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Is this the correct way to show this function is a metric?
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Is this the correct way to prove that this is metric function ?
let $d_1:Bbb Z times Bbb Z rightarrow Bbb R^+_0$
$d(m,n):=|m-n|^3$
i. $d(x,y)geq 0$ with equality iff x=y
$|m-n|geq0$ because it is an absolute value
if $m neq n$ $|m-n|>0$ if m=n $|m-n|=0$
ii.$ d(m,n)=|m-n|^3=|n-m|^3=d(n,m)$
iii.
$d(m,p)=|m-p|^3$
$d(m,n)=|m-n|^3$
$d(n,p)=|n-p|^3$
$|m-p|^3leq |m|^3+|p|^3$
$|m-n|^3leq |m|^3+|n|^3$
$|n-p|^3leq |n|^3+|p|^3$
so
$|m-n|^3+|n-p|^3 =|m-n+n-p|^3=d(m,p)$
and so $d(m,p)leq d(m,n)+d(n,p)$
metric-spaces
asked Aug 6 at 13:00
exodius
772315
 |Â
show 2 more comments
up vote
2
down vote
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Is this the correct way to prove that this is metric function ?
let $d_1:Bbb Z times Bbb Z rightarrow Bbb R^+_0$
$d(m,n):=|m-n|^3$
i. $d(x,y)geq 0$ with equality iff x=y
$|m-n|geq0$ because it is an absolute value
if $m neq n$ $|m-n|>0$ if m=n $|m-n|=0$
ii.$ d(m,n)=|m-n|^3=|n-m|^3=d(n,m)$
iii.
$d(m,p)=|m-p|^3$
$d(m,n)=|m-n|^3$
$d(n,p)=|n-p|^3$
$|m-p|^3leq |m|^3+|p|^3$
$|m-n|^3leq |m|^3+|n|^3$
$|n-p|^3leq |n|^3+|p|^3$
so
$|m-n|^3+|n-p|^3 =|m-n+n-p|^3=d(m,p)$
and so $d(m,p)leq d(m,n)+d(n,p)$
metric-spaces
asked Aug 6 at 13:00
exodius
772315
1
Take $m=-1$ and $p=1$. Then $|m-p|^3 = 2^3 > 1^3 + 1^3 = |m|^3 + |p|^3$.
– Daniel Xiang
Aug 6 at 13:04
@DanielXiang can we just say that we know $|m-p| leq |m-n|+|n-p|$ because of the triangle inequality on $Bbb R$ and then use this fact to say their cubes must also follow this ?
– exodius
Aug 6 at 13:13
No, because $(|m-n|+|n-p|)^3$ is not $(|m-n|^3+|n-p|^3)$.
– Mark
Aug 6 at 13:15
No. Suppose $m-n = -1$ and $n-p = 1$. Then $|m-p|^3 > |m-n|^3 + |n-p|^3$.
– Daniel Xiang
Aug 6 at 13:15
2
@exodius in general, it is bad form to delete a question once it has been answered. It sends the feeling that the OP does not care for the time and effort that the community puts in to answer the question. In this case, the answer was simple enough that a comment was satisfactory for you, but try not to delete your questions once you have got an answer just for the sake of removing downvotes. A few downvotes will not matter in the long run if you continue to ask good questions (formatted properly, and showing your work, etc.). All the best.
– Brahadeesh
Aug 6 at 13:49
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is this the correct way to prove that this is metric function ?
let $d_1:Bbb Z times Bbb Z rightarrow Bbb R^+_0$
$d(m,n):=|m-n|^3$
i. $d(x,y)geq 0$ with equality iff x=y
$|m-n|geq0$ because it is an absolute value
if $m neq n$ $|m-n|>0$ if m=n $|m-n|=0$
ii.$ d(m,n)=|m-n|^3=|n-m|^3=d(n,m)$
iii.
$d(m,p)=|m-p|^3$
$d(m,n)=|m-n|^3$
$d(n,p)=|n-p|^3$
$|m-p|^3leq |m|^3+|p|^3$
$|m-n|^3leq |m|^3+|n|^3$
$|n-p|^3leq |n|^3+|p|^3$
so
$|m-n|^3+|n-p|^3 =|m-n+n-p|^3=d(m,p)$
and so $d(m,p)leq d(m,n)+d(n,p)$
metric-spaces
asked Aug 6 at 13:00
exodius
772315
Is this the correct way to prove that this is metric function ?
let $d_1:Bbb Z times Bbb Z rightarrow Bbb R^+_0$
$d(m,n):=|m-n|^3$
i. $d(x,y)geq 0$ with equality iff x=y
$|m-n|geq0$ because it is an absolute value
if $m neq n$ $|m-n|>0$ if m=n $|m-n|=0$
ii.$ d(m,n)=|m-n|^3=|n-m|^3=d(n,m)$
iii.
$d(m,p)=|m-p|^3$
$d(m,n)=|m-n|^3$
$d(n,p)=|n-p|^3$
$|m-p|^3leq |m|^3+|p|^3$
$|m-n|^3leq |m|^3+|n|^3$
$|n-p|^3leq |n|^3+|p|^3$
so
$|m-n|^3+|n-p|^3 =|m-n+n-p|^3=d(m,p)$
and so $d(m,p)leq d(m,n)+d(n,p)$
metric-spaces
asked Aug 6 at 13:00
exodius
772315
asked Aug 6 at 13:00
exodius
772315
asked Aug 6 at 13:00
exodius
772315
asked Aug 6 at 13:00
exodius
772315
772315
1
Take $m=-1$ and $p=1$. Then $|m-p|^3 = 2^3 > 1^3 + 1^3 = |m|^3 + |p|^3$.
– Daniel Xiang
Aug 6 at 13:04
@DanielXiang can we just say that we know $|m-p| leq |m-n|+|n-p|$ because of the triangle inequality on $Bbb R$ and then use this fact to say their cubes must also follow this ?
– exodius
Aug 6 at 13:13
No, because $(|m-n|+|n-p|)^3$ is not $(|m-n|^3+|n-p|^3)$.
– Mark
Aug 6 at 13:15
No. Suppose $m-n = -1$ and $n-p = 1$. Then $|m-p|^3 > |m-n|^3 + |n-p|^3$.
– Daniel Xiang
Aug 6 at 13:15
2
@exodius in general, it is bad form to delete a question once it has been answered. It sends the feeling that the OP does not care for the time and effort that the community puts in to answer the question. In this case, the answer was simple enough that a comment was satisfactory for you, but try not to delete your questions once you have got an answer just for the sake of removing downvotes. A few downvotes will not matter in the long run if you continue to ask good questions (formatted properly, and showing your work, etc.). All the best.
– Brahadeesh
Aug 6 at 13:49
 |Â
show 2 more comments
1
Take $m=-1$ and $p=1$. Then $|m-p|^3 = 2^3 > 1^3 + 1^3 = |m|^3 + |p|^3$.
– Daniel Xiang
Aug 6 at 13:04
@DanielXiang can we just say that we know $|m-p| leq |m-n|+|n-p|$ because of the triangle inequality on $Bbb R$ and then use this fact to say their cubes must also follow this ?
– exodius
Aug 6 at 13:13
No, because $(|m-n|+|n-p|)^3$ is not $(|m-n|^3+|n-p|^3)$.
– Mark
Aug 6 at 13:15
No. Suppose $m-n = -1$ and $n-p = 1$. Then $|m-p|^3 > |m-n|^3 + |n-p|^3$.
– Daniel Xiang
Aug 6 at 13:15
2
@exodius in general, it is bad form to delete a question once it has been answered. It sends the feeling that the OP does not care for the time and effort that the community puts in to answer the question. In this case, the answer was simple enough that a comment was satisfactory for you, but try not to delete your questions once you have got an answer just for the sake of removing downvotes. A few downvotes will not matter in the long run if you continue to ask good questions (formatted properly, and showing your work, etc.). All the best.
– Brahadeesh
Aug 6 at 13:49
1
1
Take $m=-1$ and $p=1$. Then $|m-p|^3 = 2^3 > 1^3 + 1^3 = |m|^3 + |p|^3$.
– Daniel Xiang
Aug 6 at 13:04
Take $m=-1$ and $p=1$. Then $|m-p|^3 = 2^3 > 1^3 + 1^3 = |m|^3 + |p|^3$.
– Daniel Xiang
Aug 6 at 13:04
@DanielXiang can we just say that we know $|m-p| leq |m-n|+|n-p|$ because of the triangle inequality on $Bbb R$ and then use this fact to say their cubes must also follow this ?
– exodius
Aug 6 at 13:13
@DanielXiang can we just say that we know $|m-p| leq |m-n|+|n-p|$ because of the triangle inequality on $Bbb R$ and then use this fact to say their cubes must also follow this ?
– exodius
Aug 6 at 13:13
No, because $(|m-n|+|n-p|)^3$ is not $(|m-n|^3+|n-p|^3)$.
– Mark
Aug 6 at 13:15
No, because $(|m-n|+|n-p|)^3$ is not $(|m-n|^3+|n-p|^3)$.
– Mark
Aug 6 at 13:15
No. Suppose $m-n = -1$ and $n-p = 1$. Then $|m-p|^3 > |m-n|^3 + |n-p|^3$.
– Daniel Xiang
Aug 6 at 13:15
No. Suppose $m-n = -1$ and $n-p = 1$. Then $|m-p|^3 > |m-n|^3 + |n-p|^3$.
– Daniel Xiang
Aug 6 at 13:15
2
2
@exodius in general, it is bad form to delete a question once it has been answered. It sends the feeling that the OP does not care for the time and effort that the community puts in to answer the question. In this case, the answer was simple enough that a comment was satisfactory for you, but try not to delete your questions once you have got an answer just for the sake of removing downvotes. A few downvotes will not matter in the long run if you continue to ask good questions (formatted properly, and showing your work, etc.). All the best.
– Brahadeesh
Aug 6 at 13:49
@exodius in general, it is bad form to delete a question once it has been answered. It sends the feeling that the OP does not care for the time and effort that the community puts in to answer the question. In this case, the answer was simple enough that a comment was satisfactory for you, but try not to delete your questions once you have got an answer just for the sake of removing downvotes. A few downvotes will not matter in the long run if you continue to ask good questions (formatted properly, and showing your work, etc.). All the best.
– Brahadeesh
Aug 6 at 13:49
 |Â
show 2 more comments
1 Answer
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oldest
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up vote
4
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The last part goes completely wrong as Daniels comment suggests.
You need to show
$$ |a-b|^3 leq |a-c|^3 +|c-b|^3 $$
which is not true. Note that if $x$ close to $0$, then $x^3$ is much and much closer to $0$. An easy counter example would thus be to take $a=0$, $b=2$ and $c=1$ to get
$$ 2^3 > 1^3 + 1^3 $$
which is thus exactly the other way around.
answered Aug 6 at 13:14
Stan Tendijck
1,277110
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The last part goes completely wrong as Daniels comment suggests.
You need to show
$$ |a-b|^3 leq |a-c|^3 +|c-b|^3 $$
which is not true. Note that if $x$ close to $0$, then $x^3$ is much and much closer to $0$. An easy counter example would thus be to take $a=0$, $b=2$ and $c=1$ to get
$$ 2^3 > 1^3 + 1^3 $$
which is thus exactly the other way around.
answered Aug 6 at 13:14
Stan Tendijck
1,277110
add a comment |Â
up vote
4
down vote
accepted
The last part goes completely wrong as Daniels comment suggests.
You need to show
$$ |a-b|^3 leq |a-c|^3 +|c-b|^3 $$
which is not true. Note that if $x$ close to $0$, then $x^3$ is much and much closer to $0$. An easy counter example would thus be to take $a=0$, $b=2$ and $c=1$ to get
$$ 2^3 > 1^3 + 1^3 $$
which is thus exactly the other way around.
answered Aug 6 at 13:14
Stan Tendijck
1,277110
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The last part goes completely wrong as Daniels comment suggests.
You need to show
$$ |a-b|^3 leq |a-c|^3 +|c-b|^3 $$
which is not true. Note that if $x$ close to $0$, then $x^3$ is much and much closer to $0$. An easy counter example would thus be to take $a=0$, $b=2$ and $c=1$ to get
$$ 2^3 > 1^3 + 1^3 $$
which is thus exactly the other way around.
answered Aug 6 at 13:14
Stan Tendijck
1,277110
The last part goes completely wrong as Daniels comment suggests.
You need to show
$$ |a-b|^3 leq |a-c|^3 +|c-b|^3 $$
which is not true. Note that if $x$ close to $0$, then $x^3$ is much and much closer to $0$. An easy counter example would thus be to take $a=0$, $b=2$ and $c=1$ to get
$$ 2^3 > 1^3 + 1^3 $$
which is thus exactly the other way around.
answered Aug 6 at 13:14
Stan Tendijck
1,277110
answered Aug 6 at 13:14
Stan Tendijck
1,277110
answered Aug 6 at 13:14
Stan Tendijck
1,277110
answered Aug 6 at 13:14
Stan Tendijck
1,277110
1,277110
add a comment |Â
add a comment |Â
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1
Take $m=-1$ and $p=1$. Then $|m-p|^3 = 2^3 > 1^3 + 1^3 = |m|^3 + |p|^3$.
– Daniel Xiang
Aug 6 at 13:04
@DanielXiang can we just say that we know $|m-p| leq |m-n|+|n-p|$ because of the triangle inequality on $Bbb R$ and then use this fact to say their cubes must also follow this ?
– exodius
Aug 6 at 13:13
No, because $(|m-n|+|n-p|)^3$ is not $(|m-n|^3+|n-p|^3)$.
– Mark
Aug 6 at 13:15
No. Suppose $m-n = -1$ and $n-p = 1$. Then $|m-p|^3 > |m-n|^3 + |n-p|^3$.
– Daniel Xiang
Aug 6 at 13:15
2
@exodius in general, it is bad form to delete a question once it has been answered. It sends the feeling that the OP does not care for the time and effort that the community puts in to answer the question. In this case, the answer was simple enough that a comment was satisfactory for you, but try not to delete your questions once you have got an answer just for the sake of removing downvotes. A few downvotes will not matter in the long run if you continue to ask good questions (formatted properly, and showing your work, etc.). All the best.
– Brahadeesh
Aug 6 at 13:49