All of the following sets are subsets of positive integers.

$A = \{x\mid x\ \text{is divisible by 2}\} \\ B = \{x\mid x\ \text{is divisible by 4}\} \\ C = \{x\mid x\ \text{is divisible by 6}\}$

Below you will find what I think are the answers.

Two of the following six statements (A - F) are false. What statements are false?:

A. $A \subset B$ true my reasoning: A is a true subset to B, so A is not equal to B.

B. $C \subset A$ true my reasoning: C is a true subset to B, so C is not equal to B

C. $B \setminus A = \emptyset$ FALSE my reasoning: this almost means "everything in B belongs to A"

D. $A \cap B = B$ true my reasoning: the elements in A and B are the same as the elements in B, in A and B they are all multiples of 2, which they are in B as well.

E. $C \setminus B = \emptyset$ FALSE my reasoning: it means that all the elements in C are in B; in B they are multiples of 4, whilst in C they are multiples of 2 and 3, thus not true.

F. $B \cap C = \{x\mid x\ \text{is divisible by 12}\}$ true

Are my answers correct or not? If not, what are the correct answers, and why?

  • $\begingroup$ First, you should use set symbols around your definitions for $A$, $B$, and $C$, $A=\{x:x\text{ is divisible by }2\}$. Note also that it helps to use a colon instead of $|$ for such that when you're talking about divisibility to avoid confusion. It might help to explain why you believe that $A$ is true. $\endgroup$ – Michael Burr May 20 '15 at 17:08
  • $\begingroup$ Could you explain to us your reasoning behind your answers? This will help us look at your thought process and see if you have the right ideas. For all we know now, you may have just guessed on these. $\endgroup$ – TomGrubb May 20 '15 at 17:08
  • $\begingroup$ You really had to make the options (A), (B),(C) when the sets have those names? Ugh. Your answer to (A) means that every even number is divisible by $4$. $\endgroup$ – Thomas Andrews May 20 '15 at 17:09
  • $\begingroup$ If you want to prove $B \setminus A \neq \emptyset$, then you should find some $x$ that is in $B$, but not in $A$. $\endgroup$ – Arthur May 20 '15 at 17:12

B is a true subset of A: All integers divisible by 4 are also divisible by 2 but not the other way around.

C is a true subset to A as anything that is divisible by 6 must also be divisible by 2

As B is a subset of A, B without A is the empty set.

D: your answer is correct

E: your answer is correct

F: your answer is correct

  • $\begingroup$ Now I get A, B and C! With D/E/F: correct, do you mean that my answers are correct? $\endgroup$ – Kaedos May 20 '15 at 17:44
  • $\begingroup$ i meant your answer was correct $\endgroup$ – Wolfgang Brehm May 20 '15 at 17:57


Try writing out what each of $A,B,C$ look like explicitly.

$A = \{x~:~x~\text{is a positive integer divisible by 2}\} = \{2,4,6,8,10,12,\dots\}$

$B = \{x~:~x~\text{is a positive integer divisible by 4}\} = \{4,8,12,16,\dots\}$

$C = \{x~:~x~\text{is a positive integer divisible by 6}\} = \{6,12,18,24,\dots\}$

Remember also the definition of each of the symbols:

$A\subset B \Leftrightarrow (x\in A\Rightarrow x\in B)$

$x\in (B\setminus A)\Leftrightarrow (x\in B\wedge x\notin A)$

$x\in (A\cap B)\Leftrightarrow (x\in A \wedge x\in B)$

For the first statement $A\subset B$, this will be true if everything in $A$ is also in $B$. It will not be true if there is something in $A$ that is not in $B$. (I.e. is it true that every multiple of 2 is also a multiple of 4? Is 6 a multiple of 4?)

For the later statement, try proving the following statement:

$$E\subset F \Leftrightarrow E\cap F = E$$

and the similar statement for additional practice:

$$E\subset F \Leftrightarrow E\cup F = F$$


Let's just look at the first one. Unfortunately you got that one wrong.

$A\subset B$ means that every $x\in A$ must also satisfy $x\in B$. That is, every $x$ that is a multiple of $2$ is also a multiple of $4$. This is false; for example, consider $2$ itself. $2\in A$ but $2\notin B$. Hence $A\not\subset B$.

The general principle in solving such problems is to consider the elements of the sets. They have properties, which are given as the defining properties of the sets.


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