<PRE><B>Question 715846</B>
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Presume real-valued functions and presume *[tex \LARGE x\ \in\ \text{dom}(r)] and *[tex \LARGE \text{dom}(s)]


a)  Yes.  The sum of two functions at a value is the sum of the two values of the functions.  The set of real numbers is commutative under addition.


b)  No.  The difference of two functions at a value is the difference of the two values of the function.  But the set of real numbers is NOT commutative under addition.  Counterexample: *[tex \LARGE 0\ -\ 1\ \neq\ 1\ -\ 0]


c)  No.  r(0) is the value of the function at zero.  If *[tex \LARGE r(x) = x + 1], then *[tex \LARGE r(0) = 0 + 1\ \neq\ 0]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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