Answer :

tury160391

Answer:

a) [tex]\neg \forall x :x>5 \equiv \forall x:x\leq 5[/tex]

b) [tex]\neg [x^2+2x+1=0]\equiv x^2+2x+1\neq0[/tex] or the set [tex]\{x:x\neq-1\}[/tex]

Step-by-step explanation:

First, notice that in both cases we have to sets:

a) is the set of all real numbers which are higher than 5 and in

b) the set is the solution of the equation [tex]x^2+2x+1=0[/tex] which is the set [tex]x=-1[/tex]

De Morgan's Law for set states:

[tex]\overline{\rm{A\cup B}} = \overline{\rm{A}} \cap \overline{\rm{B}}\\[/tex], being [tex]\overline{\rm{A}}[/tex] and [tex]\overline{\rm{B}}[/tex] are the complements of the sets [tex]A[/tex] and [tex]B[/tex]. [tex]\cup[/tex] is the union operation and [tex]\cap[/tex] the intersection.

Thus for:

a) [tex]\neg \forall x :x>5 \equiv \forall x:x\leq 5[/tex]. Notice that [tex]\forall x:x\leq 5[/tex] is the complement of the given set.

b) [tex]\neg [x^2+2x+1=0]\equiv x^2+2x+1\neq0[/tex] which is the set [tex]B = \{x:x\neq-1\}[/tex]

Other Questions