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Homeinequalities for real numbers

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# inequalities for real numbers

Suppose $a$ is a real number.

1. If $a<0$ then $a$ is a

*negative number*.2. 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 argument, $0=0\cdot(-a)<(-a)(-a)=a^{2}$. ∎

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 $a\geq b$.

## Mathematics Subject Classification

54C30*no label found*26-00

*no label found*12D99

*no label found*

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## Recent Activity

new correction: Error in proof of Proposition 2 by alex2907

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