B Absolute Value Inequality, |x|>|x-1|....where's my mistake?

[mishima]

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

 

[Charles Link]

On the right side of the equation, the absolute value of $x-1$ can change signs around the point $x=1$. For a problem like this, I like to break it into segments where the absolute values will always keep the same sign=e,g. $-\infty$ to $0$, $0$ to $+1$ , and $+1$ to $+\infty$, and evaluate the inequality for these 3 separate regions, and determining for each whether $|x|=+x$, or $|x|=-x$, and similarly for $|x-1|$.

 

[member 587159]

Your mistake is that you use a rule that is valid for a fixed value. You identify a with $|x-1|$ but this is not a fixed value, since it depends on x. Therefore, the rule you listed cannot be used. See @Charles Link answer to see how you need to tackle this problem.

 

[mishima]

How can |x-1| ever be not be > 0? Aren't absolute values always greater than 0?

Ah, I get it. You're right it could be the case that |x-1|=0, which would indeed break the stipulations of the rule. Thanks.

 

[Charles Link]

How can |x-1| ever be not be > 0? Aren't absolute values always greater than 0?

Ah, I get it. You're right it could be the case that |x-1|=0, which would indeed break the stipulations of the rule. Thanks.

It's not just a case of $|x-1|=0$. The problem is that $|x-1|$ can be equal to $x-1$ or it can be equal to $1-x$. $\\$ $|x-1 |$ is not simply some positive constant, so the rule does not apply. You can't just remove the absolute value signs on the right side of the inequality, and think that it won't have any effect. The rule you are trying to impose only had absolute value signs on one side of the inequality.