Question

Consider the sets A={n e P:n is odd} B = {n e P:n is prime) C = {4n +3:n € P} D = {x e R : x2 - 8x + 15 = 0} Which of these sets are subsets of which? Consider all 16 possibilities.

Answer 1

Answer:

$A\subset A, B\subset B,C\subset C,D\subset D$,$A\subset B$, $D\subset A$ and $D\subset B$.

Step-by-step explanation:

The given sets are

A={n ∈ P:n is odd}

B = {n ∈ P:n is prime)

C = {4n +3:n ∈ P}

D = {x ∈ R : x² - 8x + 15 = 0}

P is the set of prime numbers and R is the set of real numbers.

$x^2 - 8x + 15 = 0$

$x^2 - 5x-3x + 15 = 0$

$x(x-5)-3(x-5) = 0$

$(x-5)(x-3) = 0$

So, the elements of all sets are

A = {3, 5, 7, 11, 13, 17, 19, 23, ...}

B = {2, 3, 5 , 7, 11, 13, 17, 19, 23, ...}

C = {11, 15, 23, ...}

D = {3,5}

Each sets is a subset of it self. So,

$A\subset A, B\subset B,C\subset C,D\subset D$

All the elements of A lie in set B, so A is a subset of B.

$A\subset B$

Since $15\notin A$ and $15\notin B$, So, C is not the subset of A and B.

D has two elements, 3 and 5, Since $3,5\in A$ and $3,5\in B$, therefore

$D\subset A$ and $D\subset B$