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+---- Thread: Intersection and union of sets....so confused (/Thread-Intersection-and-union-of-sets-so-confused)
Intersection and union of sets....so confused - ironheadjack - 08-22-2013
So I'm going through Aleks College algebra doing good, but then I get to this problem, and I have been stumped, perplexed, confused, and lost for the past 3 days trying to figure this out. I have read every article Bing gave me, tried using purplemath, mathway, and I am still lost. Something is just not clicking for me.
Could someone please walk me through this problem, before I pull all of my hair out.
Sets C and D are defined as:
C={y |y≥2}
D={y |y>6}
Write C∪D and C∩D, using interval notation. If set is empty, write Ã.
Any help is much appreciated.
Intersection and union of sets....so confused - Lindagerr - 08-22-2013
The union of two or more sets is a set containing all of the numbers in those sets So {1,2,3)u{3,4,5} ={1,2,3,4,5}
The intersection of two or more sets is a set containing only the members contained in every set. So {1,2,3}(intersection) {3,4,5}= {3}
Sorry too lazy to find symbols
So you set would have union y= or more then 2
and intersection y = or more then 7
I hope this makes sense
Intersection and union of sets....so confused - ironheadjack - 08-23-2013
Thank you but I'm still not getting it.
So your saying the union is y≥2
and the intersection is y≥7
Where did the 7 come from? Is the above answer written using interval notation? :confused:
Intersection and union of sets....so confused - Lindagerr - 08-23-2013
Set C ={y (The straight line means such that) y≥2} so the set C =y such that y≥ 2 so that set includes all numbers from 2 up
Set D= {y such that y>6} so the set D = includes all numbers above 6 ( I used seven but that does not account for intergers so it should be >6)
So CUD = {y≥2} because that set includes all of the numbers in C and all of the numbers that are in D but each number is only listed once no matter how many sets you have .
So C∩D ={y>6} because that set includes only the numbers that are in both sets.
If we had C= {y such that y≥2 and y≤9} that set would be {2,3,4,5,6,7,8,9} Assuming we are using only whole numbers.
If we had D= {y such that y>6 and y≤10} that set would be {7,8,9,10}
So with those C and D the CUD would be {y=2,3,4,5,6,7,8,9,10} because this is all the numbers that are in either set just written once.
and C∩D would be {y=7,8,9} because those numbers are in both sets
I hope this explains it better I found most of the keystroke shortcuts