Problem 11 - Contemporary Precalculus

Question

Use mathematical induction to prove that each of the given statements is true for every positive integer \(n .\) $$3^{n} \geq 3 n$$

Step-by-Step Solution

Answer

In conclusion, using mathematical induction, we have proved that the inequality $3^n \geq 3n$ holds for all positive integers $n$. We first established the base case, where the inequality holds for $n=1$. Then, by assuming the inductive hypothesis $3^k \geq 3k$ for an arbitrary positive integer $k$, we showed that the inequality also holds for $k + 1$. Thus, the given inequality $3^n \geq 3n$ is proven to be true for all positive integers $n$.

Step 1: Prove the base case

For the base case, we are considering \(n = 1\). We want to prove that the statement \(3^n \geq 3n\) holds for \(n = 1\). $$3^1 \geq 3(1)$$ $$3 \geq 3$$ The inequality is true for the base case, as \(3 \geq 3\).

Step 2: Inductive hypothesis

Now, we'll assume that the statement is true for an arbitrary positive integer \(k\). This is our inductive hypothesis: $$3^k \geq 3k$$

Step 3: Prove the statement for \(k + 1\)

We want to prove that the statement also holds for \(k + 1\): $$3^{k+1} \geq 3(k + 1)$$ To do so, we can start by looking at the expression for \(3^{k+1}\): $$3^{k+1} = 3^k \cdot 3$$ Now, from the inductive hypothesis, we know that \(3^k \geq 3k\). Thus, we can say that: $$3^{k+1} = 3^k \cdot 3 \geq (3k) \cdot 3$$ $$3^{k+1} \geq 9k$$ We need to show that \(3^{k+1} \geq 3(k + 1)\). However, \(9k\) is greater than \(3(k + 1)\): $$3(k+1) = 3k + 3$$ $$9k \geq 3k + 3$$ $$k \geq 1$$ Since \(k \geq 1\) (as we are considering positive integers), we can see that \(3^{k+1} \geq 3(k + 1)\), proving the statement for \(k + 1\).

Step 4: Conclusion

We have proved the statement \(3^n \geq 3n\) for the base case (\(n = 1\)) and for \(n = k + 1\) (given the statement holds for \(n = k\)). By the principle of mathematical induction, the statement is true for all positive integers \(n\).

Key Concepts

Proof Techniques in Mathematics

Proof techniques are the bread and butter of mathematics. They are logical methods used to demonstrate the truthfulness of a mathematical statement. There are several proof techniques, such as direct proofs, proof by contradiction, proof by counterexample, and the focus of our exercise, mathematical induction.

Mathematical induction is a powerful technique particularly used to prove statements about integers. It functions like a domino effect. The process begins by proving the 'base case', a fundamental step that examines the validity of the statement at the starting point, often for the smallest integer in its domain. Once the base case is established, the 'inductive step' involves assuming the statement is true for some integer, and then showing this implies it is also true for the next integer.

These two steps combined show that if the statement is true for an arbitrary integer, it will also hold for all subsequent integers. It's akin to proving once you have pushed the first domino and shown that any domino will knock over the next, that a whole infinite line of dominos will fall.

Inequality Proofs

Proofs involving inequalities are frequently encountered in mathematics. They are used to demonstrate the relationship between two expressions, typically by showing that one is greater than or equal to the other. The inequality proof in our exercise involves showing that for all positive integers, the power of three to an integer is greater or equal to three times the integer itself.

The key to inequality proofs lies in manipulation and logical deductions. You begin by establishing a starting point known as the base case. The next step, much like in mathematical induction, is assuming the inequality holds for an arbitrary positive integer. Using algebraic manipulations, you draw logical conclusions to show that if the statement is true for a certain case, it's true for the next.

Always be mindful of the properties of inequalities while manipulating them, such as when multiplying or dividing by negative numbers (which reverses the inequality) or when adding the same quantity to both sides (which maintains the inequality).

Inductive Reasoning in Precalculus

Inductive reasoning in precalculus is the process of observing patterns and making generalizations. This form of reasoning is the backbone of mathematical induction and allows us to move from specific cases to general truths. While sometimes mistaken with inductive reasoning in the scientific sense, where conclusions are probabilistic, mathematical induction provides certainty.

For example, when we observe a pattern like the one in our exercise with the inequality involving powers of three, we can apply inductive reasoning to hypothesize that a certain relationship seems to hold true for all integers. We then rigorously prove this hypothesis with mathematical induction. Precalculus students often encounter inductive reasoning when dealing with sequences and series, where understanding and proving the general form is critical.

Precalculus sets the foundation for higher-level mathematics. Here, mathematical induction is a stepping stone that fosters strong logical thinking skills necessary for more abstract concepts encountered in calculus and beyond.