Absolute value notation removal

AI Thread Summary
The discussion focuses on rewriting absolute value inequalities, specifically |x| < 1 and |x| > 1. It is confirmed that |x| < 1 can be rewritten as -1 < x < 1. However, there is confusion about |x| > 1, with incorrect assumptions made about its representation. The correct interpretation involves recognizing that |x| > 1 translates to two separate inequalities: x < -1 or x > 1. The conversation emphasizes understanding the logical structure of inequalities when eliminating absolute value notation.
kalpalned
Messages
2
Reaction score
0

Homework Statement


Rewrite |x| < 1 and |x| > 1 by eliminating the absolute value sign

Homework Equations


|x| < 1 = -1 < x < 1
|x| > 1 = ?

The Attempt at a Solution


I know that |x| < 1 can be rewritten as -1 < x < 1 but I'm not sure about |x| > 1. Am I right to assume that |x| > 1 = -1 > x > 1?
 
Physics news on Phys.org
kalpalned said:
|x| > 1 = -1 > x > 1?
Not quite.
kalpalned said:
|x| < 1 = -1 < x < 1
You did this one correctly.
If you can't think about |x|>1 separately,then you can use your previous answer to find this one.

Plot the number line.
Any number on this line will either satisfy:
|x|>1 or |x| <1
There can't be any number not falling into any of these 2 classes.
So,if a number ain't in the first class,it's surely going to be in the second class (neglecting x=1,of course).
Getting what to do now?
 
kalpalned said:
Am I right to assume that |x| > 1 = -1 > x > 1?
The way you have this written is that -1 is greater than x AND x is greater than 1. Is that even possible?
Remember when you negate an AND statement, like -1<x<1 which is read -1 is less than x AND x is less than 1, you will get an OR statement.
 
Thread closed, as a previous account of the OP's was permanently banned.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Back
Top