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Inequalities

Introduction

In an algebra course, students are introduced to the concept of order on the real numbers. This is a relation between the real numbers, denoted by the symbol \( > \), which is often read as "greater than". We have the following 4 order axioms for real numbers:

Axiom 1: [Trichotomy] If \( a, b \) are real numbers, then one and only one of the following is true: \( a >b \), \( a = b \), \( b > a \).

Axiom 2: [Transitivity] If \( a, b, c \) are real numbers and \( a > b , b > c \), then \( a > c \).

Axiom 3: If \( a, b, c \) are real numbers and \( a>b \), then \( a + c > b + c \).

Axiom 4: If \( a, b, c \) are real numbers such that \( a>b, c > 0 \), then \( ac > bc \).

With this, we can formally define a positive and negative number. A positive number is a number \( a \) that satisfies \( a > 0 \), while a negative number is a number \( a \) that satisfies \( 0 > a \). By the trichotomy axiom, a real number is either positive, negative, or zero. However, this does not appear to tell us much. In particular, is 1 a positive number?

An important inequality is the trivial inequality which states that the square of any real number is non-negative. This is easily shown by considering Trichotomy cases:

  1. If \( f = 0 \), then \( f^2 = 0 \).

  2. If \( f > 0 \), then multiplying both sides by \( f \) gives \( f^2 > 0 \), where we keep the sign since \( f \) is positive.

  3. If \( 0 > f \), then \( -f > 0 \) (Axiom 3) and multiplying both sides by \( -f \) gives \( 0 > -f^2 \) (Axiom 4), and adding \( f^2 \) to both sides yields \( f^2 > 0 \) (Axiom 3).

With the trivial inequality, we now know that \( 1 = (1) ^2 \) is a positive number.

Polynomial inequalities can be solved by understanding how the graph behaves. Because the graph is continuous (naively this means that the graph can be drawn without lifting off your pen), we only need to find the roots of the polynomial and test one value in each of the corresponding regions to determine if they are positive or negative.

Application and Extensions

Determine the region in which \( x^2 < 1 \).

The proper approach would be to shift the terms to one side, and then factorize, to get \( x^2 - 1 < 0 \Leftrightarrow (x-1)(x+1) < 0 \).

This has roots \( 1 \) and \( -1 \), so we test the 3 different regions ( i.e. \( x < -1, -1 < x < 1, 1<x \), to conclude that the inequality is satisfied for \( -1 < x < 1 \).

Note: Similar to the equality case \( x^2 = 1 \), the first instinct of many students is to simply take square roots on the inequality and conclude that \( x < 1 \). This likewise leads to the wrong answer.

 

Determine the region in which \( \frac {x^2+3x} {x+5} \leq \frac {x^2 + 5x +2 }{x+5} \).

To avoid dividing by 0, we want to ensure that the denominator is never 0. As such, we have to exclude \( x=-5\) as a possibility. From Axiom 4, we can only multiply by a quantity with a fixed sign.

Since \( x+5 \) is negative when \( x < -5 \), we cannot simply multiply by \( x+5 \). Instead, we can multiply by \( (x+5)^2 \), which we know is always non-negative, to obtain \( ( x^2 + 3x) (x+5) \leq (x^2 + 5x + 2 ) ( x+5 ). \)

At this point, it is tempting to cancel the term \( x+5 \) on both sides, but this is equivalent to multiplying by \( \frac {1}{x+5} \). We may not do so, we choose to factorize instead. Shifting all the terms to one side and factorizing, we get \( 0 \leq 2 (x+1) ( x+ 5). \)

We easily understand the graph of this quadratic equation, and (by using the number line) can conclude that the inequality is satisfied when \( x < -5 \) (Recall that \( x = -5\) is excluded) or \( x \geq -1 \).

 

Show that \( x^2 + 2x + 2 \geq 0 \) for all real values of \( x \).

If we tried to calculate the roots to the quadratic equation \( x^2 + 2x +2 = 0\), since the discriminant is negative, we know that the roots are complex numbers. Hence there is only 1 region on the number line. Testing \( x = 0 \) shows that the quantity is always positive.

Another approach is to rewrite \( x^2 + 2x + 2 = (x+1) ^2 + 1 \) by Completing the square. Since both terms are squares, they are non-negative, and hence their sum is non-negative (axiom 3).

 

Show that the product of 2 positive numbers is positive.

If \( X \) and \( Y \) are positive numbers, then \( X > 0, Y > 0 \). Using axiom 4, (with \( a = X, b = 0, c = Y \)), we get that \( X \cdot Y > 0 \cdot Y = 0 \). Hence, the product \( XY \) is positive.

Note by Calvin Lin
3 years, 6 months ago

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You can also add that inequalities are not valid for \(\text{complex numbers}\).
Example : We cannot compare which one is greater between \( 2 + 3i\) and \( 1+ 3i\). \(( i = \sqrt{-1}\))

Akhil Bansal - 2 years, 1 month ago

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You can discuss Wavy curve method also. :-)

Sachin Vishwakarma - 2 years, 1 month ago

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