Prove that x<-a or x>a for abs(x)>a where a>0.
Is my proof acceptable?
abs(x)>a
x^2>a^2
x^2-a^2>0
(x-a)(x+a)>0
Therefore, x<-a or x>a
Hi. Thanks for your reply.
What if the question is phrased this way :
Prove that abs(x)>a if and only if x>a or x<-a. (where a is any positive number).
Will the phrasing of the question in such a way change the required proof?
Hi. This is the provide solution. Is this correct? I feel that (i) should be : prove | x | > a instead of | x | > 0 . The same goes for (ii).
i) Prove | x | > 0 => x > a or x < a for any positive number, a.
For x greater or equal 0, | x | = x. | x | > a => x > a.
For x less than 0, | x | = -x. | x | > a => -x > a => x < -a.
ii) Prove x > a or x < a => | x | > 0 for any positive number, a.
a > 0 and x > a => | x | = x. So x > a | x | > a.
For a > 0, -a < 0 and x < -a => x < 0 => | x | = -x. So x < -a => -x > a => | x | > a.