inequalities for real numbers

Suppose $a$ is a real number.

1.

If $a<0$ then $a$ is a negative number.
2.

If $a>0$ then $a$ is a positive number.
3.

If $a\leq 0$ then $a$ is a non-positive number.
4.

If $a\geq 0$ then $a$ is a non-negative number.

The first two inequalities are also called strict inequalities.
The second two inequalities are also called loose inequalities.

Properties

Suppose $a$ and $b$ are real numbers.

1.

If $a>b$ , then $-a<-b$ . If $a<b$ , then $-a>-b$ .
2.

If $a\geq b$ , then $-a\leq-b$ . If $a\leq b$ , then $-a\geq-b$ .

Lemma 1.

$0<a$ iff $-a<0$ .

Proof.

If $0<a$ , then adding $-a$ on both sides of the inequality gives $-a=-a+0<-a+a=0$ . This process can also be reversed. ∎

Lemma 2.

For any $a\in\mathbbmss{R}$ , either $a=0$ or $0<a^{2}$ .

Proof.

Suppose $a\neq 0$ , then by trichotomy, we have either $0<a$ or $a<0$ , but not both. If $0<a$ , then $0=0\cdot a<a\cdot a=a^{2}$ . On the other hand, if $-(-a)=a<0$ , then $0<-a$ by the previous lemma. Then repeating the previous , $0=0\cdot(-a)<(-a)(-a)=a^{2}$ . ∎

Three direct consequences follow:

Corollary 1.

$0<1$

Corollary 2.

For any $a\in\mathbbmss{R}$ , $0<1+a^{2}$ .

Corollary 3.

There is no real solution for $x$ in the equation $1+x^{2}=0$ .

Inequality for a converging sequence

Suppose $a_{0},a_{1},\ldots$ is a sequence of real numbers converging to a real number $a$ .

1.

If $a_{i}<b$ or $a_{i}\leq b$ for some real number $b$ for each $i$ , then $a\leq b$ .
2.

If $a_{i}>b$ or $a_{i}\geq b$ for some real number $b$ for each $i$ , then $a\geq b$ .

Title	inequalities for real numbers
Canonical name	InequalitiesForRealNumbers
Date of creation	2013-03-22 13:58:16
Last modified on	2013-03-22 13:58:16
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	12
Author	mathcam (2727)
Entry type	Definition
Classification	msc 54C30
Classification	msc 26-00
Classification	msc 12D99
Related topic	SummedNumeratorAndSummedDenominator
Defines	strict inequality
Defines	inequality