S4 E Maths Textbook Solutions >> New Discovering Mathematics 4A Chapter 1 Solutions >>
Chapter 1 Try it yourself 1-19
Please email me if you spot any mistakes or have any questions.
Solutions
Click to display or to hide
(a)
\begin{align} C & = \{ \text{s, t, u, d, e, n} \} \end{align}
(b)
\begin{align} n(C) & = 6 \end{align}
(c)(i)
$$ \text{Yes} $$
(c)(ii)
$$ \text{No} $$
(a)
\begin{align} B & = \{ \text{Jan, Feb, Mar, Apr, May, Jun} \} \end{align}
(b)
\begin{align} n(B) & = 6 \end{align}
(c)
\begin{align} \text{June} \in B \end{align}
(a)
\begin{align} R & = \{ 3, 5, 7, 11, 13, 17, 19 \} \end{align}
(b)
\begin{align} S & = \{ 5, 10, 15, 20 \} \\ \\ n(S) & = 4 \end{align}
(c)
\begin{align} \text{Yes, since 5 is a prime number and a multiple of 5} \end{align}
\begin{align} C & = \{ \text{s, t, r, e, d } \} \\ \\ D & = \{ \text{d, e, s, r, t } \} \\ \\ \therefore C & = D \end{align}
(a)
\begin{align}
x^2 - 2x - 3 & = 0 \\
(x + 1)(x - 3) & = 0
\end{align}
\begin{align}
x + 1 & = 0 && \text{ or } & x - 3 & =0 \\
x & = -1 &&& x & = 3
\end{align}
\begin{align}
P & = \{ -1, 3 \}
\end{align}
(b)(i)
\begin{align} Q & \not \subset P \end{align}
(b)(ii)
\begin{align} P & \subset R \end{align}
(c)(i)
\begin{align} S & = \{ -1, 0, 1, 2, 3, 4 \} \end{align}
(c)(ii)
\begin{align} n(S) & = 6 \end{align}
\begin{align} M & = \{ 4, 8, 12, 16, 20, 24, ... \} \\ \\ N & = \{ 12, 24, ... \} \\ \\ \text{Every multiple of 12 is} & \text{ a multiple of 4 (but not vice versa)} \\ \\ \text{Since every element in} & \text{ } N \text{ is in } M \text{ and there are more elements in } M \text{ than in } N, N \subset M \end{align}
(a)
\begin{align} \xi & = \{ x: x \text{ is a tool} \} \end{align}
(b)
\begin{align} \xi & = \{ x: x \text{ is an odd number} \} \end{align}
(a)
\begin{align} & \text{Since } \emptyset \text{ has no elements, it is a proper subset of } B \\ \\ & \text{Since every element in } B \text{ is in } \xi \text{ and } B \ne \xi, B \subset \xi \\ \\ & \phantom{000000} \therefore \emptyset \subset B \subset \xi \end{align}
(b)
\begin{align} \{ \}, \{ 1 \}, \{ 2 \}, \{ 4 \}, \{ 1, 2 \}, \{ 1, 4 \}, \{ 2, 4 \} \end{align}
(a)(i)
\begin{align} P & = \{ 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 \} \end{align}
(a)(ii)
\begin{align} Q & = \{ 6, 12, 18 \} \end{align}
(b)(i)
\begin{align} \text{True, since there are no elements in } \emptyset \end{align}
(b)(ii)
\begin{align} \text{False, since every element in } Q \text{ is in } P \text{ and } Q \ne P \end{align}
(b)(iii)
\begin{align} \text{True, since } P & = \{ 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 \} \end{align}
(b)(iv)
\begin{align} \text{True, since every element in } \{ 12, 18 \} \text{ is in } Q \text{ and } \{ 12, 18 \} \ne Q \end{align}
(a)
\begin{align} B & = \{ 6, 9, 12 \} \\ \\ B' & = \{ 4, 5, 7, 8, 10, 11, 13, 14 \} \end{align}
(b)
(c)
\begin{align} B' \text{ is the set of integers between 4 and 14 inclusive and are not multiples of 3} \end{align}
(a)
\begin{align} R & = \{ 25, 36 \} \phantom{000000} [5^2 , 6^2] \\ \\ S & = \{ 26, 28, 30, 32, 34, 36 \} \end{align}
(b)
\begin{align} \text{No, since } 25 \notin S \end{align}
(c)(i)
\begin{align} R' & = \{ 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 \} \end{align}
(c)(ii)
\begin{align} S' & = \{ 25, 27, 29, 31, 33, 35 \} \end{align}
(a)
\begin{align} P \cap Q & = \{ \} \end{align}
(b)
\begin{align} P \cap R & = \{ 6, 8 \} \end{align}
(c)
\begin{align} Q \cap R & = \{ 4 \} \end{align}
(a)
(b)(i)
(b)(ii)
\begin{align} P' \cap Q & = \{ e \} \end{align}
(c)
$$ \therefore P' \cap Q \ne P \cap Q' $$
(a)
\begin{align} C \cup D & = \{ \text{b, r, e, a, t, h, d, i, w} \} \end{align}
(b)
\begin{align} \text{Yes, as all elements in } D \text{ is in } C \cup D \text{ and } D \ne (C \cup D) \end{align}
(a)
(a)
$$ \therefore A \cup B' \ne A' \cup B $$
(a)
\begin{align} ( A \cup B)' \end{align}
(b)
\begin{align} P \cup Q' \end{align}
(a)
\begin{align} \xi & = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \} \\ \\ M & = \{ 2, 3, 5, 7, 11 \} \\ \\ N & = \{ 1, 2, 3, 5, 6, 10 \} \end{align}
(b)
(c)(i)
\begin{align} M' \cap N & = \{ 1, 6, 10 \} \end{align}
(c)(ii)
\begin{align} M \cup N' & = \{ 2, 3, 5, 7, 11, 4, 8, 9 \} \end{align}