Mixing
Prove the triangle inequality in R^2
$begingroup$
I need to prove that $d(y,z) + d(x,y) geq d(x,z)$.
With $d(x,y) = sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}$.
I'm struggling to figure out how to work with the square roots especially when they're over an addition problem. Any help is appreciated.
Edit: to make this simpler. I think all I need to do is figure out how to add $sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}$ + $sqrt{(y_1-z_1)^2 + (y_2-z_2)^2}$. Then I can do the a + b proof method. But I just can't figure out how to add those equations so it ends up equalling $sqrt{(x_1-z_1)^2 + (x_2-z_2)^2}$.
real-analysis metric-spaces
asked Jan 18 at 2:04
user580909user580909
225
$endgroup$
|
show 3 more comments
$begingroup$
I need to prove that $d(y,z) + d(x,y) geq d(x,z)$.
With $d(x,y) = sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}$.
I'm struggling to figure out how to work with the square roots especially when they're over an addition problem. Any help is appreciated.
Edit: to make this simpler. I think all I need to do is figure out how to add $sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}$ + $sqrt{(y_1-z_1)^2 + (y_2-z_2)^2}$. Then I can do the a + b proof method. But I just can't figure out how to add those equations so it ends up equalling $sqrt{(x_1-z_1)^2 + (x_2-z_2)^2}$.
real-analysis metric-spaces
asked Jan 18 at 2:04
user580909user580909
225
$endgroup$
$begingroup$
Euclid gave a very nice proof in his Elements: en.wikipedia.org/wiki/Triangle_inequality.
$endgroup$
– Chris Custer
Jan 18 at 2:20
$begingroup$
In one dimension, the formula states that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides.
$endgroup$
– herb steinberg
Jan 18 at 2:21
$begingroup$
@herbsteinberg I need to do it in two dimensions though. That's my problem. With one dimension I was able to easily see that (x - y) + (y-z) = (x-z) but having to use the distance formula for two dimensions is making it much more difficult to show that as true.
$endgroup$
– user580909
Jan 18 at 2:28
$begingroup$
@user580909 I shouldn't have said one dimension. I simply meant the triangle inequality, which is best proven geometrically not algebraically.
$endgroup$
– herb steinberg
Jan 18 at 2:37
$begingroup$
@herbsteinberg I don't think I made this clear but I don't have a problem proving the triangle inequality. My problem is that my proof rests on d(x,y) + d(y,z) = d(x,z) in the second dimension. That's what I need to prove so that I can then use a proof involving abs(a) + abs(b) = abs(a+b)
$endgroup$
– user580909
Jan 18 at 2:41
|
show 3 more comments
$begingroup$
I need to prove that $d(y,z) + d(x,y) geq d(x,z)$.
With $d(x,y) = sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}$.
I'm struggling to figure out how to work with the square roots especially when they're over an addition problem. Any help is appreciated.
Edit: to make this simpler. I think all I need to do is figure out how to add $sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}$ + $sqrt{(y_1-z_1)^2 + (y_2-z_2)^2}$. Then I can do the a + b proof method. But I just can't figure out how to add those equations so it ends up equalling $sqrt{(x_1-z_1)^2 + (x_2-z_2)^2}$.
real-analysis metric-spaces
asked Jan 18 at 2:04
user580909user580909
225
$endgroup$
I need to prove that $d(y,z) + d(x,y) geq d(x,z)$.
With $d(x,y) = sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}$.
I'm struggling to figure out how to work with the square roots especially when they're over an addition problem. Any help is appreciated.
Edit: to make this simpler. I think all I need to do is figure out how to add $sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}$ + $sqrt{(y_1-z_1)^2 + (y_2-z_2)^2}$. Then I can do the a + b proof method. But I just can't figure out how to add those equations so it ends up equalling $sqrt{(x_1-z_1)^2 + (x_2-z_2)^2}$.
real-analysis metric-spaces
real-analysis metric-spaces
asked Jan 18 at 2:04
user580909user580909
225
asked Jan 18 at 2:04
user580909user580909
225
edited Jan 18 at 3:21
user580909
asked Jan 18 at 2:04
user580909user580909
225
asked Jan 18 at 2:04
user580909user580909
225
asked Jan 18 at 2:04
user580909user580909
225
225
$begingroup$
Euclid gave a very nice proof in his Elements: en.wikipedia.org/wiki/Triangle_inequality.
$endgroup$
– Chris Custer
Jan 18 at 2:20
$begingroup$
In one dimension, the formula states that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides.
$endgroup$
– herb steinberg
Jan 18 at 2:21
$begingroup$
@herbsteinberg I need to do it in two dimensions though. That's my problem. With one dimension I was able to easily see that (x - y) + (y-z) = (x-z) but having to use the distance formula for two dimensions is making it much more difficult to show that as true.
$endgroup$
– user580909
Jan 18 at 2:28
$begingroup$
@user580909 I shouldn't have said one dimension. I simply meant the triangle inequality, which is best proven geometrically not algebraically.
$endgroup$
– herb steinberg
Jan 18 at 2:37
$begingroup$
@herbsteinberg I don't think I made this clear but I don't have a problem proving the triangle inequality. My problem is that my proof rests on d(x,y) + d(y,z) = d(x,z) in the second dimension. That's what I need to prove so that I can then use a proof involving abs(a) + abs(b) = abs(a+b)
$endgroup$
– user580909
Jan 18 at 2:41
|
show 3 more comments
$begingroup$
Euclid gave a very nice proof in his Elements: en.wikipedia.org/wiki/Triangle_inequality.
$endgroup$
– Chris Custer
Jan 18 at 2:20
$begingroup$
In one dimension, the formula states that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides.
$endgroup$
– herb steinberg
Jan 18 at 2:21
$begingroup$
@herbsteinberg I need to do it in two dimensions though. That's my problem. With one dimension I was able to easily see that (x - y) + (y-z) = (x-z) but having to use the distance formula for two dimensions is making it much more difficult to show that as true.
$endgroup$
– user580909
Jan 18 at 2:28
$begingroup$
@user580909 I shouldn't have said one dimension. I simply meant the triangle inequality, which is best proven geometrically not algebraically.
$endgroup$
– herb steinberg
Jan 18 at 2:37
$begingroup$
@herbsteinberg I don't think I made this clear but I don't have a problem proving the triangle inequality. My problem is that my proof rests on d(x,y) + d(y,z) = d(x,z) in the second dimension. That's what I need to prove so that I can then use a proof involving abs(a) + abs(b) = abs(a+b)
$endgroup$
– user580909
Jan 18 at 2:41
$begingroup$
Euclid gave a very nice proof in his Elements: en.wikipedia.org/wiki/Triangle_inequality.
$endgroup$
– Chris Custer
Jan 18 at 2:20
$begingroup$
Euclid gave a very nice proof in his Elements: en.wikipedia.org/wiki/Triangle_inequality.
$endgroup$
– Chris Custer
Jan 18 at 2:20
$begingroup$
In one dimension, the formula states that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides.
$endgroup$
– herb steinberg
Jan 18 at 2:21
$begingroup$
In one dimension, the formula states that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides.
$endgroup$
– herb steinberg
Jan 18 at 2:21
$begingroup$
@herbsteinberg I need to do it in two dimensions though. That's my problem. With one dimension I was able to easily see that (x - y) + (y-z) = (x-z) but having to use the distance formula for two dimensions is making it much more difficult to show that as true.
$endgroup$
– user580909
Jan 18 at 2:28
$begingroup$
@herbsteinberg I need to do it in two dimensions though. That's my problem. With one dimension I was able to easily see that (x - y) + (y-z) = (x-z) but having to use the distance formula for two dimensions is making it much more difficult to show that as true.
$endgroup$
– user580909
Jan 18 at 2:28
$begingroup$
@user580909 I shouldn't have said one dimension. I simply meant the triangle inequality, which is best proven geometrically not algebraically.
$endgroup$
– herb steinberg
Jan 18 at 2:37
$begingroup$
@user580909 I shouldn't have said one dimension. I simply meant the triangle inequality, which is best proven geometrically not algebraically.
$endgroup$
– herb steinberg
Jan 18 at 2:37
$begingroup$
@herbsteinberg I don't think I made this clear but I don't have a problem proving the triangle inequality. My problem is that my proof rests on d(x,y) + d(y,z) = d(x,z) in the second dimension. That's what I need to prove so that I can then use a proof involving abs(a) + abs(b) = abs(a+b)
$endgroup$
– user580909
Jan 18 at 2:41
$begingroup$
@herbsteinberg I don't think I made this clear but I don't have a problem proving the triangle inequality. My problem is that my proof rests on d(x,y) + d(y,z) = d(x,z) in the second dimension. That's what I need to prove so that I can then use a proof involving abs(a) + abs(b) = abs(a+b)
$endgroup$
– user580909
Jan 18 at 2:41
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Show this first when one of the points is the origin. Then reduce all other cases to this case by using a suitable translation.
answered Jan 18 at 2:55
Sathya RengaswamiSathya Rengaswami
1
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 3:00
add a comment |
$begingroup$
The OP is attempting to define a metric on the set of points in the Cartesian plane, $mathbb R times mathbb R$. It can be safely assumed that before arriving at this point, they have been introduced to the idea of calculating the length of vectors.
In any case, I doubt there is direct algebraic technique showing
$tag 1 sqrt{(x_1-z_1)^2 + (x_2-z_2)^2} le sqrt{(x_1-y_1)^2 + (x_2-y_2)^2} + sqrt{(y_1-z_1)^2 + (y_2-z_2)^2} $
without taking a journey upon which you discover the Cauchy–Schwarz inequality.
So when the OP states
I think all I need to do is figure out...
they are bound for disappointment.
But what an opportunity to reflect on this amazing mathematical material. The inequality $text{(1)}$ is true for all $x_1, x_2,y_1,y_2,z_1,z_2 in mathbb R$, and it is still true if we 'only live on the line'. But so much abstract modern mathematical thought must be expended to bring it all (i.e. Cartesian Coordinate Space = Euclidean Space) to life.
ANSWER:
Using the C-S inequality,
$tag 2 (u_1 v_1 + u_2 v_2)^2 le (u_{1}^{2}+ u_{2}^{2}),(v_{1}^{2}+ v_{2}^{2})$
among other arguments, is the way to go if you want to show that $d(u,v)$ satisfies the triangle inequality.
answered Jan 18 at 14:45
CopyPasteItCopyPasteIt
4,2031628
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add a comment |
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2 Answers
2
active
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votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Show this first when one of the points is the origin. Then reduce all other cases to this case by using a suitable translation.
answered Jan 18 at 2:55
Sathya RengaswamiSathya Rengaswami
1
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 3:00
add a comment |
$begingroup$
Show this first when one of the points is the origin. Then reduce all other cases to this case by using a suitable translation.
answered Jan 18 at 2:55
Sathya RengaswamiSathya Rengaswami
1
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 3:00
add a comment |
$begingroup$
Show this first when one of the points is the origin. Then reduce all other cases to this case by using a suitable translation.
answered Jan 18 at 2:55
Sathya RengaswamiSathya Rengaswami
1
$endgroup$
Show this first when one of the points is the origin. Then reduce all other cases to this case by using a suitable translation.
answered Jan 18 at 2:55
Sathya RengaswamiSathya Rengaswami
1
answered Jan 18 at 2:55
Sathya RengaswamiSathya Rengaswami
1
answered Jan 18 at 2:55
Sathya RengaswamiSathya Rengaswami
1
answered Jan 18 at 2:55
Sathya RengaswamiSathya Rengaswami
1
1
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Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
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– dantopa
Jan 18 at 3:00
add a comment |
$begingroup$
Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 3:00
$begingroup$
Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 3:00
$begingroup$
Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 3:00
add a comment |
$begingroup$
The OP is attempting to define a metric on the set of points in the Cartesian plane, $mathbb R times mathbb R$. It can be safely assumed that before arriving at this point, they have been introduced to the idea of calculating the length of vectors.
In any case, I doubt there is direct algebraic technique showing
$tag 1 sqrt{(x_1-z_1)^2 + (x_2-z_2)^2} le sqrt{(x_1-y_1)^2 + (x_2-y_2)^2} + sqrt{(y_1-z_1)^2 + (y_2-z_2)^2} $
without taking a journey upon which you discover the Cauchy–Schwarz inequality.
So when the OP states
I think all I need to do is figure out...
they are bound for disappointment.
But what an opportunity to reflect on this amazing mathematical material. The inequality $text{(1)}$ is true for all $x_1, x_2,y_1,y_2,z_1,z_2 in mathbb R$, and it is still true if we 'only live on the line'. But so much abstract modern mathematical thought must be expended to bring it all (i.e. Cartesian Coordinate Space = Euclidean Space) to life.
ANSWER:
Using the C-S inequality,
$tag 2 (u_1 v_1 + u_2 v_2)^2 le (u_{1}^{2}+ u_{2}^{2}),(v_{1}^{2}+ v_{2}^{2})$
among other arguments, is the way to go if you want to show that $d(u,v)$ satisfies the triangle inequality.
answered Jan 18 at 14:45
CopyPasteItCopyPasteIt
4,2031628
$endgroup$
add a comment |
$begingroup$
The OP is attempting to define a metric on the set of points in the Cartesian plane, $mathbb R times mathbb R$. It can be safely assumed that before arriving at this point, they have been introduced to the idea of calculating the length of vectors.
In any case, I doubt there is direct algebraic technique showing
$tag 1 sqrt{(x_1-z_1)^2 + (x_2-z_2)^2} le sqrt{(x_1-y_1)^2 + (x_2-y_2)^2} + sqrt{(y_1-z_1)^2 + (y_2-z_2)^2} $
without taking a journey upon which you discover the Cauchy–Schwarz inequality.
So when the OP states
I think all I need to do is figure out...
they are bound for disappointment.
But what an opportunity to reflect on this amazing mathematical material. The inequality $text{(1)}$ is true for all $x_1, x_2,y_1,y_2,z_1,z_2 in mathbb R$, and it is still true if we 'only live on the line'. But so much abstract modern mathematical thought must be expended to bring it all (i.e. Cartesian Coordinate Space = Euclidean Space) to life.
ANSWER:
Using the C-S inequality,
$tag 2 (u_1 v_1 + u_2 v_2)^2 le (u_{1}^{2}+ u_{2}^{2}),(v_{1}^{2}+ v_{2}^{2})$
among other arguments, is the way to go if you want to show that $d(u,v)$ satisfies the triangle inequality.
answered Jan 18 at 14:45
CopyPasteItCopyPasteIt
4,2031628
$endgroup$
add a comment |
$begingroup$
The OP is attempting to define a metric on the set of points in the Cartesian plane, $mathbb R times mathbb R$. It can be safely assumed that before arriving at this point, they have been introduced to the idea of calculating the length of vectors.
In any case, I doubt there is direct algebraic technique showing
$tag 1 sqrt{(x_1-z_1)^2 + (x_2-z_2)^2} le sqrt{(x_1-y_1)^2 + (x_2-y_2)^2} + sqrt{(y_1-z_1)^2 + (y_2-z_2)^2} $
without taking a journey upon which you discover the Cauchy–Schwarz inequality.
So when the OP states
I think all I need to do is figure out...
they are bound for disappointment.
But what an opportunity to reflect on this amazing mathematical material. The inequality $text{(1)}$ is true for all $x_1, x_2,y_1,y_2,z_1,z_2 in mathbb R$, and it is still true if we 'only live on the line'. But so much abstract modern mathematical thought must be expended to bring it all (i.e. Cartesian Coordinate Space = Euclidean Space) to life.
ANSWER:
Using the C-S inequality,
$tag 2 (u_1 v_1 + u_2 v_2)^2 le (u_{1}^{2}+ u_{2}^{2}),(v_{1}^{2}+ v_{2}^{2})$
among other arguments, is the way to go if you want to show that $d(u,v)$ satisfies the triangle inequality.
answered Jan 18 at 14:45
CopyPasteItCopyPasteIt
4,2031628
$endgroup$
The OP is attempting to define a metric on the set of points in the Cartesian plane, $mathbb R times mathbb R$. It can be safely assumed that before arriving at this point, they have been introduced to the idea of calculating the length of vectors.
In any case, I doubt there is direct algebraic technique showing
$tag 1 sqrt{(x_1-z_1)^2 + (x_2-z_2)^2} le sqrt{(x_1-y_1)^2 + (x_2-y_2)^2} + sqrt{(y_1-z_1)^2 + (y_2-z_2)^2} $
without taking a journey upon which you discover the Cauchy–Schwarz inequality.
So when the OP states
I think all I need to do is figure out...
they are bound for disappointment.
But what an opportunity to reflect on this amazing mathematical material. The inequality $text{(1)}$ is true for all $x_1, x_2,y_1,y_2,z_1,z_2 in mathbb R$, and it is still true if we 'only live on the line'. But so much abstract modern mathematical thought must be expended to bring it all (i.e. Cartesian Coordinate Space = Euclidean Space) to life.
ANSWER:
Using the C-S inequality,
$tag 2 (u_1 v_1 + u_2 v_2)^2 le (u_{1}^{2}+ u_{2}^{2}),(v_{1}^{2}+ v_{2}^{2})$
among other arguments, is the way to go if you want to show that $d(u,v)$ satisfies the triangle inequality.
answered Jan 18 at 14:45
CopyPasteItCopyPasteIt
4,2031628
edited Jan 18 at 23:16
answered Jan 18 at 14:45
CopyPasteItCopyPasteIt
4,2031628
answered Jan 18 at 14:45
CopyPasteItCopyPasteIt
4,2031628
answered Jan 18 at 14:45
CopyPasteItCopyPasteIt
4,2031628
4,2031628
add a comment |
add a comment |
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$begingroup$
Euclid gave a very nice proof in his Elements: en.wikipedia.org/wiki/Triangle_inequality.
$endgroup$
– Chris Custer
Jan 18 at 2:20
$begingroup$
In one dimension, the formula states that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides.
$endgroup$
– herb steinberg
Jan 18 at 2:21
$begingroup$
@herbsteinberg I need to do it in two dimensions though. That's my problem. With one dimension I was able to easily see that (x - y) + (y-z) = (x-z) but having to use the distance formula for two dimensions is making it much more difficult to show that as true.
$endgroup$
– user580909
Jan 18 at 2:28
$begingroup$
@user580909 I shouldn't have said one dimension. I simply meant the triangle inequality, which is best proven geometrically not algebraically.
$endgroup$
– herb steinberg
Jan 18 at 2:37
$begingroup$
@herbsteinberg I don't think I made this clear but I don't have a problem proving the triangle inequality. My problem is that my proof rests on d(x,y) + d(y,z) = d(x,z) in the second dimension. That's what I need to prove so that I can then use a proof involving abs(a) + abs(b) = abs(a+b)
$endgroup$
– user580909
Jan 18 at 2:41