Does $d (x,y)= (x-y)^2$ define metric on a set of real numbers?
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It is a question in functional analysis by writer Erwin Kryzic
metric-spaces
edited Jan 22 '17 at 6:25
Fernando Revilla
3,282420
asked Jan 22 '17 at 6:21
alishba fatima
161
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up vote
3
down vote
favorite
It is a question in functional analysis by writer Erwin Kryzic
metric-spaces
edited Jan 22 '17 at 6:25
Fernando Revilla
3,282420
asked Jan 22 '17 at 6:21
alishba fatima
161
1
Let's say I want to travel from the point $0$ to the point $5$. If I go straight there, it will cost me $d(0,5)=25$ units. On the other hand, if I go one bit at a time, it will only cost me $d(0,1)+d(1,2)+dotsb+d(4,5)=5$ units.
– Akiva Weinberger
Jan 22 '17 at 8:41
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
It is a question in functional analysis by writer Erwin Kryzic
metric-spaces
edited Jan 22 '17 at 6:25
Fernando Revilla
3,282420
asked Jan 22 '17 at 6:21
alishba fatima
161
It is a question in functional analysis by writer Erwin Kryzic
metric-spaces
metric-spaces
edited Jan 22 '17 at 6:25
Fernando Revilla
3,282420
asked Jan 22 '17 at 6:21
alishba fatima
161
edited Jan 22 '17 at 6:25
Fernando Revilla
3,282420
asked Jan 22 '17 at 6:21
alishba fatima
161
edited Jan 22 '17 at 6:25
Fernando Revilla
3,282420
edited Jan 22 '17 at 6:25
Fernando Revilla
3,282420
edited Jan 22 '17 at 6:25
Fernando Revilla
3,282420
3,282420
asked Jan 22 '17 at 6:21
alishba fatima
161
asked Jan 22 '17 at 6:21
alishba fatima
161
asked Jan 22 '17 at 6:21
alishba fatima
161
161
1
Let's say I want to travel from the point $0$ to the point $5$. If I go straight there, it will cost me $d(0,5)=25$ units. On the other hand, if I go one bit at a time, it will only cost me $d(0,1)+d(1,2)+dotsb+d(4,5)=5$ units.
– Akiva Weinberger
Jan 22 '17 at 8:41
add a comment |
1
Let's say I want to travel from the point $0$ to the point $5$. If I go straight there, it will cost me $d(0,5)=25$ units. On the other hand, if I go one bit at a time, it will only cost me $d(0,1)+d(1,2)+dotsb+d(4,5)=5$ units.
– Akiva Weinberger
Jan 22 '17 at 8:41
1
1
Let's say I want to travel from the point $0$ to the point $5$. If I go straight there, it will cost me $d(0,5)=25$ units. On the other hand, if I go one bit at a time, it will only cost me $d(0,1)+d(1,2)+dotsb+d(4,5)=5$ units.
– Akiva Weinberger
Jan 22 '17 at 8:41
Let's say I want to travel from the point $0$ to the point $5$. If I go straight there, it will cost me $d(0,5)=25$ units. On the other hand, if I go one bit at a time, it will only cost me $d(0,1)+d(1,2)+dotsb+d(4,5)=5$ units.
– Akiva Weinberger
Jan 22 '17 at 8:41
add a comment |
2 Answers
2
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oldest
votes
up vote
5
down vote
In order for it to be a metric it must follow these properties by definition,
$$d(x,y) geq 0$$
$$d(x,y)=0 iff x=y$$
$$d(x,y)=d(y,x)$$
$$d(x,z) leq d(x,y)+d(y,z)$$
Does it?
answered Jan 22 '17 at 6:25
Ahmed S. Attaalla
14.7k12049
Hint: focus on the last property.
– vadim123
Jan 22 '17 at 6:29
1
Suppose $xlt ylt z$; does $d(x,z)le d(x,y)+d(y,z)$ ever hold for that funny definition of $d(x,y)?$
– bof
Jan 22 '17 at 6:34
add a comment |
up vote
1
down vote
Ok. The problem is the triangle inequality, then:
Let $x,y,z in mathbb{R}$, so $$d(x,z) = (x+y-y+z)^2$$ $$d(x,z) = ((x-y)+(y-z))^2$$ $$qquad d(x,z) = d(x,y)+2(x-y)(y-z)+d(y,z)$$ $$d(x,z) geq d(x,y)+d(y,z)quad iffquad (x-y)(y-z)geq0$$
Then, suppose what $x=3$, $z=1$, $y=2$, that which does not meet the triangular inequality, therefore, $d$ is not a metric
answered Nov 13 at 1:34
Franck Panche
112
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
In order for it to be a metric it must follow these properties by definition,
$$d(x,y) geq 0$$
$$d(x,y)=0 iff x=y$$
$$d(x,y)=d(y,x)$$
$$d(x,z) leq d(x,y)+d(y,z)$$
Does it?
answered Jan 22 '17 at 6:25
Ahmed S. Attaalla
14.7k12049
Hint: focus on the last property.
– vadim123
Jan 22 '17 at 6:29
1
Suppose $xlt ylt z$; does $d(x,z)le d(x,y)+d(y,z)$ ever hold for that funny definition of $d(x,y)?$
– bof
Jan 22 '17 at 6:34
add a comment |
up vote
5
down vote
In order for it to be a metric it must follow these properties by definition,
$$d(x,y) geq 0$$
$$d(x,y)=0 iff x=y$$
$$d(x,y)=d(y,x)$$
$$d(x,z) leq d(x,y)+d(y,z)$$
Does it?
answered Jan 22 '17 at 6:25
Ahmed S. Attaalla
14.7k12049
Hint: focus on the last property.
– vadim123
Jan 22 '17 at 6:29
1
Suppose $xlt ylt z$; does $d(x,z)le d(x,y)+d(y,z)$ ever hold for that funny definition of $d(x,y)?$
– bof
Jan 22 '17 at 6:34
add a comment |
up vote
5
down vote
up vote
5
down vote
In order for it to be a metric it must follow these properties by definition,
$$d(x,y) geq 0$$
$$d(x,y)=0 iff x=y$$
$$d(x,y)=d(y,x)$$
$$d(x,z) leq d(x,y)+d(y,z)$$
Does it?
answered Jan 22 '17 at 6:25
Ahmed S. Attaalla
14.7k12049
In order for it to be a metric it must follow these properties by definition,
$$d(x,y) geq 0$$
$$d(x,y)=0 iff x=y$$
$$d(x,y)=d(y,x)$$
$$d(x,z) leq d(x,y)+d(y,z)$$
Does it?
answered Jan 22 '17 at 6:25
Ahmed S. Attaalla
14.7k12049
answered Jan 22 '17 at 6:25
Ahmed S. Attaalla
14.7k12049
answered Jan 22 '17 at 6:25
Ahmed S. Attaalla
14.7k12049
answered Jan 22 '17 at 6:25
Ahmed S. Attaalla
14.7k12049
14.7k12049
Hint: focus on the last property.
– vadim123
Jan 22 '17 at 6:29
1
Suppose $xlt ylt z$; does $d(x,z)le d(x,y)+d(y,z)$ ever hold for that funny definition of $d(x,y)?$
– bof
Jan 22 '17 at 6:34
add a comment |
Hint: focus on the last property.
– vadim123
Jan 22 '17 at 6:29
1
Suppose $xlt ylt z$; does $d(x,z)le d(x,y)+d(y,z)$ ever hold for that funny definition of $d(x,y)?$
– bof
Jan 22 '17 at 6:34
Hint: focus on the last property.
– vadim123
Jan 22 '17 at 6:29
Hint: focus on the last property.
– vadim123
Jan 22 '17 at 6:29
1
1
Suppose $xlt ylt z$; does $d(x,z)le d(x,y)+d(y,z)$ ever hold for that funny definition of $d(x,y)?$
– bof
Jan 22 '17 at 6:34
Suppose $xlt ylt z$; does $d(x,z)le d(x,y)+d(y,z)$ ever hold for that funny definition of $d(x,y)?$
– bof
Jan 22 '17 at 6:34
add a comment |
up vote
1
down vote
Ok. The problem is the triangle inequality, then:
Let $x,y,z in mathbb{R}$, so $$d(x,z) = (x+y-y+z)^2$$ $$d(x,z) = ((x-y)+(y-z))^2$$ $$qquad d(x,z) = d(x,y)+2(x-y)(y-z)+d(y,z)$$ $$d(x,z) geq d(x,y)+d(y,z)quad iffquad (x-y)(y-z)geq0$$
Then, suppose what $x=3$, $z=1$, $y=2$, that which does not meet the triangular inequality, therefore, $d$ is not a metric
answered Nov 13 at 1:34
Franck Panche
112
add a comment |
up vote
1
down vote
Ok. The problem is the triangle inequality, then:
Let $x,y,z in mathbb{R}$, so $$d(x,z) = (x+y-y+z)^2$$ $$d(x,z) = ((x-y)+(y-z))^2$$ $$qquad d(x,z) = d(x,y)+2(x-y)(y-z)+d(y,z)$$ $$d(x,z) geq d(x,y)+d(y,z)quad iffquad (x-y)(y-z)geq0$$
Then, suppose what $x=3$, $z=1$, $y=2$, that which does not meet the triangular inequality, therefore, $d$ is not a metric
answered Nov 13 at 1:34
Franck Panche
112
add a comment |
up vote
1
down vote
up vote
1
down vote
Ok. The problem is the triangle inequality, then:
Let $x,y,z in mathbb{R}$, so $$d(x,z) = (x+y-y+z)^2$$ $$d(x,z) = ((x-y)+(y-z))^2$$ $$qquad d(x,z) = d(x,y)+2(x-y)(y-z)+d(y,z)$$ $$d(x,z) geq d(x,y)+d(y,z)quad iffquad (x-y)(y-z)geq0$$
Then, suppose what $x=3$, $z=1$, $y=2$, that which does not meet the triangular inequality, therefore, $d$ is not a metric
answered Nov 13 at 1:34
Franck Panche
112
Ok. The problem is the triangle inequality, then:
Let $x,y,z in mathbb{R}$, so $$d(x,z) = (x+y-y+z)^2$$ $$d(x,z) = ((x-y)+(y-z))^2$$ $$qquad d(x,z) = d(x,y)+2(x-y)(y-z)+d(y,z)$$ $$d(x,z) geq d(x,y)+d(y,z)quad iffquad (x-y)(y-z)geq0$$
Then, suppose what $x=3$, $z=1$, $y=2$, that which does not meet the triangular inequality, therefore, $d$ is not a metric
answered Nov 13 at 1:34
Franck Panche
112
edited Nov 13 at 1:46
answered Nov 13 at 1:34
Franck Panche
112
answered Nov 13 at 1:34
Franck Panche
112
answered Nov 13 at 1:34
Franck Panche
112
112
add a comment |
add a comment |
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1
Let's say I want to travel from the point $0$ to the point $5$. If I go straight there, it will cost me $d(0,5)=25$ units. On the other hand, if I go one bit at a time, it will only cost me $d(0,1)+d(1,2)+dotsb+d(4,5)=5$ units.
– Akiva Weinberger
Jan 22 '17 at 8:41