This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MAT 125a, HW3 Solutions 1(a) Consider the set R n of ntuples of real numbers with d ( x , y ) = n X i =1  x i y i  . We wish to verify that d defines a metric on R n : (1) Since d ( x , y ) is a sum of nonnegative numbers we know that d ( x , y ) ≥ 0 for all x , y ∈ R n . (2) If x = y , their components are equal, and d ( x , y ) = 0. Conversely, suppose that d ( x , y ) = 0. By the nonnegativities of the absolute values in the sum, we must have that  x i y i  = 0 for i = 1 ,...,n (if one of them were positive then the sum would be positive). This means x i = y i for i = 1 ,...,n , and x = y . (3) That d ( x , y ) = d ( y , x ) follows from the symmetry of the “vanilla” absolute value on R . (4) To prove the triangle inequality for this metric, recall the triangle inequality for the “vanilla” absolute value on R :  a b  ≤  a c  +  c b  for all real numbers a,b and c . For any points x , y , z ∈ R n we have d ( x , y ) = n X i =1  x i y i  ≤ n X i =1 (  x i z i  +  z i y i  ) = d ( x , z ) + d ( z , y ) where the inequality is just the application of the ”vanilla” triangle inequality n times....
View
Full Document
 Fall '10
 Mulase
 Topology, Real Numbers, Negative Numbers, Metric space, Topological space, Rn of ntuples

Click to edit the document details