Problem 29 - College Algebra

Question

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

Step-by-Step Solution

Answer

By mathematical induction, the statement \((ab)^n = a^n b^n\) is true for all positive integers \(n\). This is shown by proving the base case with \(n=1\), assuming the statement true for \(n=k\), then proving true for \(n=k+1\).

Step 1: Base Case (n=1)

When \(n=1\), the given statement becomes \((ab)^1 = a^1b^1\), which can be simplified as \(ab = ab\). Therefore, the statement is true for \(n=1\). This will initiate the induction method.

Step 2: Inductive Step (Assuming the statement is true for n=k)

Assume that the given statement \((ab)^n = a^n b^n\) is true for \(n=k\). This assumption is our Inductive Hypothesis, and is expressed as \((ab)^k = a^k b^k\).

Step 3: Induction Conclusion (Showing the statement is true for n=k+1)

Now show that the given statement is true for \(n=k+1\), which means \((ab)^{k+1} = a^{k+1} b^{k+1}\). \n\nIf we expand \((ab)^{k+1}\), we get \((ab)^k * ab\). From our Inductive Hypothesis, \((ab)^k = a^k b^k\), so we can replace it in our equation which then becomes: \(a^k b^k * ab = a^{k+1} b^{k+1}\), which is our original statement for \(n=k+1\). Therefore, the statement is proved to be true for \(n=k+1\) if it is true for \(n=k\).

Step 4: Conclusion

Since the statement is true for \(n=1\) (base case), and if it's true for \(n=k\) it's also true for \(n=k+1\) (inductive step), then by mathematical induction, the statement \((ab)^n = a^n b^n\) is true for all positive integers \(n\).

Key Concepts

Base Case

In mathematical induction, the base case is where it all begins. It's like the first domino in a long row that, once nudged, sets everything into action. For the proof by induction, establishing this first step is crucial.
In our exercise, the base case checks if the statement works for the smallest positive integer, usually 1. We substitute \(n = 1\) into the statement \((ab)^n = a^n b^n\), and it simplifies to \(ab = ab\). Simple, right? So, the base case holds true because both sides of the equation are equal. This step must always be verified to ensure the domino effect can start. Without proving the base case, we cannot move on to the next stages of the proof.

Inductive Step

After confirming the base case, the inductive step is like saying, "If one domino falls, then the next will fall too." In this step, we assume that the statement is true for some positive integer \(n = k\). This assumption is called the Inductive Hypothesis.
The hypothesis means we assume \((ab)^k = a^k b^k\) is true. Now, we need to show that if it's true for \(n = k\), then it must also be true for \(n = k+1\). We expand \((ab)^{k+1}\) to \((ab)^k \times ab\). By substituting the Inductive Hypothesis, this becomes \(a^k b^k \times ab = a^{k+1} b^{k+1}\). This shows the next step in the chain reaction is also true. If both the base case and this inductive step are valid, the statement can be deemed true for all positive integers.

Positive Integers

The domain of positive integers is central to mathematical induction. Positive integers are numbers like 1, 2, 3, and so forth—any whole number greater than zero. They are the numbers we consider when using mathematical induction.
Induction works because it relies on these numbers having a natural order. Starting from the base case with the smallest integer, the method shows that if a statement is true for an integer \(k\), it holds for \(k+1\) as well. This property wouldn't work if we had negative numbers or fractions, as their order isn't as straightforward.

• Positive integers provide a clear progression: 1, 2, 3, etc.
• They ensure that once the base case is established, each step builds on the previous, without gaps.

Thus, when dealing with proofs by induction, we always refer to these positive integers.

Algebraic Proof

An algebraic proof like the one in our exercise employs algebraic identities and operations to show that a statement is always true. Mathematical induction is a powerful tool in these proofs.
In

• the base case, we use simple substitution to confirm that the statement holds for a first integer.
• The inductive step involves assuming that the statement is true for some integer \(n = k\) and proving it true for the next integer \(n = k+1\). This often involves algebraic manipulation.

Inductive proofs hinge on clear algebraic principles and unfold step-by-step, carefully showing that the conditions hold for each subsequent integer. As a result, algebraic proofs demonstrate consistency and accuracy, which are key to proving mathematical statements like that in our exercise.