SOLUTION: 6. Given the following sets, select the statement below that is NOT true.
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
(Points : 2)
B ⊂ A
A &#
Algebra.Com
Question 476252: 6. Given the following sets, select the statement below that is NOT true.
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
(Points : 2)
B ⊂ A
A ⊂ C
B ⊆ A
C ⊆ A
C ⊆ B
thank you for your help.
Answer by Edwin McCravy(20059) (Show Source): You can put this solution on YOUR website!
Note the difference between "subset" and "proper subset".
⊆ and ⊂. Both symbols require that every member of the set
on the left of the symbol is also a member of the element on
the right of the symbol. However with ⊂, the set on the left
cannot contain ALL OF the elements of the set on the right of ⊂.
IOW, ⊆ and ⊂ are the same EXCEPT when the same set is on both
side of the symbol. "A ⊆ A" is true but "A ⊂ A" is false.
⊆ allows (but does not require) that the set on the left of ⊆
contain ALL the members of the set on the right of ⊆. The
symbol ⊂ does not allow that the set on the left contain ALL
the elements of the set on the right of it. IOW,
{t,o,p} ⊆ {p,o,t} is true
but
{t,o,p} ⊂ {p,o,t} is false
To put it another way:
The only "improper subset" is the set itself, because it's
not properly "sub" which means "less than".
---------------------------------------------------------
B ⊂ A
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
That is true because every member of of B = {r, i, s, e}
is also a member of A = {r, i, s, k, e, d} yet B is not
equal to A. So B is a PROPER subset of A
----------------------------------------------------------
A ⊂ C
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
That is false because not every member of of
A = {r, i, s, k, e, d} is a member of C = {s, i, r}, since
C has no k, e, or d. So B is a subset of A.
-----------------------------------------------------------
B ⊆ A
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
That is true because every member of of B = {r, i, s, e}
is also a member of A = {r, i, s, k, e, d}. So B is a PROPER
subset of A. B doesn't contain ALL of A
-----------------------------------------------------------
C ⊆ A
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
That is true because every member of of C = {s, i, r}
is also a member of A = {r, i, s, k, e, d}. So C is a PROPER
subset of A.
-----------------------------------------------------------
C ⊆ B
A = {r, i, s, k, e, d}, B = {r, i, s, e}, C = {s, i, r}
That is true because every member of of C = {s, i, r}
is also a member of B = {r, i, s, e}. So C is a PROPER subset
of B.
----------------------------------------------------------
So only A ⊂ C of the above statements is false.
Edwin
RELATED QUESTIONS
By how much does r - s exceed s - r ?
a (r-s)/(s-r) b 2(r-s) c -2 d 2(s-r) (answered by stanbon)
Find the union of the sets {a,b,c,d,e} and... (answered by stanbon)
P l e a s e h e l p m e s o l v e t h i s
f i n d t h e c a p a c i t y i n l i... (answered by rothauserc)
SOLVE THE FORMULA FOR THE INDICATED VARIABLE. CHOOSE FROM THE FOLLOWING ANSWERS.
I=N/R+s (answered by Fombitz)
I am working with sets. My sets are as follows:
U = {a, b, c, d, e, f, g, h, i, j, k,... (answered by solver91311)
a b c d e f g h i j k l m n o p q r (answered by ewatrrr)
file:///C:/Users/UCSD/Pictures/proof%20pic.jpg
Given angle b is congruent to angle c... (answered by ikleyn)
Prove the following three arguments to be valid using the method of Natural Deduction
A. (answered by solver91311)
Factor the following:
r(s) = 2s^2 + st - 2s -t
a. r(s) = 2(s+1)(s-1)
b. r(s) =... (answered by nabla,vleith)
I am trying to obtain the probability of randomly answering a multiple choice question... (answered by solver91311)