Proof that the sum of two odd integers is even

Preliminaries

• An integer is odd if it’s two times some integer + 1

Proof

1. Let x and y be two odd integers.
2. Then, x = 2a + 1, for some integer a.
   • Definition of odd integer
3. And y = 2b + 1, for some integer b.
   • Definition odd integer
4. So x + y = 2a + 1 + 2b + 1.
5. Thus, x + y = 2a + 2b + 2.
   • Rearranging line 6
6. And so x + y = 2(a + b + 1).
   • Factoring line 7
7. Thus x + y = two times some integer (a + b + 1) and is therefore even.
   • Per line 2

Proof that √2 is irrational

Preliminaries

• A rational number is a number that can be expressed as a fraction c/d where c and d are integers and d ≠ 0.
• An integer is even if it’s two times some integer
• An integer is odd if it’s two times some integer + 1
• The square of an odd integer is odd.
  • Suppose m is odd. We want to prove that m² is odd.
  • m = 2c + 1, for some integer c.
    • Definition of odd
  • m² = (2c + 1)(2c + 1)
  • m² = 1 + 4c + 4c²
  • m² = 1 + 2(2c + 2c²)
  • m² = 1 + even number
  • m² is odd.

Proof

• Suppose for proof by contradiction that there are integers c and d such that √2 = c/d.
• Let a/b = c/d in lowest terms.
• Thus, √2 = a/b, for some integers a and b.
• Proof that a is even:
  • √2 = a/b
  • 2 = a²/b²
  • a² = 2 b²
  • So a² is even
    • Definition of even
  • So a is even
    • If a were odd then a² would be odd
• Proof that b is even:
  • a² = 2b²
    • from above
  • a = 2k, for some integer k
    • since a is even
  • (2k)² = 2b²
  • 4k² = 2b²
  • 2k² = b²
  • So b² is even
    • Definition of even
  • So b is even
    • If b were odd then b² would be odd
• Since a and b are both even they have a common factor, 2. Therefore, a/b is not in lowest terms, contradicting the assumption above.