A proof is just a convincing argument. However, mathematicians tend to have extraordinarily high standards for what convincing means. For example, consider the Goldbach conjecture which states that “every even number greater than 2 is the sum of two primes.” This conjecture has been verified for even numbers up to \(10^{18}\) as of the time of this writing. The layman may say, “surely, this prove the result!” but the mathematician is not convinced because he (or she) requires a deductive argument, not an extrapolation based on observation. Before we begin, we need to define a few terms.
A definition is simply an abbreviation for a proposition or object. Definitions are created by mathematicians in order to more succinctly express ideas and relationships. An axiom is a proposition that we assume to be true without any justification (other than, perhaps, because our intution says so). Because axioms are unjustified propositions we try to avoid creating too many of them. A theorem is a proposition which we justify by means of a proof.
So, what are the rules of a proof from the mathematicians points of view? They can be encompassed in essentially four primary rules.
The prior results rule is vital and it is part of what makes mathematics so powerful. Instead of reproving the same result over and over, we can prove it once and for all, and then use it wherever we choose. For those of you with a background in software development, this is essentially what happens when you abstract and modularirze a piece of code into a function for reuse in other places. For this reason, you generally don't want to prove monolithic theorems. Instead, if your theorem has a lot of different pieces, break them up into smaller pieces. Alternatively, if there are a lot of steps in your proof, you can often split the proof into smaller pieces called lemmas.
The replacement rule is often useful for substituting a term (e.g., even number) with its definition (\(2k\) for some integer \(k \in \mathbb{Z}\)).
The modus ponens rule is based on the fact that \(\big( P \wedge (P \implies Q)\big) \implies Q\) is a tautology. This is probably not immediately obvious to you, so you should make a truth table to justify it to yourself.
Probably the most common kind of proof is the direct proof, which has the following structure.
If \(P\text{,}\) then \(Q\text{.}\)
Assume \(P\text{.}\)
(apply rules of proofs judiciously)
Therefore, \(Q\text{.}\)
Thus, \(P \implies Q\text{.}\)
As an example, we will prove the following proposition using a direct proof.
Let \(x\) be an integer. If \(x\) is odd, then \(x+1\) is even.
Let \(x\) be an odd integer. By Definition A.1.3, \(x = 2k+1\) for some integer \(k \in \mathbb{Z}.\) Then \(x+1 = 2k+2 = 2(k+1).\) Since the sum of two integers is an integer, \(k+1 \in \mathbb{Z}\text{,}\) and therefore, by Definition A.1.2, \(x+1 = 2(k+1)\) is even. Hence, if \(x\) is odd, then \(x+1\) is even.
In the statement of the previous theorem occurred the sentence, “let \(x\) be an odd integer.” This is not the antecedent of the implication (which is “if \(x\) is an odd integer”), but is rather something called a hypothesis, which provides the context in which the theorem takes place. In this case, we would be unable to say that \(x\) is odd (or even, for that matter) unless \(x\) is an integer.
The careful reader may have noticed that the statement we proved was not exactly a proposition in our strict sense: it is instead an open sentence because we did not pick a specific value of \(x\text{.}\) The resolution to this problem will be addressed in the next section. However, notice that our proof would have worked equally well no matter which odd integer we chose.
If either \(P\) or \(Q\text{,}\) then \(R\text{.}\)
Case 1. Assume \(P\text{,}\) …, therefore \(R\text{.}\) Hence \(P \implies R\)
Case 2. Assume \(Q\text{,}\) …, therefore \(R\text{.}\) Hence \(Q \implies R\text{.}\)
In the proof by cases above, we did not actually prove \((P \vee Q) \implies R\) directly. That is, we did not start by assuming \(P \vee Q\text{.}\) Instead, we prove two implications, namely, \(P \implies R\) and \(Q \implies R\text{.}\) This is acceptable because of the equivalence
\begin{equation*} \big( (P \vee Q) \implies R \big) \iff \big( (P \implies R) \wedge (Q \implies R) \big) \end{equation*}We may implement a proof by cases even when the antecedent does not include a disjunction. This can be achieved by inserting the Law of Excluded Middle tautology for some appropriate proposition.
Suppose \(n\) is an odd integer. Then \(n = 4j + 1\) for some integer \(j\text{,}\) or \(n = 4i - 1\) for some integer \(i\text{.}\)
Suppose \(n \in \mathbb{Z}\) is odd. Then \(n = 2k + 1\) for some integer \(k\text{.}\) The integer \(k\) is either odd or even by Parity.
Since \(k\) is even, \(k = 2j\) for some integer \(j\text{.}\) Thus
\begin{equation*} n = 2k + 1 = 2(2j)+1 = 4j + 1. \end{equation*}Since \(k\) is odd, \(k = 2i + 1\) for some integer \(i\text{.}\) Thus
\begin{equation*} n = 2k + 1 = 2(2i + 1)+1 = 4i + 3 = 4(i+1)-1. \end{equation*}Since \(i+1\) is an integer, we have proven the desired result.
Recall that Theorem 1.2.5 guarantees the conditional statement \(P \implies Q\) is equivalent to its contrapositive, namely, \((\neg Q) \implies (\neg P)\text{.}\) This leads to a new proof technique called proof by contrapositiion. In this proof technique, instead of proving the implication directly, we instead prove its contrapositive directly, and then use the stated equivalence.
If \(P\text{,}\) then \(Q\text{.}\)
Assume \(\neg Q\text{,}\) …, therefore \(\neg P\text{.}\) Hence, \((\neg Q) \implies (\neg P)\text{.}\) Therefore \(P \implies Q\text{.}\)
Let \(x \in \mathbb{Z}\text{.}\) If \(x^2\) is even, then \(x\) is even.
Suppose \(x\) is not even. By Parity \(x\) is odd. Thus \(x = 2k+1\) for some integer \(k\text{.}\) Hence
\begin{equation*} x^2 = (2k+1)^2 = (2k)^2 + 2(2k) + 1 = 2(2k^2 + 2k) + 1. \end{equation*}Since the integers are closed under addition and multiplication, \(2k^2 + 2k \in \mathbb{Z}\text{.}\) Then by the definition of odd, \(x^2\) is odd. Again, by Parity, \(x^2\) is not even. Therefore, if \(x\) is not even, then \(x^2\) is not even. Hence, by contraposition, we have proven the result.
The next proof technique is often regarded as completely strange when first encountered, but you will become used to it after a while. This proof technique essentially relies on two facts.
\(P\) is true.
Suppose \(\neg P\text{,}\) …, therefore \(Q\text{,}\) …, therefore \(\neg Q\text{.}\) Hence \(Q \wedge (\neg Q)\text{,}\) a contradiction. Thus, \(\neg \neg P\text{.}\) Therefore, \(P\text{.}\)
At this point, you should try to avoid proofs by contradiction. There are three reasons for this. First, they are conceptually harder to grasp than other proofs, both for you and for the reader. Second, many people prove things by contradiction when they really meant to use the contrapositive. Third, if you make any error in your reasoning during the proof, it can look like you've reached a contradiction and have thus proven your result, when really it was just an error.
Up until this point, we haven't mentioned how to prove a biconditional statement. However, Theorem 1.2.11 guarantees that we can just prove both directions, which we encapsulate in the following technique.
\(P\) if and only if \(Q\text{.}\)
Prove \(P \implies Q\) by any method. Prove \(Q \implies P\) by any method. Therefore, \((P \implies Q) \wedge (Q \implies P)\text{,}\) and hence \(P \iff Q\text{.}\)
Let \(x\) be an integer. Then \(x\) is even if and only if \(x^2\) is even.
Let \(x \in \mathbb{Z}\text{.}\)
Suppose \(x\) is even. Then \(x = 2k\) for some \(k \in \mathbb{Z}\text{.}\) Thus \(x^2 = (2k)^2 = 4k^2 = 2(2k^2)\text{.}\) Therefore \(x^2\) is even by definition.
This was proven in Proposition 1.4.10 using contraposition.
Thus, \(x\) is even if and only if \(x^2\) is even.
There is another way to prove “if and only if” (i.e. biconditional) statements, which is just to use a series of equivalent statements. Note, we use the useful abbreviation “iff” for “if and only if.”
\(P\) if and only if \(Q\text{.}\)
\(P\) iff \(R_1\) iff \(R_2\) … iff \(R_n\) iff \(Q\text{.}\)
Before we prove the next proposition, it will be helpful to notice that biconditional equivalence has a nice property, namely,
\begin{equation*} (P \iff Q) \iff ((\neg P) \iff (\neg Q)). \end{equation*}Let \(x\) be an integer. Then \(x\) is odd if and only if \(x^2\) is odd.
Suppose \(x\) is an integer. Then \(x\) is odd if and only if \(x\) is not even (by Parity) if and only if \(x^2\) is not even (by Proposition 1.4.13) if and only if \(x^2\) is odd.