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(Solved): (5 points) Consider the following "proof": Theorem: \pi is an algebraic number Proof: The number \p ...
(5 points) Consider the following "proof": Theorem:
\pi is an algebraic number Proof: The number
\pi obviously satisfies the quadratic polynomial
P(x)=x^(2)-\pi x. Hence, by definition,
\pi is algebraic. Q.E.D. Of course, something's wrong here because
\pi is not algebraic. Where's the logical flaw?
