude that $zin [0,1]$?
$endgroup$
– Hello_World
Jan 7 at 23:54
$begingroup$
@Hello_World I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:01
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $a,binmathbb{R}$ have no restriction is false because you can choose one of them negative, so you can't find $z$ which satisfy the request.
However, if $a,bge 0$ you can reason like this: suppose $x<y$ and take $z=frac{a}{a+b}y+frac{b}{a+b}x$.
By choice we have $zin[x,y]subseteq[0,1]$, and moreover
begin{align}
|x-z|=z-x &=frac{a}{a+b}y+frac{b}{a+b}x-x=\
&=frac{a}{a+b}y+frac{b-a-b}{a+b}x=\
&=frac{a}{a+b}y-frac{a}{a+b}x=\
&=frac{a}{a+b}(y-x)
end{align}
So $|z-x|=frac{a}{a+b}|y-x|<frac{a}{a+b}(a+b)=a$.
In the same way you can prove $|z-y|<b$.
edited Jan 8 at 0:30
Theo Bendit
17.4k12150
answered Jan 8 at 0:03
ecrinecrin
3827
$endgroup$
$begingroup$
Do you really need $x<y$?
$endgroup$
– Hello_World
Jan 8 at 0:19
$begingroup$
No, it works anyway thanks to ||, I made this assumption in order to semplify the notation
$endgroup$
– ecrin
Jan 8 at 9:03
add a comment |
$begingroup$
If $a,binmathbb{R}$ have no restriction is false because you can choose one of them negative, so you can't find $z$ which satisfy the request.
However, if $a,bge 0$ you can reason like this: suppose $x<y$ and take $z=frac{a}{a+b}y+frac{b}{a+b}x$.
By choice we have $zin[x,y]subseteq[0,1]$, and moreover
begin{align}
|x-z|=z-x &=frac{a}{a+b}y+frac{b}{a+b}x-x=\
&=frac{a}{a+b}y+frac{b-a-b}{a+b}x=\
&=frac{a}{a+b}y-frac{a}{a+b}x=\
&=frac{a}{a+b}(y-x)
end{align}
So $|z-x|=frac{a}{a+b}|y-x|<frac{a}{a+b}(a+b)=a$.
In the same way you can prove $|z-y|<b$.
edited Jan 8 at 0:30
Theo Bendit
17.4k12150
answered Jan 8 at 0:03
ecrinecrin
3827
$endgroup$
$begingroup$
Do you really need $x<y$?
$endgroup$
– Hello_World
Jan 8 at 0:19
$begingroup$
No, it works anyway thanks to ||, I made this assumption in order to semplify the notation
$endgroup$
– ecrin
Jan 8 at 9:03
add a comment |
$begingroup$
If $a,binmathbb{R}$ have no restriction is false because you can choose one of them negative, so you can't find $z$ which satisfy the request.
However, if $a,bge 0$ you can reason like this: suppose $x<y$ and take $z=frac{a}{a+b}y+frac{b}{a+b}x$.
By choice we have $zin[x,y]subseteq[0,1]$, and moreover
begin{align}
|x-z|=z-x &=frac{a}{a+b}y+frac{b}{a+b}x-x=\
&=frac{a}{a+b}y+frac{b-a-b}{a+b}x=\
&=frac{a}{a+b}y-frac{a}{a+b}x=\
&=frac{a}{a+b}(y-x)
end{align}
So $|z-x|=frac{a}{a+b}|y-x|<frac{a}{a+b}(a+b)=a$.
In the same way you can prove $|z-y|<b$.
edited Jan 8 at 0:30
Theo Bendit
17.4k12150
answered Jan 8 at 0:03
ecrinecrin
3827
$endgroup$
If $a,binmathbb{R}$ have no restriction is false because you can choose one of them negative, so you can't find $z$ which satisfy the request.
However, if $a,bge 0$ you can reason like this: suppose $x<y$ and take $z=frac{a}{a+b}y+frac{b}{a+b}x$.
By choice we have $zin[x,y]subseteq[0,1]$, and moreover
begin{align}
|x-z|=z-x &=frac{a}{a+b}y+frac{b}{a+b}x-x=\
&=frac{a}{a+b}y+frac{b-a-b}{a+b}x=\
&=frac{a}{a+b}y-frac{a}{a+b}x=\
&=frac{a}{a+b}(y-x)
end{align}
So $|z-x|=frac{a}{a+b}|y-x|<frac{a}{a+b}(a+b)=a$.
In the same way you can prove $|z-y|<b$.
edited Jan 8 at 0:30
Theo Bendit
17.4k12150
answered Jan 8 at 0:03
ecrinecrin
3827
edited Jan 8 at 0:30
Theo Bendit
17.4k12150
edited Jan 8 at 0:30
Theo Bendit
17.4k12150
edited Jan 8 at 0:30
Theo Bendit
17.4k12150
17.4k12150
answered Jan 8 at 0:03
ecrinecrin
3827
answered Jan 8 at 0:03
ecrinecrin
3827
answered Jan 8 at 0:03
ecrinecrin
3827
3827
$begingroup$
Do you really need $x<y$?
$endgroup$
– Hello_World
Jan 8 at 0:19
$begingroup$
No, it works anyway thanks to ||, I made this assumption in order to semplify the notation
$endgroup$
– ecrin
Jan 8 at 9:03
add a comment |
$begingroup$
Do you really need $x<y$?
$endgroup$
– Hello_World
Jan 8 at 0:19
$begingroup$
No, it works anyway thanks to ||, I made this assumption in order to semplify the notation
$endgroup$
– ecrin
Jan 8 at 9:03
$begingroup$
Do you really need $x<y$?
$endgroup$
– Hello_World
Jan 8 at 0:19
$begingroup$
Do you really need $x<y$?
$endgroup$
– Hello_World
Jan 8 at 0:19
$begingroup$
No, it works anyway thanks to ||, I made this assumption in order to semplify the notation
$endgroup$
– ecrin
Jan 8 at 9:03
$begingroup$
No, it works anyway thanks to ||, I made this assumption in order to semplify the notation
$endgroup$
– ecrin
Jan 8 at 9:03
add a comment |
$begingroup$
Consider the intervals $[x-a,x+a]$ and $[y-b,y+b]$. If they have a point in common then a common point $z$ will do the trick. Prove that neither interval lies completely to the left of the other , i.e. we cannot have $y+b <x-a$ or $x+a<y-b$. This proves that the intervals intersect. Consider the case $x-a leq y-bleq x+a leq y+b$. If the interval $[y-b,x+a]$ does not intersect $[0,1]$ then either $y-b >1$ or $x+a<0$ and both these are clearly false. Hence we can choose $z$ to be between $0$ and $1$. Similarly argument holds in the other case.
answered Jan 7 at 23:42
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
$begingroup$
How can I conclude that $zin [0,1]$?
$endgroup$
– Hello_World
Jan 7 at 23:54
$begingroup$
@Hello_World I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:01
add a comment |
$begingroup$
Consider the intervals $[x-a,x+a]$ and $[y-b,y+b]$. If they have a point in common then a common point $z$ will do the trick. Prove that neither interval lies completely to the left of the other , i.e. we cannot have $y+b <x-a$ or $x+a<y-b$. This proves that the intervals intersect. Consider the case $x-a leq y-bleq x+a leq y+b$. If the interval $[y-b,x+a]$ does not intersect $[0,1]$ then either $y-b >1$ or $x+a<0$ and both these are clearly false. Hence we can choose $z$ to be between $0$ and $1$. Similarly argument holds in the other case.
answered Jan 7 at 23:42
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
$begingroup$
How can I conclude that $zin [0,1]$?
$endgroup$
– Hello_World
Jan 7 at 23:54
$begingroup$
@Hello_World I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:01
add a comment |
$begingroup$
Consider the intervals $[x-a,x+a]$ and $[y-b,y+b]$. If they have a point in common then a common point $z$ will do the trick. Prove that neither interval lies completely to the left of the other , i.e. we cannot have $y+b <x-a$ or $x+a<y-b$. This proves that the intervals intersect. Consider the case $x-a leq y-bleq x+a leq y+b$. If the interval $[y-b,x+a]$ does not intersect $[0,1]$ then either $y-b >1$ or $x+a<0$ and both these are clearly false. Hence we can choose $z$ to be between $0$ and $1$. Similarly argument holds in the other case.
answered Jan 7 at 23:42
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
Consider the intervals $[x-a,x+a]$ and $[y-b,y+b]$. If they have a point in common then a common point $z$ will do the trick. Prove that neither interval lies completely to the left of the other , i.e. we cannot have $y+b <x-a$ or $x+a<y-b$. This proves that the intervals intersect. Consider the case $x-a leq y-bleq x+a leq y+b$. If the interval $[y-b,x+a]$ does not intersect $[0,1]$ then either $y-b >1$ or $x+a<0$ and both these are clearly false. Hence we can choose $z$ to be between $0$ and $1$. Similarly argument holds in the other case.
answered Jan 7 at 23:42
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
edited Jan 8 at 0:01
answered Jan 7 at 23:42
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 7 at 23:42
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 7 at 23:42
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
56.6k42159
$begingroup$
How can I conclude that $zin [0,1]$?
$endgroup$
– Hello_World
Jan 7 at 23:54
$begingroup$
@Hello_World I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:01
add a comment |
$begingroup$
How can I conclude that $zin [0,1]$?
$endgroup$
– Hello_World
Jan 7 at 23:54
$begingroup$
@Hello_World I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:01
$begingroup$
How can I conclude that $zin [0,1]$?
$endgroup$
– Hello_World
Jan 7 at 23:54
$begingroup$
How can I conclude that $zin [0,1]$?
$endgroup$
– Hello_World
Jan 7 at 23:54
$begingroup$
@Hello_World I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:01
$begingroup$
@Hello_World I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 0:01
add a comment |
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$begingroup$
As a hint about how to proceed, try to find the unique $z$ when $|x - y| = a + b$. It will lie somewhere in the interval $[x, y]$ (or $[y, x]$), so there must be some $lambda in [0, 1]$ such that $z = lambda x + (1 - lambda) y$. Try to find that $lambda$ as a function of $a$ and $b$.
$endgroup$
– Theo Bendit
Jan 7 at 23:38
$begingroup$
Also, I assume $a, b ge 0$, otherwise the result is clearly false. :-)
$endgroup$
– Theo Bendit
Jan 7 at 23:40
$begingroup$
@TheoBendit OP stated that $a,bgeq 0$. He just missed it in the title.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 23:44