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August 21st, 2018, 10:00 AM  #1 
Newbie Joined: Aug 2018 From: Utah Posts: 6 Thanks: 0  Dividing by an Absolute Value
Help. I cannot for the life of me find a tutorial on how to divide an integer by an absolute value. I am supposed to simplify this expression: 3 + ( 8 ) รท  4  
August 21st, 2018, 02:04 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
As 4 = 4, 3 + 8/4 = 3 + 2 = 5.

August 21st, 2018, 02:41 PM  #3 
Newbie Joined: Aug 2018 From: Utah Posts: 6 Thanks: 0 
It's seriously that simple? Wow. They made that look unnecessarily complicated. Thanks a bunch! So it's just the simplified answer in the bars converted into a positive integer? Is that always the case? 
August 21st, 2018, 04:05 PM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
$x < 0 \implies x = \ x > 0.$ $x \ge 0 \implies x = x \ge 0.$ That is the basic idea, but $3  y = y  3 \text { if } y > 3 \text { but } 3  y = 3  y \text { if } y \le 3.$ Notice that if y = 3, the absolute value is not a positive integer, but a nonnegative integer. More importantly, if you have an expression with a variable within the absolute value bars, you may have to deal with multiple cases. $z^2  6z  16 = z^2  6z  16 \text { if } z \le \ 2.$ $z^2  6z  16 = 16 + 6z  z^2 \text { if } \ 2 < z < 8.$ $z^2  6z  16 = z^2  6z  16 \text { if } z \ge 8.$  
August 21st, 2018, 04:18 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
As far as specific real numbers (not necessarily integers) are concerned, the bars have the effect of removing any negative sign, so 4 = 4 = 4, 1/2 = 1/2 = 1/2, and 0 = 0.


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