- #1
mishima
- 565
- 35
Rule:
Suppose a>0, then |x|>a if and only if x>a OR x<-a
So |x|>|x-1| becomes:
x>x-1 which is false (edit: or more accurately doesn't give the whole picture, it implies true for all x)
OR
x<-x+1
2x<1
x<1/2 which is false
Suppose a>0, then |x|>a if and only if x>a OR x<-a
So |x|>|x-1| becomes:
x>x-1 which is false (edit: or more accurately doesn't give the whole picture, it implies true for all x)
OR
x<-x+1
2x<1
x<1/2 which is false