Notes: Spivak’s Calculus Prologue C1 Inequalities

– $b 0$

**Proof**

If $a < 0$, $0 > a$, so $0 – a = -a \in P$. By similar reasoning, $-b \ in P$. So by P12,

$$(-a)(-b)=ab \in P $$

and we conclude, $ab > 0$. There is a special consequence if $a^2 > 0 $, for $a \neq 0$

In fact, since $a^2 > 0$, $\forall a \neq 0$, we have the useful result: $1^2 = 1$, so $1>0$. **(!)**

##Absolute values

Given $-a<0$ if $a > 0$, we define the **absolute value**,

\[
|a| =
\begin{cases}
a, &a \geq 0 \\
-a, &a \leq 0
\end{cases}

\]

Noting that $|a|$ is always positive, except when $a = 0$, which is neither positive nor negative. For example, $|-3| = 3$.

##The Triangle Inequality

**Theorem**
$$|a+b| \leq |a| + |b|$$

**Proof (by cases)**

1. $a \geq 0$, $b \geq 0$

If $a + b \geq 0 $, then it follows $$|a+b| = a+b = |a | + |b| $$

2. $a \leq 0, b \leq 0$

then $a+b \leq 0$, and again,

$$|a+b| = -(a+b) = (-a) + (-b) = |a | + |b|$$

3. $a \geq 0, b \leq 0$

We must prove that $|a+b| \leq a-b$. So suppose $a + b \geq 0$, then we need to prove $$a + b \leq a – b$$ Or in other words, $b \leq -b$. But this is clearly true, since $b \leq 0$.

Otherwise, if $a+b \leq 0$, we need to prove that $$-a – b \leq a – b$$or more simply, $-a \leq a$. But this is also true, since we assumed $a \geq 0$, so $-a \leq 0$.

The final case is achieved by interchanging $a$ and $b$ in the final proof.

**Proof (a bit shorter, and more slick)**

If we denote $\sqrt{x}$ as the strictly positive square root of $x$, then $$|a| = \sqrt{x^2}$$ is an equivalent definition of the absolute value, then

\begin{align*}
(|a+b|)^2 &= (a+b) \\
&= a^2 +2ab + b^2 \\
&\leq a^2 +2|a|\cdot|b| + b^2 \\
&= |a|^2 + 2|a\cdot|b| + b^2 \\
&= (|a|+|b|)^2
\end{align*}
Taking the square root of both sides, we conclude $|a+b| \leq |a| + |b| $ since we know that $x^2 < y^2 \implies x < y$, however the proof of this is left as exercise. *As a sidenote, this proof reminds me of the total unnecessary use of proof by cases. Fuck proof by cases, someone please prove that if there exists a proof by cases, greater abstraction allows for a proof-not-by-cases.* Closer examination finds $|a+b| = |a| + |b|$, when the parity of $a$ and $b$ are similar (or if one of $a$ and $b$ is zero). If $a$ and $b$ have opposite parity, then it will be the case that $$|a+b| < |a| + |b|$$ ##Further notes on $1+1$ We proved that $a-b = b-a \implies a=b$, by establishing $a\cdot(1+1) = b \cdot (1_1$, and dividing through by $(1+1)$. Why, though, do we assume $1+1 \neq 0$? This cannot be established with all axioms up the P12, which we now proceed to resolve. **Proposition: $1+1 \neq 0$** **Proof** We have seen that $1 > 0$, so it follows immediately from order axioms that $1+1 > 0$, and thus by definition, in particular, $1+1 \cancel{>} 0 $.

#Final notes on basic properties of (real) numbers

– Seems absurd and trite, but let’s remember that it’s difficult to justify, rigorously, (since we appeal to familiarity), the statements of P1 – P12.

– We work with numbers all the time, however just *what* numbers are, remains philosophically and precisely vague.

– We do know, however, that regardless of what numbers are (at least the ones we are used to dealing with, *cough* $\mathbb{R}$, *cough*), that they obey axioms 1 through to 12.

##A further note during studying

After finishing studiously writing these notes on a notepad, I flipped through the next $25$ questions in Spivak’s Preface, and promptly had scholastic, intellectual heart attack.

A cautious look at the next chapter resulted in an exponential increase in the intensity of said heart attack. Yeah, this should be fun.

I should stick to the ANU Analysis I course notes…they’re easier.