- #1

- 532

- 31

## Main Question or Discussion Point

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