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Oct3-06, 12:10 AM
P: 19
Quote Quote by Galileo
You've shown that: |x+2|<1 <=> -5<x-2<-3
This implies that |x-2|<5, simply because x-2<-3 means that surely x-2<5, so that -5<x-2<5, or |x-2|<5.
Alright this kinda clarifies my question, so -5 is like the min? Which would grant -5<x-2<5--since epsilon can never be E<0 - the absolute value is necessary? Sorry if I come off a bit slow, but it's just that I'm trying to justify every step so I completely understand the concept.