Derivatives and Integrals: An Annotated Discourse

Suppose \(n\) is even.
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Then there exists an \(m\) so that \(n = 2m\text{.}\)
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Then \(n\) is a prime number.
 #paired
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Then there exists an \(m\) so that \(n = 2m + 1\text{.}\)
 #paired
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Click the heels of your ruby slippers together three times.
 #distractor
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So we have the displayed equation:

\begin{equation*}
n = 2m + 0\text{.}
\end{equation*}
 
This is a superfluous second paragraph in this block.
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Thus \(n\equiv 0\mod 2\text{.}\)

n = 250
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primes = []
candidates = list(range(2,n))
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candidates = []
primes = list(range(2,n)) #paired
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primes = candidates + [p] #distractor
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while candidates:
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p = candidates[0]
primes.append(p)
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for nonprime in range(p, n, p):
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if nonprime in candidates:
candidates.remove(nonprime)
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print(primes)