GR9768 Q 56
GR9768 Q 56
Ques: For every set S and every metric d on s, which of the following is metric on S ?
A) 4+d
B) e^d 1
C) d  d
D) d^2
E) sqrt (d)
I could rule out A,B,C because it did not follow one of these conditions :
d(x,y)>=0 and d(x,y) = 0 iff x=y
d(x,y) = d(y,x)
It turned out for me that ans is d^2, but the correct ans is E
Can somebody explain ?
Thanks
A) 4+d
B) e^d 1
C) d  d
D) d^2
E) sqrt (d)
I could rule out A,B,C because it did not follow one of these conditions :
d(x,y)>=0 and d(x,y) = 0 iff x=y
d(x,y) = d(y,x)
It turned out for me that ans is d^2, but the correct ans is E
Can somebody explain ?
Thanks

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: GR9768 Q 56
Let S be points in R^2 and d be common distance, then d^2 does not satisfy the triangular inequality.
Re: GR9768 Q 56
speedychaos4 wrote:Let S be points in R^2 and d be common distance, then d^2 does not satisfy the triangular inequality.
I tried the triangular inequality but couldn't figure out why d^2 does not satisfies it.
Are we saying (xz)^2 > (xy)^2 + (yz)^2
Can you please explain ?
Thanks

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: GR9768 Q 56
by common distance in R^2 I mean $$d[(x_1,y_1),(x_2,y_2)]=\sqrt{(x_1x_2)^2+(y_1y_2)^2}$$mathQ wrote:speedychaos4 wrote:Let S be points in R^2 and d be common distance, then d^2 does not satisfy the triangular inequality.
I tried the triangular inequality but couldn't figure out why d^2 does not satisfies it.
Are we saying (xz)^2 > (xy)^2 + (yz)^2
Can you please explain ?
Thanks
therefore$$d^2[(x_1,y_1),(x_2,y_2)]=(x_1x_2)^2+(y_1y_2)^2$$
then, if the 3 points of R^2 happen to be the vertice of a rt. tri. it will not satisfy the inequality.
p.s. I'm not sure about my ans either since I'm new to topology.
Re: GR9768 Q 56
speedychaos4 wrote:by common distance in R^2 I mean $$d[(x_1,y_1),(x_2,y_2)]=\sqrt{(x_1x_2)^2+(y_1y_2)^2}$$mathQ wrote:speedychaos4 wrote:Let S be points in R^2 and d be common distance, then d^2 does not satisfy the triangular inequality.
I tried the triangular inequality but couldn't figure out why d^2 does not satisfies it.
Are we saying (xz)^2 > (xy)^2 + (yz)^2
Can you please explain ?
Thanks
therefore$$d^2[(x_1,y_1),(x_2,y_2)]=(x_1x_2)^2+(y_1y_2)^2$$
then, if the 3 points of R^2 happen to be the vertice of a rt. tri. it will not satisfy the inequality.
p.s. I'm not sure about my ans either since I'm new to topology.
The inequality for metric space says :
d(x,z) < = d(x,y) + d(y,z)
Well, if it rt. tr, the inequality will be satisfied. so d^2 should be the metric space. (for x, y z as 3 vertices of the triangle)
correct me if I am wrong

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: GR9768 Q 56
Oh!!! I'm sorry that I'm totally wrong about the concept...
The inequality for metric space says :
d(x,z) < = d(x,y) + d(y,z)
Well, if it rt. tr, the inequality will be satisfied. so d^2 should be the metric space. (for x, y z as 3 vertices of the triangle)
correct me if I am wrong
What about d(x,y)=xy defined on the real line?
then d^2(1,1)=4>2=d^2(1,0)+d^2(0,1) ?
is this one ok?
thanks for pointing out the mistakes.
Re: GR9768 Q 56
Np.speedychaos4 wrote:Oh!!! I'm sorry that I'm totally wrong about the concept...
The inequality for metric space says :
d(x,z) < = d(x,y) + d(y,z)
Well, if it rt. tr, the inequality will be satisfied. so d^2 should be the metric space. (for x, y z as 3 vertices of the triangle)
correct me if I am wrong
What about d(x,y)=xy defined on the real line?
then d^2(1,1)=4>2=d^2(1,0)+d^2(0,1) ?
is this one ok?
thanks for pointing out the mistakes.
yeah, this one looks right. However for it to satisfy sqrt(d) we need to take d .... in the choice it does not say Sqrt (d)

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: GR9768 Q 56
isn't d always nonnegative?mathQ wrote:Np.speedychaos4 wrote:Oh!!! I'm sorry that I'm totally wrong about the concept...
The inequality for metric space says :
d(x,z) < = d(x,y) + d(y,z)
Well, if it rt. tr, the inequality will be satisfied. so d^2 should be the metric space. (for x, y z as 3 vertices of the triangle)
correct me if I am wrong
What about d(x,y)=xy defined on the real line?
then d^2(1,1)=4>2=d^2(1,0)+d^2(0,1) ?
is this one ok?
thanks for pointing out the mistakes.
yeah, this one looks right. However for it to satisfy sqrt(d) we need to take d .... in the choice it does not say Sqrt (d)

 Posts: 10
 Joined: Thu Aug 05, 2010 2:29 pm
Re: GR9768 Q 56
Let D represent the candidate new metric:
A) D = 4+d
d(x, x) = 0 => D(x, x) = 4, so D is clearly not a metric as a point has a distance from itself.
B) e^d 1
Let d be the regular Euclidean metric and consider x = (0, 0), y = (0, 1), and z = (0, 2). It should be clear that going directly from x to z is "longer" than going from x to y to z, violating the triangle inequality.
C) d  d
Suppose x and y are distinct. Then d(x, y) = m > 0. But D(x, y) = m  m = 0, so two distinct points have no distance separating them, so D is not a metric.
D) d^2
Consider the same environment as in B. Going directly from 0 to 2 is a distance of 4, whereas going in two discrete steps is a distance of 2.
E) sqrt (d)
Fairly easy to convince yourself that this is reasonable and the triangle inequality probably holds by considering the graph.
A) D = 4+d
d(x, x) = 0 => D(x, x) = 4, so D is clearly not a metric as a point has a distance from itself.
B) e^d 1
Let d be the regular Euclidean metric and consider x = (0, 0), y = (0, 1), and z = (0, 2). It should be clear that going directly from x to z is "longer" than going from x to y to z, violating the triangle inequality.
C) d  d
Suppose x and y are distinct. Then d(x, y) = m > 0. But D(x, y) = m  m = 0, so two distinct points have no distance separating them, so D is not a metric.
D) d^2
Consider the same environment as in B. Going directly from 0 to 2 is a distance of 4, whereas going in two discrete steps is a distance of 2.
E) sqrt (d)
Fairly easy to convince yourself that this is reasonable and the triangle inequality probably holds by considering the graph.
Re: GR9768 Q 56
Take triangle inequalities in special position d_1=1 d_2=1 d_3=2 then d_1+d_2=d_3
Then check th triagle inequality condition for a) b) c) d) e) It's easy to see that only c) dde) sqrt(d) satisfy the triangle condition, and c) dd=0 for all elements hence is not a metric.
Then check th triagle inequality condition for a) b) c) d) e) It's easy to see that only c) dde) sqrt(d) satisfy the triangle condition, and c) dd=0 for all elements hence is not a metric.