anonymous
  • anonymous
Which of the following has a solution set of {x | x = 0}? (x + 1 < -1) ∩ (x + 1 < 1) (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) (x + 1 < 1) ∩ (x + 1 > 1) i got c
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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zzr0ck3r
  • zzr0ck3r
lets look at the first one x+1<-1 implies x<-2 and x+1<1 implies x<0 if we take the intersection we know 0 will not be there because it is no in both of our sets
zzr0ck3r
  • zzr0ck3r
do the same thing for part b x+1<=1 implies x<=0 x+1>=1 implies x>=0 what is the intersection here?
anonymous
  • anonymous
is it c tho

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zzr0ck3r
  • zzr0ck3r
x+1<1 implies x<0 so we know already it cant be c because we are taking the intersection of this set and another one, so if we want the solution {x|x=0} = {0} then 0 must be in both sets
ybarrap
  • ybarrap
You want the intersection where x=0 is the only possibility. Taking the intersection of the left inequality and the right individually by drawing it or listing out some numbers should make it clearer. If I told you that you that a ball as high or higher than the table and as low or lower than the table, what is the intersection of those two possibilities? The answer to this question is logically exactly like this description. Notice also if the endpoints of the range are included or not. If they are, then they might be included in the intersection. In the table analogy, I included the table's distance itself; otherwise, there would be no solution because the table can't be higher than it is and lower than it is at the same time.
phi
  • phi
(x + 1 < 1) ∩ (x + 1 > 1) is the same as (x < 0) ∩ (x>0) which says: all numbers that are *at the same time* less than 0 and greater than 0. In other words: all numbers that are *at the same time* both positive and negative.... Guess what? There are no such numbers.

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