Prove that √3 is irrational (outline the usual contradiction method).
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Assume √3 = p/q in lowest terms. Then p² = 3q² so p is divisible by 3; write p = 3k. Then 9k² = 3q² ⇒ q² = 3k², so q is also divisible by 3. This contradicts lowest terms. Hence √3 is irrational.
The same pattern works for √2, √5, √7.
