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===Answer 1===
 
===Answer 1===
<math>
+
 
 
No, because a set must have unique elements; sin(t+pi/2) is basically cos(t).
 
No, because a set must have unique elements; sin(t+pi/2) is basically cos(t).
 
The union of both sets is a set with elements from both S1 and S2.
 
The union of both sets is a set with elements from both S1 and S2.
 
S1 U S2 = {sin(t),cos(t),sin(t/2),sin(t+pi/2)}
 
S1 U S2 = {sin(t),cos(t),sin(t/2),sin(t+pi/2)}
<math>
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===Answer 2===
 
===Answer 2===
 
Write it here.
 
Write it here.

Revision as of 21:15, 9 January 2013

Practice Problemon set operations


Consider the following sets:

$ \begin{align} S_1 &= \left\{ \sin (t), \cos (t)\right\}, \\ S_2 & = \left\{ \sin (\frac{t}{2}), \sin (t+\frac{\pi}{2})\right\}. \\ \end{align} $

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


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Answer 1

No, because a set must have unique elements; sin(t+pi/2) is basically cos(t). The union of both sets is a set with elements from both S1 and S2. S1 U S2 = {sin(t),cos(t),sin(t/2),sin(t+pi/2)}

Answer 2

Write it here.

Answer 3

Write it here.


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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

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