SOLUTION: set p consists of all the prime numbers among the first 100 positive integers. set e consists of all the even numbers among the first 100 positive integers. which of the following

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Question 1202601: set p consists of all the prime numbers among the first 100 positive integers. set e consists of all the even numbers among the first 100 positive integers. which of the following must be true?
1. Set P contains more numbers than Set E.
2, The intersection of the two sets contains exactly one number.
3. The union of the two sets contains exactly one number.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52905)   (Show Source): You can put this solution on YOUR website!
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Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Answers:
  1. False
  2. True
  3. False
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Explanation for Problem 1)

There are 100 values in the set {1,2,3,...,98,99,100}
Cut that in half to find there are 100/2 = 50 even numbers {2,4,6,...,96,98,100}
There are 50 elements in set E.

The other 50 elements are odd numbers {1,3,5,...,95,97,99} some of those are prime.
1 is not prime
3 is prime
5 is prime
7 is prime
9 is not prime
11 is prime
13 is prime
15 is not prime
17 is prime
19 is prime
21 is not prime
etc

Since we're kicking out values that aren't prime, we go from 50 elements to something smaller. Even when considering the even prime 2, the set of primes between 1 and 100 will not exceed 50 values.

This is enough to prove that set P has fewer items than set E.
Statement 1 is proven false.

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Explanation for Problem 2)

This statement is true because the intersections of these two sets consists of the singleton set {2}

Nearly everything in set P is odd. The single exception is the value 2.

The value 2 is both prime and an even number. It is the only even number prime.
Every other even number is composite because 2 is a factor.

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Explanation for Problem 3)

Consider this example
A = {1,2,3}
B = {3,7,8,9}

Then
A union B = {1,2,3,7,8,9}
We combine the two sets into one big set. Toss any duplicates.

Now we must find the union of:
E = {2,4,6,...,96,98,100}
P = set of primes between 1 and 100

It should be fairly clear that we cannot obtain a set of 1 item when unioning sets E and P
The number of items in set E u P will be at least 50 values. That 50 referring to the number of items in set E.
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