Quiz 1 (Solution)

Date: MATH 1090-4 - Fall 2003

1. Answer the following statements with True or False (No justification is required!).
   1. $3\in \{1,0,-4,7\}$.

      False. $3$ is not an element of $\{1,0,-4,7\}$.

   2. $3\in \{x:x$    is an odd number$\}$.

      True. $3$ is an odd number and, so, belongs to $\{x:x$    is an odd number$\}$.

   3. $\ensuremath{\mathbb{Q}}$ is the set of all real numbers.

      False. $\ensuremath{\mathbb{Q}}$ is the set of all rational numbers. The set of all real numbers is $\ensuremath{\mathbb{R}}$.

   4. $\{1,4,\pi\}\subset \{-1,1,-2,3,\pi\}$.

      False. To be a subset ($\subset$) $4$ should be an element of $\{-1,1,-2,3,\pi\}$, and this is not the case.

   5. $\{1,2,3\}\cap\{-1,-2,3,4\} = \{3\}$, where $\cap$ denotes intersection.

      True. $3$ is the only element that is common to both sets.

   6. $\{1,2,3\}\cup\{-1,-2,3,4\} = \{3\}$, where $\cup$ denotes union.

      False. The union consists of all elements that belong to at least one of the sets. Thus, $\{1,2,3\}\cup\{-1,-2,3,4\} = \{1,2,3,-1,-2,4\}$.

2. Insert the proper sign ($<$,$>$,$=$):
   1. $\vert-7-4\vert\quad \framebox[1cm]{$=$} \quad \vert-7\vert + \vert 4\vert$.

      The left hand side is $\vert-7-4\vert = \vert-11\vert=11$ and the right hand side is $\vert-7\vert+4 = 7+4 =11$ and they agree.

   2. $(-2)\cdot (3-5) \quad \framebox[1cm]{$>$} \quad (-2) \cdot 3 -5$.

      The left hand side is $(-2)\cdot (3-5) = (-2)\cdot (-2) = 4$ and the right hand side is $(-2) \cdot 3 -5 = -6-5 = -11$. So, the corresponding sign is $>$.

3. Evaluate

   $$\frac{(-3)^2-2\cdot3+6}{4-2^2+3}$$

   
   

   $$\frac{(-3)^2-2\cdot3+6}{4-2^2+3} = \frac{9-6+6}{4-4+3} = \frac{9}{3} = 3.$$

   
   

4. Use your calculator to find $15.3 \cdot (1.23)^{11}$.

   From the calculator $(1.23)^{11} = 9.7489137$ so that $15.3 \cdot (1.23)^{11} = 149.1583796$.