Mathematics, 13.10.2020 23:01 vanleervanleer

Which of the following statements is true for all sets A, B and C? Give a proof or a counter example.

(a) A ⊆ ((A∩B)∪C).

(b) (A∪B)∩C = (A∩B)∪C.

(c) (A\B)∩C = (A∩C) \ (B∩C).

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Answer from: flamingtacos4026

(a) and (b) are not true in general. Refer to the explanations below for counterexamples.

It can be shown that (c) is indeed true.

Step-by-step explanation:

This explanation will use a lot of empty sets $\phi$ just to keep the counterexamples simple.

(a)

Note that $A \cap B$ can well be smaller than . It should be alarming that the question is claiming $A\!$ to be a subset of something that can be smaller than $\! A$ . Here's a counterexample that dramatize this observation:

Consider:

$A = \left\lbrace 1 \right\rbrace$ . $B = \phi$ (an empty set, same as $\left\lbrace \right\rbrace$ .) $C = \phi$ (another empty set.)

The intersection of an empty set with another set should still be an empty set:

$A \cap B = \left\lbrace 1\right\rbrace \cap \left\lbrace\right\rbrace = \left\lbrace\right\rbrace$ .

The union of two empty sets should also be an empty set:

$((A \cap B) \cup C) = \left\lbrace\right\rbrace \cup \left\lbrace\right\rbrace = \left\lbrace\right\rbrace$ .

Apparently, the one-element set $A = \left\lbrace 1 \right\rbrace$ isn't a subset of an empty set. $A \not \subseteq ((A\cap B) \cup C)$ . Contradiction.

(b)

Consider the same counterexample

$A = \left\lbrace 1 \right\rbrace$ . $B = \phi$ (an empty set, same as $\left\lbrace \right\rbrace$ .) $C = \left\lbrace 2 \right\rbrace$ (another empty set.)

Left-hand side:

$(A \cup B) \cap C = \left(\left\lbrace 1 \right\rbrace \cup \left\lbrace \right\rbrace\right) \cap \left\lbrace 2 \right\rbrace\right = \left\lbrace 1 \right\rbrace \cap \left\lbrace 2 \right\rbrace = \left\lbrace \right\rbrace$ .

Right-hand side:

$(A \cap B) \cup C = \left(\left\lbrace 1 \right\rbrace \cap \left\lbrace \right\rbrace\right) \cup \left\lbrace 2 \right\rbrace\right = \left\lbrace \right\rbrace \cup \left\lbrace 2 \right\rbrace = \left\lbrace 2 \right\rbrace$ .

Apparently, the empty set on the left-hand side $\left\lbrace \right\rbrace$ is not the same as the $\left\lbrace 2 \right\rbrace$ on the right-hand side. Contradiction.

(c)

Part one: show that left-hand side is a subset of the right-hand side.

Let be a member of the set on the left-hand side.

$x \in (A \backslash B) \cap C$ .

$\implies x\in A \backslash B$ and $x \in C$ (the right arrow here reads "implies".)

$\implies x \in A$ and $x \not\in B$ and $x \in C$ .

$\implies (x \in A\cap C)$ and $x \not\in B \cap C$ .

$\implies x \in (A \cap C) \backslash (B \cap C)$ .

Note that $x \in (A \backslash B) \cap C$ (set on the left-hand side) implies that $x \in (A \cap C) \backslash (B \cap C)$ (set on the right-hand side.)

Therefore:

$(A \backslash B) \cap C \subseteq (A \cap C) \backslash (B \cap C)$ .

Part two: show that the right-hand side is a subset of the left-hand side. This part is slightly more involved than the first part.

Let be a member of the set on the right-hand side.

$x \in (A \cap C) \backslash (B \cap C)$ .

$\implies x \in A \cap C$ and $x \not\in B \cap C$ .

Note that $x \not\in B \cap C$ is equivalent to:

$x \not \in B$ , OR $x \not\in C$ , ORboth $x \not\in B$ AND $x \not \in C$ .

However, $x \in A \cap C$ implies that $x \in A$ AND $x \in C$ .

The fact that $x \in C$ means that the only possibility that $x \not\in B \cap C$ is $x \not \in B$ .

To reiterate: if $x \not \in C$ , then the assumption that $x \in A \cap C$ would not be true any more. Therefore, the only possibility is that $x \not \in B$ .

Therefore, $x \in (A \backslash B)\cap C$ .

In other words, $x \in (A \cap C) \backslash (B \cap C) \implies x \in (A \backslash B)\cap C$ .

$(A \cap C) \backslash (B \cap C) \subseteq (A \backslash B)\cap C$ .

Combine these two parts to obtain: $(A \backslash B) \cap C = (A \cap C) \backslash (B \cap C)$ .

Answer from: Quest

Okay so for every triangle has three angles that always add up to 180 degrees. and to tell the difference between whether they are secure, right or obtuse is that acute: has one angle that is less than 90 degrees. right triangle means there is at least one angle that is exactly 90 degrees. and an obtuse angle has an angle that is bigger than 180 degrees. i’ll do #13 as an example: 13. 40 and 50 so if every triangle has 180 degrees 40+50= 90 degrees then180-90= 90 degrees so the third angle is 90 degrees and because this triangle has a 90 degree angle it means it is a right triangle

Answer from: Quest

ian gonna lie that is way to much

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