- #1
happyg1
- 308
- 0
Hello,
I'm working on this problem:
If t(P,Q)=max(|x1 - x2|,|y1 - y2|), show that t is a metric for the set of all ordered pairs of real numbers.
I have proved the first three parts of the definition of a metric
1) t(P,Q) >0
2) t(P,Q) =0 IFF P=Q
3) t(P,Q) = t(Q,P)
all not so hard.
I'm having trouble getting started on the 4th part:
4) t(P,Q) <= t(P,R) + t(R,Q)
My confusion lies in where, exactly, I'm trying to go.
I wrote: Let R=|z1-z2|
It appears that I need to go for something that looks like this:
max(|x1 - x2|,|y1 -y2|)<=max(|x1 - x2|,|z1-z2|)+max(|y1 -y2|,|z1-z2|)
So I did this:
t(P,Q)=max(|x1 - x2|,|y1 -y2|)=max(|x1-z1+z1-x2|,|y1+z2-z2+y2|)
now, I could group x1-z1 and z1-x2, together and do the same thing with the second part, but I don't see how this is going to get me any closer to proving that this is a metric. I know I'm just having a block on something. What I have done doesn't seem right, and I don't know which way to go.
Any nudges in the right direction will be greatly appreciated.
CC
I'm working on this problem:
If t(P,Q)=max(|x1 - x2|,|y1 - y2|), show that t is a metric for the set of all ordered pairs of real numbers.
I have proved the first three parts of the definition of a metric
1) t(P,Q) >0
2) t(P,Q) =0 IFF P=Q
3) t(P,Q) = t(Q,P)
all not so hard.
I'm having trouble getting started on the 4th part:
4) t(P,Q) <= t(P,R) + t(R,Q)
My confusion lies in where, exactly, I'm trying to go.
I wrote: Let R=|z1-z2|
It appears that I need to go for something that looks like this:
max(|x1 - x2|,|y1 -y2|)<=max(|x1 - x2|,|z1-z2|)+max(|y1 -y2|,|z1-z2|)
So I did this:
t(P,Q)=max(|x1 - x2|,|y1 -y2|)=max(|x1-z1+z1-x2|,|y1+z2-z2+y2|)
now, I could group x1-z1 and z1-x2, together and do the same thing with the second part, but I don't see how this is going to get me any closer to proving that this is a metric. I know I'm just having a block on something. What I have done doesn't seem right, and I don't know which way to go.
Any nudges in the right direction will be greatly appreciated.
CC