You know that the discrete metric only takes values of 1 and 0. Now suppose it comes from some norm ||.||. Then for any α in the underlying field of your vector space and x,y∈X, you must have that
∥α(x−y)∥=|α|∥x−y∥.
But now ||x−y|| is a fixed number and I can make α arbitrarily large and consequently the discrete metric does not come from any norm on X.
Step-by-step explanation:
hope this helps
I think its 162 but i dont know so hope this helps sorry if it dont
The answer would be L because it is away from the rest ( outlier)
