How to Prove Inequality for Equivalent Distances?

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SUMMARY

This discussion focuses on proving the relationship between two equivalent distances, d_1 and d_2, under the condition of trichotomy. It establishes that if d_1(a,b) < d_1(c,d), then it must follow that d_2(a,b) < d_2(c,d) due to the equivalence definition. The proof hinges on the logical implications of the trichotomy principle, which states that for any two real numbers, one of three conditions must hold: equality or one being less than the other. The discussion emphasizes the necessity of understanding these implications to solidify the proof.

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mnb96
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Hello,
two distances d_1 and d_2 are said to be equivalent if for any two pairs (a,b) and (c,d)

d_1(a,b)=d_1(c,d) \Leftrightarrow d_2(a,b)=d_2(c,d)

How can I (dis)prove that:

d_1(a,b)&lt;d_1(c,d) \Rightarrow d_2(a,b)&lt;d_2(c,d)
 
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Use "trichotomy". For any two real numbers, x and y, one and only one of these must apply:
1) x= y
2) x< y
3) y< x.

If d_1(a, b)&lt; d_1(c, d) then it is NOT possible that d_1(a,b)= d_1(c, d) nor that d_1(c, d)&gt; d_1(a, b) which implies the same for d_2(a, b) and d_2(c, d).
 
There is still something bugging my mind.
I think the part of your proof that is giving me troubles is this:
HallsofIvy said:
...which implies the same for d_2(a, b) and d_2(c, d).

How did you prove that "it implies the same for d2(a,b) and d2(c,d)"

Thanks in advance.
 
Last edited:

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