October 29th, 2016, 03:50 PM  #1 
Member Joined: Oct 2016 From: New York Posts: 49 Thanks: 13 Math Focus: Analysis and Differential Geometry  Question about a proof
I am having difficulty understand why 0<=t<1. The proof says, "positive real numbers t such that.." Earlier (when it states the theorem) it also says x >0, so how is it possible that t is 0? Shouldn't it be 0<t<1? By the way this is my first time asking a question on here, so I don't know if my file got attached correctly. 
October 30th, 2016, 01:54 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,216 Thanks: 493 
You are correct, but it doesn't matter as far as the proof is concerned.

October 31st, 2016, 04:49 AM  #3 
Member Joined: Oct 2016 From: New York Posts: 49 Thanks: 13 Math Focus: Analysis and Differential Geometry 
Okay, thank you. I understand the proof still works. I just wanted clarification on that part.


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