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From looking at the properties of both what makes a function a metric and a function a norm, I'd gather that I'd have to create a metric that would not satisfy the scalar multiplication property of a norm (i.e. $||ax|| \ne |a|||x||$). So, I went with $d(x,y) = \sqrt{|x-y|}$, since then $\sqrt{|a||x-y|} \ne |a|\sqrt{|x-y|}$. Am I correct in my going about of doing this? If not, I'd prefer a hint as to how I should think about creating such a metric.

zodross
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Yes, this is correct. All metrics induced by a norm have to be homogeneous (scalar property) and hence counterexamples only work if they are not homogeneous themselves. Your example works just fine, you just need to verify the triangle inequality which works out nicely.

Also, the discrete metric is a standard example for a metric which is not induced by a norm. I recommend also the answers to this question: Not every metric is induced from a norm

user190080
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    Thank you very much sir/ma'm. I was actually looking over that just now, and the discrete metric was a very clear example of this. – zodross Sep 01 '18 at 02:14