Theorem: Suppose a, b, c are natural numbers. If a divides b and b divides c, then a divides c.
- Bad Proof: Suppose the hypothesis. We can substitute a times something for b in the equation b times something equals c. The two somethings together multiply a to c so we’re done.[way too vague; might be logically sound but difficult even to assess that]
- Deceptive Proof: Suppose a, b, c are natural numbers such that a divides b and b divides c. All of a’s prime factors must also be prime factors of b, and likewise for b and c. Therefore any prime factor of a must be a prime factor of c, so a divides c.
[seems convincing, but the prime factor business is without its own proof here, so preliminary work is missing. Even with that, would need repeated prime factors to occur sufficiently many times, so the argument would need cleaning up.] - Good Proof: Recall a divides b if and only if there is an integer k such that ak = b. Suppose, then, that ak_1 = b and bk_2 = c for k_1, k_2 natural numbers. By substitution, ak_1k_2 = c. Since the product k_1k_2 must be some natural number k, ak=c and a divides c.
Multiple proof techniques for one statement:
Quick definition: integer n is even if, for some integer k, n=2k, and odd if, for some integer k, n = 2k+1.
Claim: If n is even, n+1 is odd.
- Direct proof: Let n be even and k be such that n=2k. Then n+1 = 2k+1, so by definition n+1 is odd.
- Contrapositive proof: Suppose n+1 is not odd. Then it must be even, so for some k n+1 = 2k. That means n = 2k-1 = 2k-2+1 = 2(k-1)+1. Since k is an integer, so is k-1, so n is odd by definition. Therefore, if n is even, n+1 must be odd.
- Contradiction proof: Suppose n is even but n+1 is not odd. Then for some integers k, l, n=2k and n+1=2l. However, then n+1 = 2k+1 = 2l, so l=k+1/2. Since k and l are both integers, this is impossible, so whenever n is even n+1 must be odd.
Proof using multiple methods together:
Theorem: The polynomial p(x) = x^3 + x – 1 has exactly one real root.
Proof: First we show that p(x) has at least one real root. Note that p(0) = -1 and p(1) = 1. By the Intermediate Value Theorem, since 0 lies between -1 and 1, there must be a real number c between 0 and 1 such that p(c) = 0. [Direct]
Now suppose for a contradiction that for c_1 ≠ c_2, p(c_1) = p(c_2) = 0. Then by the Mean Value Theorem, there must be some b between c_1 and c_2 such that 
Case proof:
Claim: If n is an integer, then n^2+n is even.
Proof: Case 1: n is even. Then n^2 and n are both even, and the sum of two even numbers is even.
Case 2: n is odd. Then n^2 and n are both odd, and the sum of two odd numbers is even.
Proof of equivalence using two implications:
Theorem: Let a be an integer. Then a is nondivisible by 3 if and only if a^2-1 is divisible by 3.
Proof: (->) Suppose a is nondivisible by 3. Then a = 3k + r for some integer k, and r equal to 1 or 2. Substituting, a^2-1 = (3k+r)^2-1 = 9k^2+6kr+r^2-1. Clearly this is divisible by 3 exactly when r^2-1 is. Since r=1 gives r^2-1 = 0 and r=2 gives r^2-1 = 3, both of which are divisible by 3, a^2-1 is divisible by 3 for any a nondivisible by 3.
(<-) Suppose a^2-1 is divisible by 3. Since a^2-1 = (a-1)(a+1) and 3 is prime, 3 must divide at least one of a-1 or a+1. However, in either case 3 cannot also divide a itself.
The key to a (one-part) “if and only if” proof is in the -> half:
Proof: Any integer a may be written as 3k + r for some integer k with r equal to 0, 1, or 2; a is divisible by 3 if in this form its r is 0. By substitution, a^2-1 = (3k+r)^2-1 = 9k^2+6kr+r^2-1, which is divisible by 3 exactly when r^2-1 is. We may plug in each of our values of r: r=0 gives r^2-1 = -1, r=1 gives r^2-1 = 0, and r=2 gives r^2-1 = 3. The latter two, corresponding to a nondivisible by 3, are divisible by 3, and the first one, corresponding to a divisible by 3, is not divisible by 3, giving the desired correspondence.
Depending on context, these might qualify as “bad” proofs since they take for granted the division algorithm and the fact that if a prime number divides m it must divide at least one of the terms in any factorization of m.