ε={integer x:0<x≤12}\varepsilon = \{\text{integer } x : 0 < x \le 12\} A={prime numbers}A = \{\text{prime numbers}\} B={numbers that have at least 2 distinct factors}B = \{\text{numbers that have at least 2 distinct factors}\}

  1. (a)

    不列出元素,解释为什么 A 是 B 的真子集。

    答案

    ...

  2. (b)

    绘制维恩图并列出元素以说明上述信息。

    答案

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完整解法

17(a)

1. Show that all elements of $A$ are also elements of $B$.

A prime number has exactly two distinct factors, namely 11 and itself. Therefore every prime number has at least 22 distinct factors, so every element of AA is in BB. Hence A⊆BA \subseteq B.

2. Show that $A$ is not equal to $B$, so the subset is proper.

There are numbers in BB that are not prime, for example 44 has at least 22 distinct factors but is not prime. Hence BB contains at least one element not in AA, so A≠BA \ne B and AA is a proper subset of BB.

答案

Every prime number has exactly two distinct factors, 11 and itself, so every element of AA is in BB. Also, there is at least one number in BB that is not prime, e.g. 44 has at least two distinct factors but is not prime. Hence AA is a proper subset of BB.

17(b)

1. Write the universal set and identify the elements of $A$.

The universal set is
ε={1,2,3,4,5,6,7,8,9,10,11,12}.\varepsilon=\{1,2,3,4,5,6,7,8,9,10,11,12\}.
The prime numbers in this set are
A={2,3,5,7,11}.A=\{2,3,5,7,11\}.

2. Place elements correctly in the Venn diagram according to $A \subset B$.

Since AA is a proper subset of BB, draw AA entirely inside BB.

Numbers in BB but not in AA have at least 22 distinct factors but are not prime:
B∖A={4,6,8,9,10,12}.B\setminus A=\{4,6,8,9,10,12\}.
The only element not in BB is
ε∖B={1}.\varepsilon\setminus B=\{1\}.
So place 2,3,5,7,112,3,5,7,11 inside AA; place 4,6,8,9,10,124,6,8,9,10,12 inside BB but outside AA; place 11 outside BB but inside ε\varepsilon.

答案

Draw a rectangle labelled ε\varepsilon with a circle BB inside it and a smaller circle AA entirely inside BB.

A={2,3,5,7,11}A=\{2,3,5,7,11\}

B∖A={4,6,8,9,10,12}B\setminus A=\{4,6,8,9,10,12\}

ε∖B={1}\varepsilon\setminus B=\{1\}

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