**Proposition.** For 𝑛 ∈ ℕ ,If √𝑛 ∉ ℕ, then √𝑛 ∉ ℚ.
**Proof.** Suppose that √𝑛 ∉ ℕ but √𝑛 ∈ ℚ. Let 𝑘 ∈ ℕ be such that 𝑘 < √𝑛 < k+1 and let
ℓ := 𝑚𝑖𝑛 { m ∈ ℕ : 0 <m, m√n ∈ ℕ }.
Then 0 < ℓ (√n - k) < ℓ and ℓ (√n - k) ∈ ℕ. This is a contradiction to the minimality of ℓ. □
