Proofs¶
As a CS70 student, you will be taught how to prove logical statements. Proofs guarantee a statement is true, and can be very powerful. They are a finite sequence of steps called logical deductions which establishes the truth of a desired statement. It uses finite means to guarantee the truth of a statement with infinitely many cases.
A proof is often structured as follows. First, there are certain statements called axioms that we accept without proof. Starting from these, a proof consists of a sequence of logical deductions: simple steps that apply the rules of logic. Each statement follows from previous statements.
This note will first cover each proofing technique, then give an example.
Preamble
In this section, we will use the following notation and basic mathematical facts.
Let \(\mathbb{Z}\) denote the set of integers (\(\mathbb{Z}=\{\cdots,-1,0,1,2, \cdots\}\)).
Let \(\mathbb{N}\) denote the set of natural numbers (\(\mathbb{N}=\{0,1,2, \cdots\}\)).
The sum or product of two integers is an integer; we say the set of integers is closed under addition and multiplication.
The same is true for natural numbers.
Given integers \(a\) and \(b\), we say \(a\) divides \(b\) (denoted \(a|b\)) iff there exists an integer \(q\) such that \(b=aq\). (For example, \(2|10\)).
\(:=\) is used to indicate a definition. (For example, \(q:=6\) defines the variable \(q\) as having the value 6).
Direct Proof¶
A direct proof proceeds as follows. For each \(x\), the proposition we are trying to prove is of the form \(P(x)\implies Q(x)\). A direct proof of this starts by assuming \(P(x)\) for a generic value of \(x\) and eventually concludes \(Q(x)\) through a chain of implications.
Goal: To prove \(P\implies Q\).
Approach:
Assume \(P\)
\(\cdots\)
Therefore \(Q\).
Theorem
For any \(a, b, c \in\mathbb{Z}\), if \(a|b\) and \(a|c\), then \(a|(b+c)\).
Proof
Assume that \(a|b\) and \(a|c\). There exists integers \(q_1\) and \(q_2\) such that \(b=q_1a\) and \(c=q_2a\). Then, \(b+c=q_1a+q_2a=(q_1+q_2)a\). Since \(\mathbb{Z}\) is closed under addition, we conclude \((q_1+q_2)\in\mathbb{Z}\), so \(a|(b+c)\) as desired.
The key insight in the above proof is that we did not assume any specific values for \(a, b\) and \(c\), so our proof holds for arbitrary values of \(a, b, c\in \mathbb{Z}\)!.
Proof by Contraposition¶
Recall from the previous topic that \((P\implies Q)\equiv(\neg Q\implies\neg P)\). Sometimes, the latter can be easier to prove than the former.
Goal: To prove \(P\implies Q\).
Approach:
Assume \(\neg Q\)
\(\cdots\)
Therefore \(\neg P\).
Theorem
Let \(n\) be a psoitive integer and let \(d\) divide \(n\). If \(n\) is odd, then \(d\) is odd.
Proving this via the technique of direct proof seems difficult, since we have no easy way to proceed after assuming \(n\) is odd. But the contraposition technique is easier!
Proof
Assume \(d\) is even. By definition, \(d=2k\) for some \(k\in\mathbb{Z}\). Because \(d|n, n=dl\) for some \(l\in\mathbb{Z}\). Combining these two statements, \(n=dl= (2k)l=2(kl)\). We conclude that \(n\) is even, thus by contraposition, the statement is true.
Proof by Contradiction¶
The idea in proof by contradiction is to assume the claim you wish to prove is false. Then you show this leads to a conclusion which is utter nonsense: a contradiction; hence, the claim must in fact have been true.
Proof by contradiction is a good technique for proving something you think is obviously true.
Goal: To prove \(P\).
Approach:
Assume \(\neg P\).
\(\cdots\)
\(R\)
\(\cdots\)
\(\neg R\)
Conclusion: \(\neg P\implies\neg R\wedge R\), which is a contradiction, therefore \(P\) holds.
Theorem
There exist no integers \(a\) and \(b\) for which \(18a+6b=1\).
The credit for this proof goes to Dr. Jonathan Senning at Gordon College, from Math 231: Transition to Higher Mathematics class from Fall 2014.
Proof
Assume there exist integers \(a\) and \(b\) for which \(18a+6b=1\).
Dividing by 6, we obtain \(3a+b=\frac{1}{6}\).
This is a contradiction, since by the closure properties, \(3a+b\) is an integer, but \(\frac{1}{6}\) is not. Therefore, it must be that no integers \(a\) and \(b\) can exist for which \(18a+6b=1\).
Proof by Cases¶
The idea behind proof by cases is that sometimes, when we wish to prove a claim, we don’t know which of a set of possible cases is true, but we know at least one of the cases is true. We can instead prove all the cases; then, clearly, the claim must hold.
Theorem
There exist irrational numbers \(x\) and \(y\) such that \(x^y\) is rational.
Proof
We proceed by cases. The statement is quantified by an existential quantifier, thus it is sufficient to demonstrate a single example, \(x\) and \(y\), where the statement is true. To do so, let \(x=\sqrt{2}\) and \(y=\sqrt{2}\).
Let us divide our proof into two cases, exactly one of which must be true:
\(\sqrt{2}^\sqrt{2}\) is irrational, or
\(\sqrt{2}^\sqrt{2}\) is rational
Assume that \(\sqrt{2}^\sqrt{2}\) is rational. Our first guess for \(x\) and \(y\) was not quite right, but now we have a new irrational number, \(\sqrt{2}^\sqrt{2}\). Let \(x=\sqrt{2}^\sqrt{2}\) and \(y=\sqrt{2}\). Then,
We started with two irrational numbers \(x\) and \(y\) and obtained a rational \(x^y\) in this case. In the other case, our claim is immediately yielded. Thus, since one of the two cases must hold, we conclude the theorem must be true.
The above is an example of a \(non-constructive\) proof. We do not know which of the two pairs of values we set \(x\) and \(y\) to is actually true: we’ve proven some object \(X\) exists, without explicitly revealing what \(X\) is!