## GR9768 Q 56

Forum for the GRE subject test in mathematics.
mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

### 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

speedychaos4
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.

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

### 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 (x-z)^2 > (x-y)^2 + (y-z)^2

Thanks

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: GR9768 Q 56

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 (x-z)^2 > (x-y)^2 + (y-z)^2

Thanks
by common distance in R^2 I mean $$d[(x_1,y_1),(x_2,y_2)]=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$
therefore$$d^2[(x_1,y_1),(x_2,y_2)]=(x_1-x_2)^2+(y_1-y_2)^2$$
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.

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

### Re: GR9768 Q 56

speedychaos4 wrote:
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 (x-z)^2 > (x-y)^2 + (y-z)^2

Thanks
by common distance in R^2 I mean $$d[(x_1,y_1),(x_2,y_2)]=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$
therefore$$d^2[(x_1,y_1),(x_2,y_2)]=(x_1-x_2)^2+(y_1-y_2)^2$$
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

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: GR9768 Q 56

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
Oh!!! I'm sorry that I'm totally wrong about the concept...
What about d(x,y)=|x-y| 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.

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

### Re: GR9768 Q 56

speedychaos4 wrote:

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
Oh!!! I'm sorry that I'm totally wrong about the concept...
What about d(x,y)=|x-y| 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.
Np.

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|)

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: GR9768 Q 56

mathQ wrote:
speedychaos4 wrote:

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
Oh!!! I'm sorry that I'm totally wrong about the concept...
What about d(x,y)=|x-y| 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.
Np.

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|)
isn't d always nonnegative?

Mathemagician
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.

Rus
Posts: 1
Joined: Thu Sep 01, 2011 2:33 am

### 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) d-|d|e) sqrt(d) satisfy the triangle condition, and c) d-|d|=0 for all elements hence is not a metric.