00:01
Let sets a and b be defined as follows.
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A is a set of numbers negative 23, negative 22, negative 21, negative 20, negative 19, negative 18, negative 17 and negative 16.
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B is a set of integers greater than or equal to negative 2 and less than or equal to 0.
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In part a we find the cardinologies of a and b and in part b we select true or false for the statements negative 12 belongs to a, negative 1 belongs to b, negative 20 belongs to a and 4 does not belong to b.
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So let's start with the cardinologies of a and b but before doing that we see here that set a is defined by extension given each and every element of the set a.
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But set b is defined with a property that is by compression.
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So set b must be defined by extension which you can do in this case because we have a finite number of elements.
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The set is defined as a set of integers.
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That's the first thing we got to take into account.
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We have all the integers numbers in set b and any element of b is greater than or equal to negative 2.
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That is the numbers on the line, the integer numbers on the line, the real line that are to the right of negative 2 including negative 2.
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But at the same time the numbers on set b are less than or equal to 0.
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So we have integer numbers between negative 2 and 0 including both negative 2 and both and 0.
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So we do a simple graph here.
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We have the integer numbers.
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We have negative 3, negative 2, that keep on to the left here, negative 1, 0, 1 and that keep to the right.
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As you can see here these are the only three numbers satisfying this property defined in set b.
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They are integer numbers negative 2, negative 1 and 0 and they are greater than or equal to negative 2.
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Negative 2 is equal to negative 2.
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Negative 1 is greater than negative 2 and 0 is greater than negative 2.
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And at the same time the three numbers are less than or equal to 0.
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Negative 2 is less than 0.
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Negative 1 is less than 0 and 0 is equal to 0.
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So we have three elements negative 2, negative 1 and 0 in the set b.
02:50
So set b by extension is negative 2, negative 1 and 0.
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And having both sets a and b given explicitly by extension then we can solve easily both parts a and b...