Section Chapter 1: Logic and Proofs
Subsection 1.1 Mathematical Statements
Doing math is about deciding what statements are true. But what makes a statement true?
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A statement is a sentence that is either true or false. For example, "2 + 2 = 4" is a statement, and it is true. "2 + 2 = 5" is also a statement, but it is false. "What time is it?" is not a statement, because it is neither true nor false.
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The most basic types of statements are called atomic statements; they cannot be broken down into simpler statements. To determine whether an atomic statement is true or false, we must know the meaning of words in the statement. The meaning of the words and the “state of the world” (whether the words agree with reality) determine the truth value of the statement.
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Where logic comes in is that we can combine statements to make new more complicated statements. We can call these molecular statements. The truth value of a molecular statement is determined by the truth values of the atomic statements that make it up, and the way they are combined.
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The ways we combine statements are called logical connectives. The most common logical connectives are “and”, “or”, and “not”. We can also use “if ... then ...” to make a statement that is true if the first part is false, or the second part is true. Since these two-part statements have a truth value depending on the truth values of the two parts, we can define the meaning of these molecular statements by listing what happens in all four cases.
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The Statements and the Search for Truth activity has some practice determining the truth values of molecular statements.
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The second part of the activity considers quantified statements. Here the truth value depends on something else. The goal of the second part of the activity is to figure out what that is.
Subsection 1.2 Implications
Recall the truth table for the implication (“if ... then ...”) connective:
| \(P\) | \(Q\) | \(P \imp Q\) |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
Some of the four cases are easy to understand: if the hypothesis is true, then the conclusion better be true. That’s just restating the implication. So the first two rows should produce no argument.
The real question comes when the hypothesis is false. How can we make sense of this? One option: just accept this as a definition. The book has a lot more discussion about this.
Consider an example statement: “If a playing card has a blue back, then it is a face card.” How do we know this is false in general?
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Place four cards on the table: one with a blue back, one with a red back, one that is a face card, and one that is not a face card. Which cards do you need to turn over to see if the statement is true or false?
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Crucially, turning over a red backed card gives you no information. Similarly, turning over a face card gives you no information. The only way to prove the statement false is to find a blue backed card that is not a face card. This is the only case that makes the statement false.
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That means you need to turn over the blue backed card (to make sure it is a face card) and the non-face card (to make sure it does not have a blue back). If both of those cards check out, then the statement is true. At least, it is true of the four cards.
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Of course, the statement is NOT true of all playing cards. This brings us to the concept of quantifiers. When we say, “if a playing card has a blue back then it is a face card,” we are really saying, “for all playing cards, if it has a blue back then it is a face card.” The truth value of this statement depends on all playing cards, not just the four we have on the table.
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We claimed this statement was false: our evidence was a single card that both has a blue back and is not a face card. So to negative a universal statement, we need to find a single counterexample. This is the only way to prove a universal statement false.
Subsection 1.3 Truth Tables and Deduction
To explore truth tables and deductions, see the Truth Tables and Deductions activity
Subsection 1.5 Proofs about Discrete Structures
Consider another question like the stamps making 72 cents problem: suppose you keep your socks in a drawer all loose. You have lots of each type of 5 different styles of socks. How many socks do you have to pull out before you guarantee that you have a matching pair? Prove your answer.
Another version of the sock problem: if you pour out M&Ms, how many do you need to get before you can be sure that you have 10 of the same color?
Another version: prove that for any finite sets \(A\) and \(B\text{,}\) if \(f:A \to B\) is injective, then \(|A| \le |B|\text{.}\)
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What? We need to first make sense of this statement before we can try to prove it. For that, we need to look at the definitions of the terms used.
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A function \(f:A \to B\) is injective provided every element of the codomain \(B\) is the image of at most one element of the domain \(A\text{.}\)In other words, if two elements \(a_1\) and \(a_2\) of the domain are different, then their images \(f(a_1)\) and \(f(a_2)\) are different. If \(a_1 \ne a_2\text{,}\) then \(f(a_1) \ne f(a_2)\text{.}\)
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Given a set \(A\text{,}\) we write \(|A|\) for the cardinality of \(A\text{,}\) by which we mean the number of elements in \(A\text{.}\)
What it means for sets \(A\text{,}\) \(B\text{,}\) and \(C\) to have \(A \subseteq B\) and for \(|B \cap C| = |C|\text{.}\) Then we decided whether these two things happening would imply that \(A \subseteq C\text{.}\)
We looked at examples of each of these (see section 1.5 for the actual definitions). We then proved that for any two finite sets \(A\) and \(B\text{,}\) if \(|A \cap B| = |B|\) then \(B \subseteq A\text{.}\)
Let’s prove something else about sets. Prove that if \(A \subseteq B\text{,}\) then for any set \(C\text{,}\) \(A \cup C \subseteq B \cup C\text{.}\)
Another: Let \(f: X \to Y\) be a function. Let \(A, B \subseteq X\) be subsets of the domain. If \(A \subseteq B\) then \(f(A) \subseteq f(B)\text{.}\)
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What does \(f(A)\) even mean? It is a subset of the codomain consisting of all the images of elements in \(A\text{.}\) That is,\begin{equation*} f(A) = \{y \in Y \st y = f(a) \text{ for some } a \in A\}\text{.} \end{equation*}
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The picture here helps. But then we can actually write down the proof.
Here is another interesting proof idea: suppose you join a party with five other people. Everyone at the party either shakes hands or fist-bumps.
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Prove that you either shook at least three hands or fist-bumped at least three fists.
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Must there be some group of three people who all mutually shook hands or all mutually fist-bumped?
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This question can be modeled with a graph. Show how.
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A very interesting fact is that 6 is the smallest number to guarantee a group of 3 mutual interactions. 18 is the smallest number to guarantee a group of 4 mutual interactions. The smallest number for 5 mutual interactions is not known: it is somewhere between 43 and 46.
