– Book is driven by a ‘desire to present calculus not as a prelude, but a first real encounter with mathematics’.

– Precision and rigour shouldn’t be deterrents – they are the natural medium for formulating mathematical questions

– 29 Chapters, with 2 ‘starred’ as optional

– Presented as an *evolution of ideas*, rather than disjointed topics.

– 625 problems total (from 1st edition), additional 160 with 2nd edition, and even more with the third.

>Read: I will not be finishing this entire book in a semester with concurrent study…

#Chapter 1: Basic Properties of Real Numbers

**not a review, a condensation!**

Addition can be thought of as an operation performed on a pair of numbers.

\begin{equation}\label{eq:P1}

a+(b+c) = (a+b)+c

\end{equation}

we may construct all of addition, for any length sum (finite or infinite) by considering a pair of numbers being added at a time, and “piling them up”. For example, we can define the sum $a+b+c+d$ five different ways;

\begin{align*}

& ((a+b)+c)+d \\

& (a+(b+c))+d \\

& a+((b+c)+d) \\

& a+(b+(c+d)) \\

& (a+b)+(c+d)

\end{align*}

Similarly, there are $14$ different possible ways of summing $5$ numbers…

\begin{equation}\label{eq:P2}

a+0=0+a=a

\end{equation}

\begin{equation}\label{eq:P3}

a+(-a)=(-a)+a=0

\end{equation}

>Or in other words, $0$ is important.

If a number $x$ satisfies $a+x = a$, then $x=0$.

\proof

\begin{align*}

a+x &= a \\

(-a) + (a+x) &= (-a) + (a) \\

((-a)+a)+x &=0 \\

0 + x &= 0 \\

x & = 0

\end{align*}

\qed

Moreover, subtraction is simply addition rephrased: $a-b \equiv a + (-b)$

\begin{equation}

\label{eq:P4}

a+b = b+a

\end{equation}

This shouldn’t be taken for granted for all operations. Indeed, not all are so

well behaved.

\likeremark{Example}

\begin{align*}

b-a \neq a=-b

\end{align*}

in general. In particular, $a-b = b-a$ iff $a=b$. Curiously, we need

multiplication to fully explore this.

\qef

Now onwards to multiplication!

\likeremark{Remark}

\begin{equation}

\label{eq:P5}

a\cdot (b \cdot c) = (a \cdot b) \cdot c

\end{equation}

\begin{equation}

\label{eq:P6}

a \cdot 1 = 1 \cdot a = a

\end{equation}

and of course, $1 \neq 0$.

\qef

>We prove that in abstract algebra actually. It’s pretty easy, you just get a

contradiction somewhere.

A few more essential axioms for multiplication…

$\forall a \neq 0, \exists a^{-1}$ such that

\begin{equation}

\label{eq:P7}

a \cdot a^{-1} = a^{-1} \cdot a = 1

\end{equation}

\likeremark{Remark}

We actually do need $a \neq 0$, since most importantly $\cancel{\exists} 0

^{-1}$ such that $0 \cdot 0^{-1} = 1$. This has very important consequences for

division, and further on, limits.

\qef

>and commutative multiplication, because we’re not crazy.

\begin{equation}

\label{eq:P8}

a \cdot b = b \cdot a

\end{equation}

Just as subtraction is rephrased addition, so to is division rephrased

multiplication: $a/b = a \cdot b^{-1}$. Now as in our remark, since there

doesn’t exist $0^{-1}$, we now see that $a/0$ cannot mean anything, by

definition!

Now from \ref{eq:P7}, we find the **cancellation law**,

Suppose $a \cdot b = a \cdot c$, and $a \neq 0$,

\begin{align*}

a^{-1}(a\cdot b) &= a^{-1}(a \cdot c) \\

(a^{-1}a)b &= (a^{-1}a)c \\

1b &= 1c \\

b &= c

\end{align*}

also, if $a \cdot b = 0$, then it follows from the above that $a = 0$ or $b =

0$.

>In abstract algebra, rings which have this property are known as Euclidean

domains, in which there are no zero divisors.

\proof

Suppose $a \cdot b = 0$, and w/o loss, $a \neq 0$. Then

\begin{align*}

a^{-1}\cdot(a\cdot b) &= 0 \\

(a^{-1}a)b &= 0 \\

1b &= 0 \\

b&= 0

\end{align*}

\qed

This is massively important for solving equations.

\begin{example}

Consider $(x-1)(x-2)=0$, then by the previous property, it must be the case

that either $x-1 =0$ or $x-2 =0$, so we find $x=1$ or $x=2$.

\end{example}

Now we’re getting into the swing of things, we can begin combining addition and

multiplication with one more axiom…

\begin{equation}

\label{eq:P9}

a \cdot (b+c) = a \cdot b + a \cdot c

\end{equation}

Now we may truly prove when it is that $a=b = b-a$!

\proof

Suppose $a-b = b-a$. Then

\begin{align*}

(a-b)+b &= (b-a)+b

a &= b+ b -a \\

a+ a &= (b+b-a)+a \\

&= b+b \\

a \cdot (1+1)&=b \cdot (1+1)\\

a &= b

\end{align*}

\qed

Furthermore, we may now also assert that $a \cdot 0 = 0$.

\proof

\begin{align*}

a \cdot 0 + a \cdot 0 &= a\cdot (0+ 0) \\

&= a \cdot 0 \\

a \cdot + a\cdot 0 – (a\cdot 0 ) &= a \cdot 0 – (a \cdot 0) \\

a \cdot 0&= 0

\end{align*}

\qed

Yay! That’s a useful thing to have isn’t it?

We may also assert some reasoning about negative numbers.

\begin{theorem}

$(-a)\cdot b = -(a \cdot b)$

\end{theorem}

\proof

\begin{align*}

(-a)\cdotb + a \cdot b &= [(-1)+a]\cdot b \\

&= 0 \cdot b \\

-(a \cdot b)+(-a) \cdot b + a \cdot b &= 0 – (a \cdot b) \\

(-a \cdot b) &= -(a \cdot b)

\end{align*}

\qed

>and now, a drumroll…why two negatives make a positive!

\proof

\begin{align*}

(-a)\cdot (-b)+[-(a\cdot b)] &= (-a) \cdot (-b) + (-a) \cdot b\\

&= (-a) \cdot [(-b)+b] \\

&= (-a) \cdot 0 \\

&= 0 \\

\end{align*}

and now add $(a \cdot b)$,

$(a \cdot b) + (-a)\cdot (-b) – (a \cdot b) = (a \cdot b)$

Therefore, $(-a) \cdot (-b) = (a \cdot b)$

\qed

>Hurrah! Now all those annoying questions from junior high school have been satisfied!

…just don’t ask where we got those axioms k?

\ref{eq:P9}, the distributive law, is key to factorization and multiplication of

arabic numerals.

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