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Chapter 1: Problem 36

$$ \text { If } a>b \geq 0, \text { prove that } a^{2}>b^{2} $$

Short Answer

Expert verified
If \(a > b \text{ and } b \text{鈮� 0}\), then \(a^{2} > b^{2}\).

Step by step solution

01

- Understand the Given Inequality

Given the inequality, we have two numbers where \(a > b \text{ and } b \text{ is non-negative (} b \text{ 鈮� 0)}\). Thus, between the two numbers, \(a\) is always larger than or equal to \(b\), and both numbers are non-negative.
02

- Square Both Sides of the Inequality

When you square both sides of an inequality, the direction of the inequality remains the same provided both sides are non-negative numbers. Since \(a > b\), squaring both sides results in:
03

- Conclusion

Since \(a\) is greater than \(b\), and \(b\) is non-negative, squaring both sides maintains the inequality. Therefore, \(a^{2}\) is greater than \(b^{2}\).

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Key Concepts

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

Squaring inequalities
Squaring inequalities can be a bit tricky.
To simplify, if we have two numbers and we know one is greater than the other, squaring can help maintain that relationship. Consider the inequality given: \( a > b \).
We know that \( a \) is greater than \( b \), but both \( a \) and \( b \) are non-negative numbers. This part is crucial because squaring negative numbers would invert the inequality.
For non-negative numbers, when you square them, the inequality holds. That means:
  • If \( a > b \), then \( a^2 > b^2 \)
  • This is true only if both \( a \) and \( b \) are non-negative.

It's straightforward when dealing with positive numbers. When squaring an inequality, always ensure the numbers involved are clear and check their properties.
Non-negative numbers
Non-negative numbers are numbers that are either positive or zero.
This means:
  • Any number greater than or equal to zero.
In the context of our problem, since \( b \geq 0 \), we know \( b \) is non-negative.
Non-negative numbers play a special role in inequalities as they ensure the inequality holds after squaring. This is vital because:
  • Squaring a negative number would flip the inequality.
  • With non-negative numbers, however, the direction remains unchanged.
For example:
Given \( a > b \geq 0 \), we know
That \( a^2 > b^2 \).
This holds true because both \( a \) and \( b \) are non-negative. Non-negative numbers thus provide a safe environment for operations like squaring without altering the nature of the inequality.
Steps in proof
In mathematical proofs, breaking down the problem into smaller steps helps in understanding and solving it efficiently. Let's look at the steps for our inequality problem:
  • Step 1: Understand the Given Inequality: We are given that \( a > b \geq 0 \). This implies \( a \) is greater than or equal to \( b \), and \( b \) is a non-negative number.
  • Step 2: Square Both Sides of the Inequality: Since both \( a \) and \( b \) are non-negative, squaring will keep the inequality direction the same. Hence, squaring both \( a \) and \( b \) gives us \( a^2 > b^2 \).
  • Step 3: Conclusion: Our final statement shows that \( a^2 > b^2 \), which was what we needed to prove. This is true because squaring non-negative numbers keeps the original inequality intact.
These steps help systematically solve and understand the problem. They ensure each part of the proof is logically derived and easy to follow.

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Most popular questions from this chapter

In Exercises 11 through 32 , find the solution set of the given inequality and illustrate the solution on the real number $$ x^{3}+1>x^{2}+x $$

In Exercises 7 through 12, the functions \(f\) and \(g\) are defined. In each problem define the following functions and determine the domain of the resulting function: (a) \(f+g ;\) (b) \(f-g ;\) (c) \(f \cdot g ;\) (d) \(f / g ;\) (e) \(g / f\); (f) \(f \circ g ;(\mathrm{g}) g \circ f\). $$ f(x)=\frac{x+1}{x-1} ; g(x)=\frac{1}{x} $$

In Exercises 11 through 34, the function is the set of all ordered pairs \((x, y)\) satisfying the given equation. Find the domain and range of the function, and draw a sketch of the graph of the function. $$ H: y= \begin{cases}x^{2}-4 & \text { if } x<3 \\ 2 x-1 & \text { if } 3 \leq x\end{cases} $$

In Exercises 5 through 10, find an equation of the circle satisfying the given conditions. Center is at \((-2,5)\) and tangent to the line \(x=7\).

In Exercises 7 through 12, the functions \(f\) and \(g\) are defined. In each problem define the following functions and determine the domain of the resulting function: (a) \(f+g ;\) (b) \(f-g ;\) (c) \(f \cdot g ;\) (d) \(f / g ;\) (e) \(g / f\); (f) \(f \circ g ;(\mathrm{g}) g \circ f\). $$ f(x)=\sqrt{x^{2}-1} ; g(x)=\sqrt{x-1} $$
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