– 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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