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<span style="color:red">Instructor's note: This is really the second answer presented. It would be better if we could keep the first answer "as is", and put the correction as a second answer. Mistakes are nothing to be ashamed of! Making mistakes makes you learn! -pm</span>
 
<span style="color:red">Instructor's note: This is really the second answer presented. It would be better if we could keep the first answer "as is", and put the correction as a second answer. Mistakes are nothing to be ashamed of! Making mistakes makes you learn! -pm</span>
  
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[[User:Green26|(alec green)]]
  
 
All elements in the following union are distinct, therefore the union is a set.
 
All elements in the following union are distinct, therefore the union is a set.

Latest revision as of 12:48, 13 February 2013

Practice Problem on set operations


Consider the following sets:

$ \begin{align} S_1 &= \left\{ \frac{1}{2}, 1, 1.4, 2 \right\}, \\ S_2 & = \left\{ 0.\bar{9}, 1.40, \frac{42}{21}, 17\right\}. \\ \end{align} $

Write $ S_1 \cup S_2 $ explicitely. Is $ S_1 \cup S_2 $ a set?


Share your answers below

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Answer 1

Instructor's note: This is really the second answer presented. It would be better if we could keep the first answer "as is", and put the correction as a second answer. Mistakes are nothing to be ashamed of! Making mistakes makes you learn! -pm

(alec green)

All elements in the following union are distinct, therefore the union is a set.

$ S_1 \cup S_2 = \{ \frac{1}{2}, 0{\color{red}\not}.\bar{9}, 1, 1.4, 2, 17 \} $

Lecture 3.PNG ($ S_1 \cup S_2 $ represented by colored region.)

WOW! That's a VERY nicely written answer. Great work. You only missed one little (somewhat tricky) detail. Can you guess what it is? MATH MAJORS: Can you help him? -pm
Okay, answer above edited to account for the following:
$ \frac{1}{9} = 0.\bar{1} $
$ \frac{1}{9} * 9 = 0.\bar{9} $
$ \frac{1}{9} * 9 = 1 $
$ \therefore 0.\bar{9} = 1 $
Instructor's comment: There you go! -pm

Answer 2

The union of S1 and S2 is all the elements in the Venn diagram: in S1, S2, and in both S1 and S2.


Answer 3

Write it here.


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