- #1
bonfire09
- 249
- 0
The problem is "For every real number x, there exists integers a and b such that a≤x≤b
and b-a=1"
I am stuck on the first part of the proof. So in my proof I let a=x and b=x+1. Then x+1-x=1=b-a. But what I don't get why is that is safe for me assume here that a and b are not always going to be integers since x is a real number? I was thinking that even when the hypothesis is false the implication always true vacuously.
and b-a=1"
I am stuck on the first part of the proof. So in my proof I let a=x and b=x+1. Then x+1-x=1=b-a. But what I don't get why is that is safe for me assume here that a and b are not always going to be integers since x is a real number? I was thinking that even when the hypothesis is false the implication always true vacuously.