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Chapter 4: Problem 10

Use mathematical induction to prove the following proposition: Let \(x\) be a real number with \(x>0\). Then for each natural number \(n\) with \(n \geq 2,(1+x)^{n}>1+n x\). Explain where the assumption that \(x>0\) was used in the proof.

Short Answer

Expert verified
To prove the given proposition using mathematical induction, we first showed that the base case (\(n=2\)) holds true as \(x^2 > 0\) when \(x>0\). We then assumed that the proposition is true for \(n=k\) and showed it is true for \(n=k+1\), using the assumption \(x>0\) to divide both sides of an inequality. The assumption that \(x>0\) was crucial in multiple steps of our proof, including the base case and the inductive step. Thus, we have proven that for every natural number \(n\) with \(n\geq 2\), and for any real number \(x\) with \(x>0\), we have \((1+x)^n > 1 + nx\).

Step by step solution

01

Define Base Case (\(n=2\))

Let's check if the given proposition is true for \(n=2\). Let's plug in \(n=2\) into the inequality and check if it holds. \[ (1+x)^2 > 1 + 2x \] Expanding the left side of the inequality: \[ 1 + 2x + x^2 > 1 + 2x \] Simplifying and canceling out the terms that appear on both sides of the inequality, we get: \[ x^2 > 0 \] Since \(x > 0\), we know that \(x^2 > 0\). Thus, the given proposition is true for the base case \(n=2\).
02

Assume Proposition True for \(n=k\)

Let us assume that the given proposition holds true for \(n=k\), where \(k \geq 2\). So, we have: \[ (1+x)^k > 1 + kx \]
03

Show Proposition True for \(n=k+1\)

We need to show that the given proposition also holds true for \(n=k+1\). Consider the inequality for \(n=k+1\): \[ (1+x)^{k+1} > 1 + (k+1)x \] To show this, let's multiply both sides of our assumption inequality \((1+x)^k > 1 + kx\) by \((1+x)\): \[ (1+x)^k (1+x) > (1+kx)(1+x) \] Expanding the right side of the inequality: \[ (1+kx)(1+x) = 1 + kx + x + kx^2 \] Now, let's compare the expression on the right side with the expression we want to prove: \[ 1 + kx + x + kx^2 > 1 + (k+1)x \] Check if the additional term on the left side, \(kx^2\), is greater than the additional term in the expression we want to prove, \(x\): \[ kx^2 > x \] Since we know that \(x > 0\), we can divide both sides by \(x\): \[ kx > 1 \] Our initial assumption \((1+x)^k > 1 + kx\) guarantees that the left side of the inequality above is greater than 1, ensuring that our inductive step is valid. Thus, by mathematical induction, we have proven that for every natural number \(n\) with \(n\geq 2\), and for any real number \(x\) with \(x>0\), we have \((1+x)^n > 1 + nx\).
04

Explain the Use of Assumption \(x>0\)

The assumption that \(x>0\) was utilized in multiple parts of the proof: 1. In the base case for \(n=2\), we noted that since \(x > 0\), it must be true that \(x^2 > 0\). 2. In the inductive step, to show that the proposition holds for \(n=k+1\), we divided both sides of the inequality \(kx^2 > x\) by \(x\). Dividing by \(x\) was only valid because \(x>0\). Therefore, this proof relies heavily on the assumption that \(x > 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Base Case Verification
The foundation of any mathematical induction proof starts with what's known as a 'base case verification'. This is the part of the proof where you test your proposition on the initial value, in most cases, when the variable in question is at its smallest integer value for which the proposition is claimed to be true. For example, if we're given that the inequality (1+x)^{n}>1+n x holds for each natural number greater than or equal to 2, we first check this for the smallest value, which in this case is 2.

It's essential to establish the base case because this lays the groundwork for the next stage of the proof and confirms that the initial premise is correct. When verifying the base case, we are looking to show that the inequality holds by actually plugging in the value and ensuring that our proposition evaluates as true. If the inequality doesn't hold for the base case, then the entire proposition fails. However, when the inequality holds true, like how the square of a positive number must indeed be positive ((1+x)^2 > 1 + 2x simplifies to x^2 > 0 for x>0), we're ready to move on to the next pivotal stage, the inductive step.
Inductive Step
Now that we're confident our proposition is valid for the base case, we approach the heart of a mathematical induction proof 鈥� the 'inductive step'. This stage involves taking an assumption that our proposition is true for some arbitrary natural number 'k' and then, based on that assumption, proving that the statement must also be true for the next natural number, 'k+1'.

The key in this phase is to judiciously apply the assumption to manipulate the inequalities or equalities in question, adding to them in such a way to reflect the 'k+1' scenario. In our exercise, this would mean understanding how the expression (1+x)^k (1+x) expands and exceeds (1+kx)(1+x), essentially why (1+x)^{k+1} should be greater than 1 + (k+1)x. The logical progression from assuming (1+x)^k > 1 + kx to the required inequality for 'k+1' is critical and often the most creative step of the induction proof.

One should note that each movement or manipulation of our terms, especially when inequalities are involved, must maintain the direction of the inequality and rest upon solid mathematical laws 鈥� divisibility rules, properties of exponents, and the like. Lastly, the assumption that 'k+1' is also true ensures that the truth of our proposition cascades from our base value all the way through the natural numbers, establishing our proof comprehensively.
Inequality Proof
When dealing with an 'inequality proof' in the context of a mathematical induction problem, the strategy involves a few delicate considerations. Unlike an equality proof where each step is reversible, inequalities demand careful attention to preserve the inequality's direction. Verifying the inequality involves ensuring no steps involve multiplying or dividing by negative numbers (which would invert the inequality) and being cautious with non-linear operations which might disrupt the initial relation.

In our example with real numbers and natural numbers, the assumption that 'x' must be greater than zero ((x>0) is pivotal. This condition has major implications in how we manipulate the inductive hypothesis. It allows us to multiply both sides of an inequality by 'x' without reversing the inequality sign, which is crucial in the inductive step of our proof. It is essential for the inductive step, where we show that kx^2 > x, and hence, by dividing both sides by 'x', underpinning that kx > 1, while safeguarding the inequality.

Understanding this, we can see why the assumption that 'x>0' is not just a trivial detail but a core part of the inductive process. Thus, an inequality proof in mathematical induction not only relies on the 'inductive hypothesis' but also on characteristics of the variables involved, in this case, that the real number 'x' is strictly positive.

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Most popular questions from this chapter

Evaluation of proofs See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) Let \(f_{n}\) be the \(n^{\text {th }}\) Fibonacci number, and let \(\alpha\) be the positive solution of the equation \(x^{2}=x+1 .\) So \(\alpha=\frac{1+\sqrt{5}}{2}\). For each natural number \(n, f_{n} \leq \alpha^{n-1}\) Proof. We will use a proof by mathematical induction. For each natural number \(n,\) we let \(P(n)\) be, " \(f_{n} \leq \alpha^{n-1}\) " We first note that \(P(1)\) is true since \(f_{1}=1\) and \(\alpha^{0}=1\). We also notice that \(P(2)\) is true since \(f_{2}=1\) and, hence, \(f_{2} \leq \alpha^{1}\). We now let \(k\) be a natural number with \(k \geq 2\) and assume that \(P(1)\), \(P(2), \ldots, P(k)\) are all true. We now need to prove that \(P(k+1)\) is true or that \(f_{k+1} \leq \alpha^{k}\). Since \(P(k-1)\) and \(P(k)\) are true, we know that \(f_{k-1} \leq \alpha^{k-2}\) and \(f_{k} \leq \alpha^{k-1}\). Therefore, $$ \begin{aligned} f_{k+1} &=f_{k}+f_{k-1} \\ f_{k+1} & \leq \alpha^{k-1}+\alpha^{k-2} \\ f_{k+1} & \leq \alpha^{k-2}(\alpha+1) . \end{aligned} $$ We now use the fact that \(\alpha+1=\alpha^{2}\) and the preceding inequality to obtain $$ \begin{array}{l} f_{k+1} \leq \alpha^{k-2} \alpha^{2} \\ f_{k+1} \leq \alpha^{k} \end{array} $$ This proves that if \(P(1), P(2), \ldots, P(k)\) are true, then \(P(k+1)\) is true. Hence, by the Second Principle of Mathematical Induction, we conclude that or each natural number \(n, f_{n} \leq \alpha^{n-1}\).

The quadratic formula can be used to show that \(\alpha=\frac{1+\sqrt{5}}{2}\) and \(\beta=\frac{1-\sqrt{5}}{2}\) are the two real number solutions of the quadratic equation \(x^{2}-x-1=0\). Notice that this implies that $$ \begin{array}{l} \alpha^{2}=\alpha+1, \text { and } \\ \beta^{2}=\beta+1 \end{array} $$ It may be surprising to find out that these two irrational numbers are closely related to the Fibonacci numbers. (a) Verify that \(f_{1}=\frac{\alpha^{1}-\beta^{1}}{\alpha-\beta}\) and that \(f_{2}=\frac{\alpha^{2}-\beta^{2}}{\alpha-\beta}\). (b) (This part is optional, but it may help with the induction proof in part (c).) Work with the relation \(f_{3}=f_{2}+f_{1}\) and substitute the expressions for \(f_{1}\) and \(f_{2}\) from part (a). Rewrite the expression as a single fraction and then in the numerator use \(\alpha^{2}+\alpha=\alpha(\alpha+1)\) and a similar equation involving \(\beta .\) Now prove that \(f_{3}=\frac{\alpha^{3}-\beta^{3}}{\alpha-\beta}\).(c) Use induction to prove that for each natural number \(n,\) if \(\alpha=\frac{1+\sqrt{5}}{2}\) and \(\beta=\frac{1-\sqrt{5}}{2},\) then \(f_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta} .\) Note: This formula for the \(n^{t h}\) Fibonacci number is known as Binet's formula, named after the French mathematician Jacques Binet ( \(1786-1856\) ).

There is a famous theorem in Euclidean geometry that states that the sum of the interior angles of a triangle is \(180^{\circ}\). (a) Use the theorem about triangles to determine the sum of the angles of a convex quadrilateral. Hint: Draw a convex quadrilateral and draw a diagonal. (b) Use the result in Part (a) to determine the sum of the angles of a convex pentagon. (c) Use the result in Part (b) to determine the sum of the angles of a convex hexagon. (d) Let \(n\) be a natural number with \(n \geq 3 .\) Make a conjecture about the sum of the angles of a convex polygon with \(n\) sides and use mathematical induction to prove your conjecture.

Use mathematical induction to prove that for each natural number \(n, 3 \mathrm{di}-\) vides \(n^{3}+23 n\). Compare this proof to the proof from Exercise (19) in Section 3.5

Assume that \(f_{1}, f_{2}, \ldots, f_{n}, \ldots\) are the Fibonacci numbers. Prove each of the following: \(k\) (a) For each \(n \in \mathbb{N}, f_{4 n}\) is a multiple of 3 . (b) For each \(n \in \mathbb{N}, f_{5 n}\) is a multiple of 5 . (c) For each \(n \in \mathbb{N}\) with \(n \geq 2, f_{1}+f_{2}+\cdots+f_{n-1}=f_{n+1}-1\). (d) For each \(n \in \mathbb{N}, f_{1}+f_{3}+\cdots+f_{2 n-1}=f_{2 n}\). (e) For each \(n \in \mathbb{N}, f_{2}+f_{4}+\cdots+f_{2 n}=f_{2 n+1}-1\). (f) For each \(n \in \mathbb{N}, f_{1}^{2}+f_{2}^{2}+\cdots+f_{n}^{2}=f_{n} f_{n+1}\). (g) For each \(n \in \mathbb{N}\) such that \(n \neq 0(\bmod 3), f_{n}\) is an odd integer.
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