05. Inequalities and arithmetic

Results on inequalities

Take $a,b,c\in \mathbb{R}$

1. $0<1$
2. $a<b$ if $-b<-a$
   In particular, $a>0$ if and only if $-a<0$
3. If $a<b$ then $a+c<b+c$
4. If $a<b$ and $0<c$ then $ac<bc$
5. $a^2\ge 0$, with equality if and only if $a=0$
6. $a>0$ if and only if $\frac{1}{a}>0$
7. If $a,b>0$ and $a<b$ then $\frac{1}{b}<\frac{1}{a}$

Note that 1, 2, 3 hold when using $\le$ instead of $<$

Proof

1. By trichotomy then either $0<1$ or $0=1$ or $0>1$
   By Axioms for Arithmetic (To Avoid Total Collapse) then $0\ne 1$ so either

$$0<1\text{ or }1>0$$

Suppose for a contradiction that $0>1$ then

$$(-1)\in \mathbb{P}$$

Hence by $\cdot$ and ordering

$$(-1)\cdot (-1)\in \mathbb{P}$$

However $(-1)\cdot (-1)=1\in P$ (Properties of arithmetic (12))
So $0<1$ but this contradicts trichotomy hence

$$0<1$$

2. Using properties of addition

$$\begin{array}{rl}a<b& ⟺b-a\in \mathbb{P}\\ & ⟺(-a)-(-b)\in \mathbb{P}\\ & ⟺-a>-b\end{array}$$

3. Assume that $a<b$

$$(b+c)-(a+c)=b-a>0\text{so}a+c<b+c$$

4. Assume that $a<b$ and $0<c$

$$bc-ac=b(a-c)>0(\cdot \text{ and ordering})$$

5. Naturally $a^2=0⟺a=0$ by Properties of Arithmetic (10 & 13)
   If $a\ne 0$ then by trichotomy then either $a>0$ or $-a>0$
   Either way $a^2=a\cdot a=(-a)\cdot (-a)>0$($\cdot$ and ordering)
6. Suppose for a contradiction that $a>0$ and $\frac{1}{a}<0$, hence $-\frac{1}{a}>0$
   Then

$$-1=-\left(a\cdot \frac{1}{a}\right)=a\cdot \left(-\frac{1}{a}\right)>0$$

This contradicts $(1)$ hence $a>0$ if $\frac{1}{a}>0$
Similarly if $a<0$ and $\frac{1}{a}>0$ then $-\frac{1}{a}<0$

$$-1=-\left(a\cdot \frac{1}{a}\right)=a\cdot \left(-\frac{1}{a}\right)>0$$

This contradicts $(1)$ hence

$$a>0⟺\frac{1}{a}>0$$

7. Suppose that $a,b,>0$ and $a<b$
   Then $\frac{1}{a},\frac{1}{b}>0$ by $(6)$
   So

$$a\cdot \frac{1}{a}\cdot \frac{1}{b}<b\cdot \frac{1}{a}\cdot \frac{1}{b}(\text{by }(4))$$

Hence

$$\frac{1}{b}<\frac{1}{a}$$

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01 - Bernoulli's Inequality

Bernoulli's Inequality

Let $x$ be a real number with $x>-.1$
Let $n$ be a positive integer then

$$(1+x{)}^n\ge 1+nx$$

Proof

By induction on $n$
Fix $x>-1$

Base Case: $n=1$

$$(1+x{)}^1=1+x=1+1\cdot x=1+nx$$

So statement holds true for $n=1$

Inductive Hypothesis:
Suppose the statement holds true for $n=k$ (for some $k\ge 1$) hence

$$(1+x{)}^k\ge 1+kx$$

Note that $1+x>0$, and $n{x}^2>0$ (since $n>0$ and $x^2\ge 0$ by Results on Inequalities (5))
Inductive Step: $n=k+1$

$$\begin{array}{rlrl}(1+x{)}^{n+1}& =(1+x)(1+x{)}^n& & (\text{ by defintion})\\ & \ge (1+x)(1+nx)& & (\text{by inductive hypothesis})\\ & =1+(n+1)x+n{x}^2& & (\text{properties of arithmetic})\\ & \ge 1+(n+1)x& & (\text{since }n{x}^2>0)\end{array}$$

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