Listing elements or cardinality

Question

A. List the elements of each of the following (roster method). Use '...' notation for infinite sets

Let X = {0,1,2}

B = {z : z = 2x and x element X}

B = {z : z = x + y where x element X and y element X}

B = {z : z element X and -z element X}

 

I'm confused why is there a "z" and "y"? 

 

B. Determine the cardinality 

1) A = {x : z = 2x and x element X}

 

Letter B has no conditions or the "let x be something"

 

 

Please explain how did you got those answers. :)

 

 

Answer 1

It think it's a matter of understanding the notation.

 

"B = {z : z = 2x and x element X}" means that B is a set of elements called z, where is 2 times an element of X.

 

So what are all the numbers that are equal to 2 times an element of X?  X has three elements, 1, 2, and 3.  The numbers that are 2 times some element of X are 2*1, 2*2, and 2*3 -- in other words, 2, 4, and 6.  So B = {2, 4, 6}

 

In the next one ("B = {z : z = x + y where x element X and y element X}"), where they show the y that's confusing you, B is the set of all numbers that are the sum of two elements of X.  To read it across, it's like this:

 

"B is the set of all numbers z, such that z is the sum of two numbers x and y, where x and y are both elements of X"

 

...or to shorten it...

 

"B is the set of all numbers that are the sum of two numbers x and y, where x and y are both 0, 1, or 2"

 

So what numbers are the sum of two numbers that are both 0, 1, or 2?

 

0 + 0 = 0

0 + 1 = 1

0 + 2 = 2

1 + 0 = 1

1 + 1 = 2

1 + 2 = 3

2 + 0 = 2

2 + 1 = 3

2 + 2 = 4

 

The results are all 0, 1, 2, 3, or 4 -- so that gives is that B = {0, 1, 2, 3, 4}

 

For the last one, they want to know all numbers that are elements of X (i.e., 0, 1, or 2) where the number and its negative are both in X.  Well the only candidates are 0, 1, and 2. 

 

0 and -0 are both equal to 0, and 0 is in X.  So 0 is one of the elements of B. 

1 is in X but -1 is not

2 in in X but -2 is not

 

So the only number that qualifies according to the rule for B is 0.  So B = {0}

 

Now for part B...

 

A = {x : z = 2x and x element X}

 

A is the set all the numbers in that are in X such that some number z is equal to twice that number.

 

Well, for every number in X (1, 2, and 3), there is some number z that is twice that number.  Since there are no restrictions on z, every number in X qualifies.  That means that A = {1, 2, 3} so the cardinality of A is 3.

 

However, are you sure you wrote it correctly?  Is there some other condition that puts a restriction on z, such as that z also has to be in X?  If so that would change the result.  But based on the way you wrote it, the answer is 3.