Subsection Direct Proofs
Theorems are not proved in a vacuum. To prove one theorem, we usually need to use various relevant definitions, and theorems that have already been proved. If we do not want to keep going backwards infinitely, we need to start with some objects that we use without definition, as well as some facts about these objects that are assumed without proof. Such facts are called axioms, and a body of knowledge that can be derived from a set of axioms is called an axiomatic system.
In modern abstract mathematics, we take set theory as our basis for all arguments.
As a basis for our work in the present chapter, we will make use of standard definitions and properties of the familiar number systems such as the integers, rational numbers and real numbers.
The statement of virtually every theorem, when viewed appropriately, is of the form P \to Q, or some combination of such statements. To prove theorems, we therefore need to know how to prove statements of the form P \to Q.
Definition 2.1.
The most intuitive form of proof, which we treat in this section, is called a direct proof: assume that \(P\) is true, and produce a series of steps, each one following from the previous ones, which eventually lead to \(Q\text{.}\)
That this sort of proof deserves a name is because there are other approaches that can be taken, as we will soon see.
Direct proofs, when completed, typically have the following form.
When constructing a proof, the first thing to do is specify what you are assuming, and what it is you are trying to prove. Then you pick a strategy for the proof; one such strategy is direct proof. The next stage is actually figuring out a proof, making use of your chosen strategy. If you cannot devise a proof using your chosen strategy, perhaps another strategy should be attempted. There is no fixed way of finding a proof; it requires experimentation, playing around and trying different things.
You are probably familiar with the statement “the sum of even numbers is even.” This statement can be viewed in the form P \to Q if we look at it properly, because it actually says “if n and m are even numbers, then n + m is an even number. To construct a rigorous proof of our statement (as well as the corresponding result for odd numbers), we first need precise definitions of the terms involved.
Our theorem is concerned with the integers, that is, the numbers . . . , -3, -2, -1, 0, 1, 2, 3, . . . , and so we need to assume that we know what the integers are, that we have the operations addition, subtraction, multiplication and division, and that these operations satisfy standard properties, for example the Distributive Law. Using only those standard facts about the integers, we can make the following definition, which is the basis for our theorem and its proof.
Definition 2.3.
Let \(n\) be an integer. The number \(n\) is even if there is some integer \(k\) such that \(n = 2k\text{.}\) The number \(n\) is odd if there is some integer \(j\) such that \(n =2 j + 1\text{.}\)
We are now ready to state and prove our theorem.
Theorem 2.4.
Let \(n\) and \(m\) be integers.
If \(n\) and \(m\) are both even, then \(n + m\) is even.
If \(n\) and m are both odd, then \(n + m\) is even.
If \(n\) is even and \(m\) is odd, then \(n + m\) is odd.
Proof.
Suppose that \(n\) and \(m\) are both even. Then there exist integers \(k\) and \(j\) such that \(n = 2k\) and \(m = 2 j\text{.}\) Then
\begin{equation*}
n + m = 2k + 2 j = 2(k + j).
\end{equation*}
Because \(k\) and \(j\) are integers, so is \(k + j\text{.}\) Hence \(m + n\) is even.
Combing soon!
Combing soon!
The proof of Part (1) of Theorem 2.1.3 is quite simple, but there are a few features worth mentioning, because they are typical of what is found in virtually all our subsequent proofs (and in the proofs you will need to write). First, the proof relies completely on the definition of what it means to be an even or an odd integer. In a large number of proofs, going back to the formal definitions involved is the key step; forgetting to do so is a major source of error by students who are first learning about proofs.
Second, observe that the proof is written in grammatically correct English. Complete sentences are used, with proper punctuation. Each sentence begins with a capital letter, and ends with a period, even if the end of the sentence is in a displayed equation. Mathematical formulas and symbols are parts of sentences, and are treated no differently from other words.
Definition 2.6.
Let \(a\) and \(b\) be integers. The number \(a\) divides the number \(b\) if and only if there is some integer \(q\) such that \(aq = b\text{.}\) If \(a\) divides \(b\text{,}\) we write \(a|b\text{,}\) and we say that \(a\) is a factor of \(b\text{,}\) and that \(b\) is divisible by \(a\text{.}\)
Theorem 2.8.
Let \(a\text{,}\) \(b\) and \(c\) be integers. If \(a|b\) and \(b|c\text{,}\) then \(a|c\text{.}\)
Proof.
Suppose that \(a|b\) and \(b|c\text{.}\) Hence there are integers \(q\) and \(r\) such that \(aq = b\) and \(br = c\text{.}\) Define the integer \(k\) by \(k = qr\text{.}\) Then \(ak = a(qr) = (aq)r = br = c\text{.}\) Because \(ak = c\text{,}\) it follows that \(a|c\text{.}\)
Theorem 2.9.
Any integer divides zero.
Proof.
Let \(n\) be an integer. Observe that \(n · 0 = 0\text{.}\) Hence \(n|0\text{.}\)
Exploration 2.10.
Let \(n\) be an integer. Prove that if \(n\) is even then \(n^2\) is even, and if \(n\) is odd then \(n^2\) is odd.