Absolute value notation removal

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SUMMARY

The discussion focuses on rewriting absolute value inequalities, specifically |x| < 1 and |x| > 1. The correct transformation for |x| < 1 is -1 < x < 1, while |x| > 1 can be expressed as x < -1 or x > 1. Participants clarify that the initial assumption of -1 > x > 1 is incorrect, as it misrepresents the relationship. The conversation emphasizes the importance of understanding the distinction between AND and OR statements in inequalities.

PREREQUISITES
  • Understanding of absolute value notation
  • Basic knowledge of inequalities
  • Familiarity with number line representation
  • Concept of negating statements in logic
NEXT STEPS
  • Study the properties of absolute value functions
  • Learn how to graph inequalities on a number line
  • Explore logical operators, specifically AND and OR statements
  • Practice rewriting complex inequalities involving absolute values
USEFUL FOR

Students studying algebra, educators teaching mathematical concepts, and anyone looking to improve their understanding of inequalities and absolute value notation.

kalpalned
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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?
 
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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.
 

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