Question

For natural number exponents and nonzero bases, most of the familiar laws of exponents can be proved by induction on the exponents using the facts that \(b^{0}=1\) (for \(b \neq 0\) ) and \(b^{n+1}=b \cdot b^{n}\), Assuming that \(m\) and \(n\) are natural numbers and both \(r\) and \(s\) are nonzero real numbers, prove the following: (a) \(r^{m+n}=r^{m} \cdot r^{n}\) (b) \(r^{m n}=\left(r^{m}\right)^{n}\). (c) If \(r>1,\) then \(r^{m}>r^{n}\) if and only if \(m>n\). (d) If \(n, r, s>0,\) then \(r^{n}>s^{n}\) if and only if \(r>s\).

Short answer

The statements (a), (b), (c), and (d) are all true under the given conditions.

Step-by-step solution

Prove Statement (a) using Induction

We need to prove that \(r^{m+n} = r^m \cdot r^n\). We will use mathematical induction on \(n\).**Base Case:** Let \(n = 0\). Then \(r^{m+n} = r^m = r^m \cdot 1 = r^m \cdot r^0\).**Inductive Step:** Assume \(r^{m+n} = r^m \cdot r^n\) for some \(n\). Prove for \(n+1\):\[r^{m+(n+1)} = r^{m+n+1} = r^{m+n} \cdot r = (r^m \cdot r^n) \cdot r = r^m \cdot (r^n \cdot r) = r^m \cdot r^{n+1}\]Thus, by induction, \(r^{m+n} = r^m \cdot r^n\) holds.

Prove Statement (b) using Induction

To prove \(r^{mn} = (r^m)^n\), we again use mathematical induction on \(n\).**Base Case:** For \(n = 0\), both sides equal 1 because \(r^{m \cdot 0} = r^0 = 1\) and \((r^m)^0 = 1\).**Inductive Step:** Assume \(r^{mn} = (r^m)^n\) for some \(n\). Prove for \(n+1\):\[r^{m(n+1)} = r^{mn+m} = r^{mn} \cdot r^m = (r^m)^n \cdot r^m = (r^m)^{n+1}\]Thus, by induction, \(r^{mn} = (r^m)^n\) holds.

Prove Statement (c) by Contradiction

Assume \(r > 1\) and \(r^m \leq r^n\) while \(m > n\). Then, \(r^{m-n} \leq 1\). But since \(r > 1\), any positive power of \(r\) is greater than 1, which contradicts \(r^{m-n} \leq 1\). Therefore, \(r^m > r^n\) if and only if \(m > n\).

Prove Statement (d) by Similar Reasoning

Assume \(n, r, s > 0\) and \(r^n \leq s^n\) while \(r > s\). This implies \( (r/s)^n \leq 1\). However, since \(r > s\), we have \(r/s > 1\) and so \((r/s)^n > 1\), which contradicts \((r/s)^n \leq 1\). Therefore, \(r^n > s^n\) if and only if \(r > s\).

Key concepts

Mathematical Induction

Mathematical induction is a powerful tool used in mathematics to prove statements involving natural numbers. It's like a domino effect in numbers. First, you show the statement is true for the starting point, often referred to as the "base case." Let's say the base case is for the number 0 or 1. Then, you assume it is true for some arbitrary number, say \( n \), and show it also holds for the next number \( n+1 \). By doing this, you essentially "cover" every natural number, ensuring the statement is true for all.

Imagine stacking blocks on a table: if you can place the first block and knowing if a block holds the next block, you can then endlessly add blocks. In our exercise, we applied induction to prove laws of exponents like \( r^{m+n} = r^m \cdot r^n \). Starting from simpler numbers, we extended this truth to all natural numbers using step-by-step assumptions and confirmations.

Natural Numbers

The concept of natural numbers stems from the basic need to count. They begin from 1 and extend infinitely as 1, 2, 3, 4, and so on. These numbers only include positive integers.

Natural numbers are foundational in mathematics. They form the basis for more complex notions such as integers, rational numbers, and real numbers. In the realm of powers and exponents, natural numbers serve as exponents to express repeated multiplication like \( r^n \), where \( n \) is a natural number.

They are simple yet hold immense significance in mathematical proofs and concepts. When we solve problems using exponents, we often rely on these numbers to describe cycles, count repetitions, or determine sequences.

Real Numbers

Real numbers are a broader category in the number system encompassing both rational and irrational numbers. You can think of them as points on an infinite line stretching in both the negative and positive directions, including all decimals and fractions.

Rational numbers, like \( \frac{1}{2} \) or 4, can be written as fractions, whereas irrational numbers, like \( \pi \) or \( \sqrt{2} \), cannot be expressed as clean fractions. Every real number has a position on the number line.

In the context of exponents, using real numbers allows us to consider behaviors and properties involving decimal expansions and roots. For instance, \( r \) in our equations, whether a whole number, fraction, or decimal, still follows the laws of exponents as proven using induction.

Proof By Contradiction

Proof by contradiction is an elegant method in mathematical reasoning. Instead of proving directly that something is true, you assume the opposite is true, then find inconsistencies leading to a contradiction.

This approach helps when direct proof is elusive or convoluted. It's a way of showing that some assumptions, when taken to logical conclusions, cannot all be true simultaneously.

In our exercise, we used proof by contradiction to show statements like \( r^m > r^n \) if \( m > n \) and \( r > 1 \). By considering the opposite situation, assuming \( r^m \le r^n \), and identifying the flaws that arise, we validated the statement's truth. It's like proving a friend didn't eat the last cookie by showing that their alibi contradicts various observed facts. The contradiction reveals the real truth.