Discussion
π and e are
Answer: D
Both \( \pi \) and \( e \) are transcendental: neither is a root of any polynomial equation with integer coefficients.
An algebraic number is one that is such a root - \( \sqrt{2} \) qualifies, since it satisfies \( x^{2} - 2 = 0 \). No such equation exists for \( \pi \) or \( e \); Hermite proved it for \( e \) in 1873 and Lindemann for \( \pi \) in 1882, and that second proof is what finally settled the impossibility of squaring the circle.
Every transcendental number is irrational, but the reverse fails: \( \sqrt{2} \) is irrational and still algebraic.
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