Answer to Question #287932 in Algebra for Reens

Question #287932

which of the following statements are true justify your answer



a) For any three set A,B,C in a universal set U,(A\B)×C=A×(B\C)



b) the operation of conjugation is closed on C



c) if f(x) and g(x) are polynomials for which f(0)= g(0) and f(t)= g(t) ,then f(x)=g(x)



d) the solution set of 2x= 1,y+5=x,x=y+3 is a singleton

1
Expert's answer
2022-02-07T17:28:20-0500

(a)

it is not true because A universal set is a set which contains all the elements or objects of other sets, including its own elements. It is usually denoted by the symbol ‘U’.

Suppose Set A consists of all even numbers such that, A = {2, 4, 6, 8, 10, …} and set B consists of all odd numbers, such that, B = {1, 3, 5, 7, 9, …}. The universal set U consists of all natural numbers, such that, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10,….}. Therefore, as we know, all the even and odd numbers are a part of natural numbers. Therefore, Set U has all the elements of Set A and Set B.

From the above example, we can see that the elements of sets A, B and C are altogether available in Universal set ‘U’. Also, if you observe, no elements in the universal set are repeated and all the elements are unique.


(b)

It's not closed on a complex number. It's closed in the set of complex numbers, C

C. Another way to say this is that the C (the set) is closed under the operation of complex conjugation. This property of an operation is called closure

What it means is that the result of applying the operation of conjugation to any complex number is another complex number, in the same set C

Consider the "reciprocal" operation on the set of non-zero real numbers. Again, the reciprocal of a non-zero real number is another non-zero real number, so this set is closed under the operation. But what if we took the set of non-zero integers? For any integer n

n other than 1 or −1, the reciprocal is not an integer, so this set is not closed under this operation.


(c)

No. because

Suppose f(x)= "x^2"

and g(x)= "x^3"

Thus f(0)=0; f(1)=1

g(0)=0; g(1)=1

but f(2)=4 and g(2)=8


(d)


no the given statement is not true because A singleton set is a set containing exactly one element. For example, {a}, {∅}, and { {a} } are all singleton sets (the lone member of { {a} } is {a}). The cardinality or size of a set is the number of elements it contains.


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