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# 2.5.2: Examples


Let’s look at a few examples.

Theorem 3.5.

The number $$2^{2 n}$$ − 1 is divisible by 3 for all natural numbers n.

Proof. Here, P(n) is the statement that $$2^{2 n}$$ − 1 is divisible by 3.
Base case: When $$n=0,2^{2 n}-1=2^{0}-1=1-1=0$$ and 0 is divisible by 3 (since 0 = 3 · 0.) Therefore the statement holds when n = 0.

Inductive case: We want to show that if the statement is true for n = k (where k is an arbitrary natural number), then it is true for n = k + 1 also. That is, we must prove the implication $$P(k) \rightarrow P(k+1) .$$ So we assume $$P(k),$$ that is, we assume that $$2^{2 k}$$ is divisible by $$3 .$$ This means that $$2^{2 k}-1=3 m$$ for some integer $$m .$$ We want to prove $$P(k+1),$$ that is, that $$2^{2(k+1)}-1$$ is also divisible by $$3 :$$

and from the last line we see that $$2^{2 k+1}$$ is in fact divisible by 3. (The third step-subtractin and adding 4—was done to enable us to use our inductive hypothesis.) Altogether, we have proved that P(0) holds and that, for all k, P(k) → P(k + 1) is true. Therefore, by the principle of induction, $$P(n)$$ is true for all $$n$$ in $$\mathbb{N},$$ i.e. $$2^{2 n}-1$$ is divisible by 3 for all n in N.

The principle of mathematical induction gives a method for proving P(n) for all n in the set N. It should be clear that if M is any natural number, a similar method can be used to show that P(n) is true for all natural numbers n that satisfy n M. Just start the induction with a base case of n = M instead of with a base case of n = 0. I leave the proof of this extension of the principle of induction as an exercise. We can use the extended principle of induction to prove a result that was first mentioned in Section 2.1.

Theorem 3.6.

Suppose that a compound proposition contains exactly n propositional variables, where $$n \geq 1 .$$ Then there are exactly $$2^{n}$$ different ways of assigning truth walues to the n variables.

Proof. Let P(n) be the statement “There are exactly $$2^{n}$$ different ways of assigning truth values to n propositional variables.” We will use induction to prove the P(n) is true for all n ≥ 1.

Base case: First, we prove the statement P(1). If there is exactly one variable, then there are exactly two ways of assigning a truth value to that variable. Namely, the variable can be either true or false. Since $$2=2^{1}, P(1)$$ is true.

Inductive case: Suppose that P(k) is already known to be true. We want to prove that, under this assumption, $$P(k+1)$$ is also true. Suppose that $$p_{1}, p_{2}, \ldots, p_{k+1}$$ are $$k+1$$ propositional variables. Since we are assuming that P(k) is true, we know that there are $$2^{k}$$ ways of assigning truth values to $$p_{1}, p_{2}, \dots, p_{k} .$$ But each assignment of truth values to $$p_{1}, p_{2}, \ldots, p_{k}$$ can be extended to the complete list $$p_{1}, p_{2}, \ldots, p_{k}, p_{k+1}$$ in two ways. Namely, $$p_{k+1}$$ can be assigned the value true or the value false. It follows that there are 2$$\cdot 2^{k}$$ ways of assigning truth values to $$p_{1}, p_{2}, \ldots, p_{k+1} .$$ since $$2 \cdot 2^{k}=2^{k+1},$$ thishes the proof.

2.5.2: Examples is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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