Update Sun Nov 24 11:16:32 EST 2024

Math2001/Homework/hw9.lean

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+/- Copyright (c) Heather Macbeth, 2023-4.  All rights reserved. -/
+import Mathlib.Data.Real.Basic
+import Library.Theory.InjectiveSurjective
+import Library.Basic
+import AutograderLib
+
+math2001_init
+set_option pp.funBinderTypes true
+
+open Function
+
+
+/-! # Homework 9
+
+Don't forget to compare with the text version,
+https://github.com/hrmacbeth/math2001/wiki/Homework-9,
+for clearer statements and any special instructions. -/
+
+
+/- Problem 1: prove one of these, delete the other -/
+
+@[autograded 4]
+theorem problem1a : Surjective (fun (x : ℝ) ↦ 2 * x) := by
+  sorry
+
+@[autograded 4]
+theorem problem1b : ¬ Surjective (fun (x : ℝ) ↦ 2 * x) := by
+  sorry
+
+
+/- Problem 2: prove one of these, delete the other -/
+
+@[autograded 4]
+theorem problem2a : Surjective (fun (x : ℤ) ↦ 2 * x) := by
+  sorry
+
+@[autograded 4]
+theorem problem2b : ¬ Surjective (fun (x : ℤ) ↦ 2 * x) := by
+  sorry
+
+
+/- Problem 3: prove one of these, delete the other -/
+
+@[autograded 4]
+theorem problem3a : ∀ (f : ℚ → ℚ), Injective f → Injective (fun x ↦ f x + 1) := by
+  sorry
+
+@[autograded 4]
+theorem problem3b : ¬ ∀ (f : ℚ → ℚ), Injective f → Injective (fun x ↦ f x + 1) := by
+  sorry
+
+
+/- Problem 4: prove one of these, delete the other -/
+
+@[autograded 4]
+theorem problem4a : Bijective (fun (x : ℝ) ↦ 3 - 2 * x) := by
+  sorry
+
+@[autograded 4]
+theorem problem4b : ¬ Bijective (fun (x : ℝ) ↦ 3 - 2 * x) := by
+  sorry
+
+
+/- Problem 5: prove one of these, delete the other -/
+
+@[autograded 5]
+theorem problem5a :
+    Injective (fun ((x, y, z) : ℝ × ℝ × ℝ) ↦ (x + y + z, x + 2 * y + 3 * z)) := by
+  sorry
+
+@[autograded 5]
+theorem problem5b :
+    ¬Injective (fun ((x, y, z) : ℝ × ℝ × ℝ) ↦ (x + y + z, x + 2 * y + 3 * z)) := by
+  sorry
+
+
+/- Problem 6: prove one of these, delete the other -/
+
+@[autograded 4]
+theorem problem6a : Bijective (fun ((r, s) : ℚ × ℚ) ↦ (s, r + 2 * s)) := by
+  sorry
+
+@[autograded 4]
+theorem problem6b : ¬ Bijective (fun ((r, s) : ℚ × ℚ) ↦ (s, r + 2 * s)) := by
+  sorry