If d(x,y) is euclidean distance between x and y Prove that
d(x,y)>=0
if d(x,y)=0 than x=y
and d(x,y)=d(y,x)

Hi Velma,

The formula for d(x,y) has the form of the square root of a sum of squares. For example, in 2-dimensional euclidean space with x=(x1,x2) and y=(y1,y2), d(x,y)=SQR[ (x1-y1)2 + (x2-y2)2 ]

  1. The square of a number is greater than or equal to 0, the sum of non-negative numbers is non-negative, and the square root of a non-negative number is non-negative so d(x,y) >= 0
     
  2. If d(x,y) = 0 then we have a sum of squares equal to 0 (*) But squares are always non-negative so (*) means each square must be zero. Each square in the formula for d(x,y) represent the distance between x and y when in each of the co-ordinate directions. Since the distance between x and y in all co-ordinate directions is 0 it must be that x and y are at the same location in euclidean space. Therefore, x = y
     
  3. The formulas for d(x,y) and d(y,x) are only different in the subtraction of the co-ordinates of x and y, where a negative sign appears (i.e., they are the same in absolute value). But squaring eliminates the negative so the sum of squares are the same. Therefore, d(x,y)=d(y,x)
Cheers,
Paul
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