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Chapter 3: Problem 103

Let \(x>0\) and \(n>1\) be real numbers. Prove that \((1+x)^{n}>1+n x\)

Short Answer

Expert verified
The inequality \((1+x)^n > 1 + nx\) is proven by mathematical induction, establishing it holds for the base case at \(n=2\), assuming it holds for \(n=k\), and proving it holds for \(n=k+1\).

Step by step solution

01

Base Case

Start by testing the base case, \(n=2\). Plug into the inequality \((1+x)^2 > 1 + 2x\) to find that it simplifies to \(x^2 > 0\), which is true because \(x > 0\). Therefore, the inequality holds for the base case.
02

Inductive Hypothesis

Next, assume the inequality holds for some \(k\), that is, \((1+x)^k > 1 + kx\). This is your inductive hypothesis.
03

Inductive Step

To complete the proof, you need to show the inequality also holds for \(k+1\). Start with the left side of the inequality, \((1+x)^{k+1} = (1+x)^k *(1+x)\). By the inductive hypothesis, \((1+x)^k > 1+kx\), so \((1+x)^k*(1+x) > (1+kx)(1+x) = 1+x+kx+x^2 = (1+(k+1)x) + x^2\). Since \(x > 0\) and \(x^2 >= 0\), \((1+(k+1)x) + x^2 > 1+(k+1)x\). Therefore, \((1+x)^{k+1} > 1+(k+1)x\).
04

Conclusion

The inequality \((1+x)^n > 1+nx\) has been proven to hold for all real numbers \(n > 1\) and \(x > 0\) by mathematical induction.

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Key Concepts

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

Inequality Proof
Understanding the proof of an inequality using mathematical induction involves establishing that an inequality holds for all members of a specific set, often the natural numbers or, as in our exercise, the real numbers greater than a certain value. Here's a breakdown of the components you need to keep in mind:
  • Base Case: The starting point where the inequality is tested for the smallest number the statement is meant to apply to鈥攊n our case, when 'n' is 2.
  • Inductive Hypothesis: The assumption that the inequality holds true for a certain 'k'. It's like saying, 'We'll trust that this works for this one instance.'
  • Inductive Step: Using the inductive hypothesis to show that if the inequality is valid for 'k', it must also be true for 'k+1.'
  • Conclusion: After confirming the steps above, we conclude that the inequality holds for all valid numbers following the base case.
By systematically verifying these steps, we can confidently assert the truth of the inequality without testing every possible value, which is often impractical or impossible.
Inductive Hypothesis
The heart of mathematical induction lies in the inductive hypothesis. In plain terms, it's the 'assume' step, where we take for granted that our inequality works for an arbitrary but particular stage 'k'.

The inductive hypothesis serves as a bridge from the known (our base case) to the unknown (proving the statement for all numbers). Placing our trust in this hypothesis is what enables us to proceed with the inductive step. To break it down:
  • First, we say, 'Let's assume the inequality is true when 'n' equals 'k'.'
  • Then, we rely on this assumption to prove that if the statement holds for 'k', it will hold for 'k+1' as well, pushing the boundary further by one more number.
If you find this concept challenging, imagine a row of dominoes鈥攊f one falls, the next one must follow. The inductive hypothesis confirms the fall of one domino, allowing us to predict the fall of subsequent ones endlessly.
Real Numbers
Real numbers encompass a vast array of numbers, including rational and irrational numbers, positives, negatives, and zero. Whenever we're tackling exercises with inequalities and real numbers, understanding some key properties is essential:
  • Orderliness: Real numbers can always be placed in a particular sequence, making inequalities meaningful.
  • Completeness: Every non-empty set of real numbers that's bounded above has a least upper bound; this property often underpins the logic of proving inequalities.
  • Continuity: The real numbers have no 'gaps,' which makes our inductive approach valid.
In our example, the real numbers we are considering are all greater than zero. This fact ensures that when we work with powers and products, like \(x^2 > 0\) or \(1+x > 1\), we're applying properties of positivity and multiplication without fear of unexpected behavior, such as negative numbers flipping the sign of an inequality.

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