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inequalities for real numbers
Suppose $a$ is a real number.
- If $a<0$ then $a$ is a negative number.
- If $a>0$ then $a$ is a positive number.
- If $a\le 0$ then $a$ is a non-positive number.
- If $a\ge 0$ then $a$ is a non-negative number.
The second two inequalities are also called loose inequalities.
Properties
Suppose $a$ and $b$ are real numbers.- If $a>b$ , then $-a<-b$ . If $a<b$ , then $-a>-b$ .
- If $a\ge b$ , then $-a\le -b$ . If $a\le b$ , then $-a\ge -b$ .
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
, either $a=0$ or $0<a^2$ .
, either $a=0$ or $0<a^2$ .
Proof. Suppose $a\ne 0$ , 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 argument, $0 = 0\cdot(-a) < (-a)(-a)=a^2$ . 
Three direct consequences follow:
Corollary 1 $0<1$
Corollary 2 For any
, $0<1+a^2$ .
, $0<1+a^2$ .
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 $a$ .- If $a_i < b$ or $a_i \le b$ for some real number $b$ for each $i$ , then $a\le b$ .
- If $a_i > b$ or $a_i \ge b$ for some real number $b$ for each $i$ , then $a\ge b$ .
inequalities for real numbers is owned by Cam McLeman, matte, Michael Slone, J. Pahikkala, danelray.