Inclusive or is a low hurdle. In plain English, when someone asks “Should we take our vacation to New York or Boston?” the assumption is that the answer will be “New York” or “Boston” (or “I don’t care” or “neither”). The geeky joke – sometimes serious – answer of “yes” is totally unhelpful. However, it’s not too hard to get used to inclusive or, and we do have examples in natural language. One of the best is “Would you like sugar or cream in your coffee?” Of course, even then “yes” isn’t a useful answer, since there are three possible coffee fixings that would lead to it.

Implication is a much higher bar; there’s still a part of me, even, that doesn’t think A implies B means much of anything when A is false. Implications where there is clearly no causal relationship between A and B can be helpful, since they rarely appear in plain English (outside of statistical correlations, I suppose) and thus resist natural language intuition. For teaching, in addition to that, I settled on the approach of “an implication is true unless proven false.” You can only prove that it’s false by having A be true and B be false, so in the A-false situations the implication is therefore true. This is basically the conversion of “A implies B” into the disjunction “B or not-A,” but hopefully in a way that doesn’t just shift the confusion to a different location.

I think the base of that confusion is a disconnect between allowed truth values. In plain English, a sentence can be true, false, or nonsensical. The third option is not permitted in mathematics (except in the sense of ill-formed formulas), and it is confusing that many implications that seem nonsensical or are constructed from false clauses (“if the moon is made of green cheese, then fish swim in the sea;” “if the moon is made of green cheese, then carriages turn into pumpkins at midnight”) are logically true.

In the fifteen-plus years since my bridge class, I have found only one plain English example of an implication considered true but constructed with false clauses, and in general my students were unfamiliar with it: the adage “If wishes were horses, beggars would ride.” I would love another, even though nowadays it would be purely for my own interest.

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• 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 However, p'(x) = 3x^2 + 1, which is always strictly greater than zero. Therefore there cannot be such c_1 ≠ c_2, and p has only one real root.

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.

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• explicit/specific (non-vague)
• logically sound, including complete
• lacking irrelevant statements
• understandable to the reader
• self-contained (may assume basic things; anything else needs explicit reference to previous work or must be written out in the proof)

Proof kinds

• Direct proof: Assume hypothesis and march to conclusion.
• Contradiction: If proving a single clause, assume its negation and prove something that contradicts laws of mathematics. If proving A->B, assume not B and (implicitly or explicitly) A, and prove something that contradicts one of those two assumptions or general laws of mathematics.
• Contraposition: Arguably a special case of contradiction; used only for proving implications, A->B. Assume not B and prove not A.
• Induction: Theorem asserts something about an infinite collection, such as all natural numbers. Prove it for the lowest value (starting point) and then show that if you know it for n you can prove it for n+1. Conclude it holds for all values.

Proof tips

• Bi-implication is often easiest to prove as two separate implications, possibly using different methods.
• Existence: Either show the statement holds for a specific value or kind of value (e.g. all even numbers), or draw a contradiction from the assumption that it holds for no value.
• Universal: Leave the value as a variable and use only properties common to all possible instances of the variable.
• “For all x there exists y” (AE): Universally-quantified value must be represented as a variable, existentially-quantified value may (often must) be a function of the universal one.
• “There exists x such that for all y” (EA): Universally-quantified value must still be a variable. Existentially-quantified value can be specified, but must not be a function of the universal one because the same value (not just the same function) has to work for all possible values of the variable.
• Disjunction: “A or B” is often proved as “not A -> B” or “not B -> A”, whichever is more straightforward.
• Cases: Sometimes your universally-quantified values fall into a few neat categories (odd/even, positive/negative/zero) and it is easier to use different proofs for each category.
• If it’s not coming together, two things to check: Can you work backwards? Are there any definitions you haven’t fully unpacked?

Proof warnings

• Existence of an item satisfying the given criteria may be shown by presenting an example (constructive proofs) but need not be (nonconstructive proofs). For example, the Intermediate Value Theorem says if f is continuous and f(a) ≠ f(b), for any r between f(a) and f(b) there is c between a and b such that f(c) = r. Its proof gives no clue as to where c might be found, not even in terms of a, b, and r.
• If you are trying to prove something holds of all items in a certain group you may assume only traits that hold of every item in the group. For example, if you are proving something about all rational numbers you may use the fact that they may each be written as p/q where p, q are integers, q is nonzero, and gcd(p,q)=1. Be careful not to assume rational numbers are nonintegers, integers are positive, and so forth.
• If you are asked to prove an implication, it is likely it is not actually an equivalence, and if you “prove” an equivalence it will be incorrect. This trap catches more students than you might expect.

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I’d have to be crazy to fall out of love with you.

Based on that hypothesis, he asserts

The place where I hold you is true, so I know I’m all right.

Is this a logically valid argument? That is, is still being in love a logically sound reason for Willie to assert he’s not crazy? Why or why not?

2. The King James translation of the Bible contains the following logically nontrivial statement:

Every one that hath forsaken houses, or brethren, or sisters, or father, or mother, or wife, or children, or lands, for my name’s sake, shall receive a hundredfold, and shall inherit everlasting life. [Matthew 19:29]

Write a sentence which gives the negation of that statement.

3. You are checking to see if it is always true that people who are taller than average dislike sitting in low chairs. In each of the following situations you have partial information. In which situations must you investigate further to make sure your premise is not invalidated, and what piece of information must you find out in each of those cases?
     a) The subject is 6′6″ tall.
     b) The subject is 4′11″ tall.
     c) The subject complains a lot about the low chair.
     d) The subject happily settles into the low chair.

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