Problem 30 - College Algebra

Question

In Exercises \(11-30,\) use mathematical induction to prove that each statement is true for every positive integer \(n\) $$ \left(\frac{a}{b}\right)^{n}=\frac{a^{n}}{b^{n}} $$

Step-by-Step Solution

Answer

Using the principle of mathematical induction, the statement \( \left(\frac{a}{b}\right)^{n}=\frac{a^{n}}{b^{n}} \) has been successfully proved to be true for every positive integer \(n\). The process involved a base step validating the statement for \(n = 1\), and an inductive step which showed, by assuming the statement true for an arbitrary integer \(k\), it also holds for \(k + 1\).

Step 1: Base case for \(n = 1\)

First, the base case is checked for \(n = 1\). Plug \(n = 1\) into the equation, you get \( \left(\frac{a}{b}\right)^1 = \frac{a^1}{b^1} \). Since raising anything to the power 1 gives the same number, the base case correctly simplifies to \( \frac{a}{b} = \frac{a}{b} \). Thus, the base case holds true.

Step 2: Inductive Hypothesis

For the inductive hypothesis, the assumption is made that the statement is true for an arbitrary positive integer \(k\). Hence it is assumed that \( \left(\frac{a}{b}\right)^k = \frac{a^k}{b^k} \). This assumption will be used in the inductive step.

Step 3: Inductive Step

In this step, the aim is to show the statement holds for \(k+1\). Starting with the left-hand side of the equation for \(n=k+1\): \( \left(\frac{a}{b}\right)^{k+1} = \left(\frac{a}{b}\right)^k * \left(\frac{a}{b}\right) \). Using the inductive hypothesis, it can be replaced \( \left(\frac{a}{b}\right)^k \) with \(\frac{a^k}{b^k}\), hence, \( \left(\frac{a}{b}\right)^{k+1} = \frac{a^k}{b^k} * \frac{a^1}{b^1}= \frac{a^{k+1}}{b^{k+1}} \). The left-hand side of the equation now matches the right-hand side for \(n=k+1\). Thus the inductive step holds and the statement is proved to be true for any positive integer \(n\).

Key Concepts

Proof by Induction

Proof by induction is a powerful mathematical technique used to establish the truth of a given proposition for all natural numbers. It's akin to a series of mathematical dominoes; once the first one falls (the base case), and provided that the falling of one domino causes the next one to fall (the inductive step), you ensure that all the dominoes will fall in sequence.

By demonstrating that the base case is true and that the truth of the proposition for a certain number 'k' implies its truth for the next number 'k+1', you can infer that the proposition holds for all natural numbers. This technique can be particularly valuable when dealing with infinite sequences where checking each individual case would be impossible.

Exponentiation of Fractions

Exponentiation of fractions might seem tricky at first, but it follows the same rules as exponentiation with whole numbers. When you raise a fraction \(\frac{a}{b}\) to the power of \(n\), each part of the fraction—the numerator \(a\) and the denominator \(b\)—gets raised to the power of \(n\) separately. This is mathematically expressed as \(\left(\frac{a}{b}\right)^{n} = \frac{a^{n}}{b^{n}}\).

Understanding this concept is crucial when solving problems involving the growth or decay of quantities, and in scenarios where quantities are repeatedly multiplied, such as in compound interest calculations or in the computation of probabilities.

Inductive Step

The inductive step is where you prove that if the statement is true for some integer \(k\), then it must also be true for \(k+1\). Think of it as the mechanism that orchestrates the domino effect mentioned earlier—it ensures that the initial truth (the base case) echoes through every subsequent case.

To successfully complete the inductive step, you manipulate the expression from its known form at \(k\) to its desired form at \(k+1\). This often involves algebraic manipulation, using known identities, or employing the assumption (from the inductive hypothesis) to transform the expression. If done correctly, the inductive step not only links \(k\) to \(k+1\) but it also cements the logic of the induction process.

Base Case

The base case is the starting point of mathematical induction and serves as the first 'domino' in the sequence. It typically involves proving that the proposition is true for the smallest natural number, usually \(n=1\). Establishing the base case is crucial because without it, the entire proof would crumble.

To ascertain the base case, you simply substitute \(n=1\) into the given proposition and verify its truth. For the exponentiation of fractions, you would plug \(n=1\) into the formula \(\left(\frac{a}{b}\right)^{n} = \frac{a^{n}}{b^{n}}\) to show that \(\frac{a}{b} = \frac{a}{b}\), which is evidently true. Having a valid base case paves the way for applying the inductive step to the rest of the natural numbers.