Problem 50
Question
Given that \(a > 0\) and \(b > 0,\) show that if \(a^{2} \leq b^{2}\), then \(a \leq b\).
Step-by-Step Solution
Verified Answer
With the given conditions, if \(a^{2} \leq b^{2}\), then it is shown that \(a \leq b\). This has been proven by factoring the inequality using the difference of squares and using the positivity of \(a\) and \(b\).
1Step 1: Identifying Given Information
The problem provides three pieces of information. The first two are that \(a > 0\) and \(b > 0\). This means both \(a\) and \(b\) are positive. The third piece of information is that \(a^{2} \leq b^{2}\). This will be the main inequality used throughout this exercise.
2Step 2: Simplify the Given Inequality
The given inequality \(a^{2} \leq b^{2}\) can also be written as \(b^{2} - a^{2} \geq 0\). The left side of the inequality can then be factored using the difference of squares method yielding \((b - a) (b + a) \geq 0\).
3Step 3: Analyze the Factored Inequality
We know both \(b\) and \(a\) are positive and hence \(b + a\) is positive. Therefore, for the inequality to hold, \((b - a)\) must be greater than or equal to zero. This can be simply written as \(b \geq a\) or \(a \leq b\). This is the required result.
Key Concepts
Positive Numbers PropertiesDifference of SquaresFactoring Quadratic ExpressionsSolving Inequalities
Positive Numbers Properties
Understanding the properties of positive numbers is essential when working with inequalities and equations in calculus. In essence, positive numbers are all the numbers greater than zero. As an important note, any non-zero number squared is positive because the product of a number by itself will always be non-negative. This implies that if you are dealing with variables that are constrained to be positive, as we are in the given exercise with constraints that both a and b are greater than zero, then their squares, a^2 and b^2, will also be positive.
In the context of our original exercise, taking the square roots of both sides of the inequality a^2 ≤ b^2 retaining the inequality order is valid because the square root function is increasing for positive numbers. This is why when we have a^2 ≤ b^2 and know that a and b are positive, we can confidently state that a ≤ b.
In the context of our original exercise, taking the square roots of both sides of the inequality a^2 ≤ b^2 retaining the inequality order is valid because the square root function is increasing for positive numbers. This is why when we have a^2 ≤ b^2 and know that a and b are positive, we can confidently state that a ≤ b.
Difference of Squares
The difference of squares is a quick factoring technique that applies to expressions of the form a^2 - b^2. It relies on the property that such an expression can always be factored into (a + b)(a - b). This technique is incredibly useful in simplifying equations and inequalities involving squares.
In our exercise, the inequality a^2 ≤ b^2 gets rewritten as b^2 - a^2 ≥ 0. When we factor b^2 - a^2, we apply the difference of squares and obtain the factored form (b + a)(b - a). Since we know both a and b are positive, b + a is definitely positive, leaving the sign of (b - a) to determine the truth of the inequality.
In our exercise, the inequality a^2 ≤ b^2 gets rewritten as b^2 - a^2 ≥ 0. When we factor b^2 - a^2, we apply the difference of squares and obtain the factored form (b + a)(b - a). Since we know both a and b are positive, b + a is definitely positive, leaving the sign of (b - a) to determine the truth of the inequality.
Factoring Quadratic Expressions
Factoring quadratic expressions is a technique to simplify and solve quadratic equations. A quadratic expression generally takes the form ax^2 + bx + c. When factoring, we look for two numbers that both add up to b and multiply to ac. In the case of a difference of squares, which is a specific type of quadratic expression where b = 0 and c = -a^2, the factoring becomes straightforward, splitting the expression into two binomials.
Understanding this concept is crucial in our exercise because after determining the inequality b^2 - a^2 ≥ 0, factoring allows us to break it down into simpler parts, namely (b - a) and (b + a), which we can then evaluate based on the given information that a and b are positive.
Understanding this concept is crucial in our exercise because after determining the inequality b^2 - a^2 ≥ 0, factoring allows us to break it down into simpler parts, namely (b - a) and (b + a), which we can then evaluate based on the given information that a and b are positive.
Solving Inequalities
Solving inequalities involves finding the values that make the inequality true. Unlike equations, inequalities do not have a 'single solution,' rather they define a range of possible solutions. Critical to solving inequalities is understanding the behavior of inequalities when multiplying or dividing by negative numbers, which flips the inequality sign, and recognizing how to dismantle compound inequalities into simpler parts.
In our problem, from a factored inequality (b - a)(b + a) ≥ 0, knowing that one factor (b + a) is positive allows us to focus on the sign of (b - a). Since (b - a) must also be greater than or equal to zero for the product to be non-negative, we deduce that the inequality b ≥ a must hold, indicating the solution to our original problem and completing our inequality reasoning.
In our problem, from a factored inequality (b - a)(b + a) ≥ 0, knowing that one factor (b + a) is positive allows us to focus on the sign of (b - a). Since (b - a) must also be greater than or equal to zero for the product to be non-negative, we deduce that the inequality b ≥ a must hold, indicating the solution to our original problem and completing our inequality reasoning.
Other exercises in this chapter
Problem 50
State whether the function is odd, even, or neither. $$f(x)=x^{2}+1$$
View solution Problem 50
Find the solutions \(x\) that are in the interval [0.2 \(\pi\) ]. Express your answers in radians and use four decimal place accuracy. $$\csc x=10.260$$.
View solution Problem 50
Estimate the point(s) of intersection. $$l_{1}: 2 x-3 y=5, \quad \text { circle }: x^{2}+y^{2}=4$$
View solution Problem 51
State whether the function is odd, even, or neither. $$g(x)=x(x-1)$$
View solution