For x and y in R, let d(x,y)=(|x-y|)/(1+|x-y|),prove that d defines a.bounded metric on R
1)
Positivity: d(x,y)≥0d(x,y)\ge0d(x,y)≥0 with equality if x=y
2)
Symmetry: d(x,y)=d(y,x)d(x,y)=d(y,x)d(x,y)=d(y,x)
∣x−y∣1+∣x−y∣=∣y−x∣1+∣y−x∣\frac{|x-y|}{1+|x-y|}=\frac{|y-x|}{1+|y-x|}1+∣x−y∣∣x−y∣=1+∣y−x∣∣y−x∣
3)
Triangle Inequality: d(x,y)≤d(x,z)+d(z,y)d(x,y)\le d(x,z)+d(z,y)d(x,y)≤d(x,z)+d(z,y)
d(x,y)=∣x−y∣1+∣x−y∣=∣(x−z)+(z−y)∣1+∣(x−z)+(z−y)∣≤∣(x−z)+(z−y)∣1+∣(x−z)+(z−y)∣=d(x,z)+d(z,y)d(x,y)=\frac{|x-y|}{1+|x-y|}=\frac{|(x-z)+(z-y)|}{1+|(x-z)+(z-y)|}\le \frac{|(x-z)+(z-y)|}{1+|(x-z)+(z-y)|}=d(x,z)+d(z,y)d(x,y)=1+∣x−y∣∣x−y∣=1+∣(x−z)+(z−y)∣∣(x−z)+(z−y)∣≤1+∣(x−z)+(z−y)∣∣(x−z)+(z−y)∣=d(x,z)+d(z,y)
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