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08 Jun 2011, 06:06
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Hi everyone, I had a step in a practice DS question with a step that i'm confused about...
"we have a + b + c < 0 < a + b – 1. Subtracting a + b from both parts, we have c < a + b < 1 and so c must be less than 1"
Now unless I'm missing something I thought you would get c<-a-b<-1 and then multiply by -1 which would flip the inequalities and end up with a -c in the first place. Thanks in advance for your help!
-Tom
"we have a + b + c < 0 < a + b – 1. Subtracting a + b from both parts, we have c < a + b < 1 and so c must be less than 1"
Now unless I'm missing something I thought you would get c<-a-b<-1 and then multiply by -1 which would flip the inequalities and end up with a -c in the first place. Thanks in advance for your help!
-Tom
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09 Jun 2011, 11:47
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tmccar1
You are correct.
Given: a + b + c < 0 < a + b – 1
Subtracting (a + b), you get
c < - (a + b) < -1
The point to note here is that since c is less than -1, it is certainly less than 1 too. You need to provide the question and the entire solution to see if this is what they meant.
