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\bf
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{
Exponents
}
\rm
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\centerline{ {\it
Doron 
ZEILBERGER}\footnote{$^1$}
{\eightrm  \raggedright
Department of Mathematics, Rutgers University (New Brunswick),
Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway,
NJ 08854-8019, USA.
%\break
{\eighttt zeilberg  at math dot rutgers dot edu} ,
\hfill \break
{\eighttt http://www.math.rutgers.edu/\Tilde zeilberg} .
}
}

{\bf Algebra}

In algebra we have {\bf symbols} and {\bf numbers} mixed  up.
The notation is confusing. $ab$ means $a$ times $b$,
i.e. $a \times b$, but we don't write the $\times$.
But for numbers, we have to write the $\times$, since
$23$ means ``twenty-three'' and not $2$ times $3$.

In algebra we have {\bf parantheses}, which means
{\it do it first}. They are very important.

{\bf Example:}  $3(a+b)$ , means: first add up $a$ and $b$
then multiply by $3$. They are rules for {\it getting rid} of
parentheses, called {\bf foiling}. In this case we have
$3(a)+3(b)=3a+3b$.

When we pronounce $3(a+b)$ we say ``$3$ times the quantity $a+b$.


{\bf Precedence}: Mathematicians are lazy and often they don't
write parentheses explicity, but use {\bf conventions} of
precedence.

First: All powers

Second: All multiplications and divisions

Third: all additions and subtractions


So $a+bc+d$ is really shorthand for $a+(bc)+d$. This
is not to be confused with $(a+b)(c+d)$ which means something
else completely.


{\bf Problem:} Do $(3+11)(8-4)+ 7 \cdot 2-8/2$.

{\bf Exponents:}

$a^n$ means ``$a$ times $a$ times $a$ $\dots$ times $a$'',
where there are $n$ of the $a$'s.

You can always {\bf spell-out} numerical or symbolic powers.

{\bf Examples:} $2^3=2 \times 2 \times 2=8$.

$a^4=aaaa$.

{\bf Important convention}: 
$$
a^{-n}={{1} \over {a^n}}
$$

{\bf Examples:}

$$
2^{-3}={{1} \over {2^3}}={{1} \over {8}}
$$
$$
3^{-2}={{1} \over {3^2}}={{1} \over {9}}
$$
$$
(1/4)^{-2}={{1} \over {(1/4)^2}}={{1} \over {1/16}}=16
$$


{\bf Reminder}: ``The bottom of the bottom goes to the top''

$$
{ {1} \over {{{1} \over {Whatever} } }} = 
Whatever \quad .
$$


{\bf The Spelling-Out Method}

$$
 { {3^{-2} } \over {2^{-2} } } =
 { {1/9 } \over {1/4 } } = 4/9 \quad .
$$

{\bf Watch Out from a Common Mistake}: The following is wrong!!!!
$$
 { {3^{-2} } \over {2^{-2} } } = {{3} \over {2}} \quad.
$$

You can't ``cancel-out'' the the same thing (in our case $-2$)
when they are exponents. 


To be safe, always do the numerical part of exponent expressions
by the spelling-out method.

Examples:

$5^2=5 \cdot 5=25$ , 

$(-5)^2=(-5) \cdot (-5)=25$ (remember: minus times minus is plus),

$-5^2=-(5^2)=-25$.

{\bf Watch out:} $-5^2$ is {\bf not} $(-5)^2$. By the precedence rules
you {\bf first} do the power, $5^2=25$ and only {\bf then} you do the
$-$, getting $-25$. $-5^2$ is the same as $-(5^2)$. On the
other-hand, if they want you to do the $-$ first and then
do the power, then they need  parantheses $(-5)^2$.

{\bf Do right now:}

$2^3=$

$(-2)^4=$

$-2^4=$

$-(-5)^2=$

$-0.2^3=$

$(-0.2)^3=$

$(-{{3} \over {5}})^2=$


{\bf Properties of Exponents}

{\bf Rule 1:}

$$
(ab)^n=a^n b^n \quad .
$$

{\bf Why is it true?}: For example, when $n=3$, we have
$$
(ab)^3=(ab)(ab)(ab)=ababab=aaabbb=a^3b^3 \quad.
$$


{\bf Rule 2:}
$$
x^n x^m=x^{n+m} \quad .
$$
{\bf Why is it true?}: For example, when $n=3,m=4$, we have
$$
x^3 x^4=(xxx)(xxxx)=xxxxxxx=x^7 \quad.
$$

{\bf Rule 3:}
$$
{{x^n} \over {x^m}}=x^{n-m} \quad .
$$
{\bf Why is it true?}: For example, when $n=5,m=3$, we have
$$
{{x^5} \over {x^3}}={{xxxxx} \over {xxx}}=xx=x^2 \quad .
$$

{\bf Rule 4}

$$
(x^m)^n=x^{mn} \quad .
$$
{\bf Why is it true?}: For example, when $n=3,m=4$, we have
$$
(x^3)^4=(x^3)(x^3)(x^3)(x^3)=(xxx)(xxx)(xxx)(xxx)=xxxxxxxxxxxx=x^{12} \quad .
$$


Examples:

$x^5 \cdot x^4=x^{5+4}=x^9$

$z^2 \cdot z^5=z^{2+5}=z^7$

$(3^2)^5=3^{2 \cdot 5}=3^{10}$

$$
\left ( - {{2} \over {3}} x^2 \right )^3=
(-1)^3 {{2^3} \over {3^3}} (x^2)^3=
- {{8} \over {27}} x^6 \quad .
$$

$$
5a^7(-4a^6)= (5)(-4) a^7 \cdot a^6 =-20 a^{7+6}=-20a^{13} \quad .
$$

{\bf Common Mistake (Watch Out!):}
$$
(a^6)(a^7)=a^{42} (WRONG!!!)
$$

$(a^6)^7=a^{42}$ is right!, but if $6$ and $7$ are at the {\bf same level}
then $(a^6)(a^7)=a^{6+7}=a^{13}$, i.e. you {\bf add} the exponents
{\bf not} multiply them. But in
$$
(a^6)^7 \quad
$$
the $7$ is ``above'' the $6$, so you {\bf multiply} .


{\bf Do right now!}

$-3a^2(4a^4)=$

$-3b^3a^2 (3a^5b) (2ab^2)=$

$(3^3)^4=$

$(2^5)^2 \cdot (2^2)^5 \cdot 2^4=$

{\bf Writing with positive exponents}

Remember 

$$
x^{-n}={{1} \over {x^n}} \quad .
$$

{\bf Examples:}

$$
3^{-2}={{1} \over {3^2}} ={{1} \over {9}} \quad
$$

$$
(-5)^{-2}={{1} \over {(-5)^2}} ={{1} \over {25}} \quad
$$

$$
(-2)^{-5}={{1} \over {(-2)^5}} ={{1} \over {-32}} 
=-{{1} \over {32}} \quad .
$$


$$
({{3} \over {4}})^{-2}= 
{{3^{-2}} \over {4^{-2}}}=
{{4^{2}} \over {3^{2}}}={{16} \over {9}} \quad .
$$


{\bf Do Right Now!:}
$$
\left ( {{2} \over {3}} \right )^{-3}= 
$$
$$
\left ( {{1} \over {3}} \right )^{-2} + \left ({{1} \over {2}} \right ) ^{-3} =
$$
$$
\left ({{1} \over {4}} \right ) ^{-3} +
\left ( {{1} \over {4}} \right )^{-2} =
$$

{\bf Simplifying complicated expressions}

{\bf Example}

$$
(5y^{4})^{-3}(2y^{-2})^3=
(5^{-3} 2^3) y^{-12} \cdot y^{-6}=
{{8} \over {125}} y^{-18}=
{{8} \over {125y^{18}}}  \quad .
$$

{\bf Do right now!}
$$
(3z^{-2})^{3}(4z^{-3})^2=
$$
$$
(5w^{-4})^{2}(4w^{-5})^3=
$$

{\bf Dividing Exponents}

Remember:

$$
{{x^m} \over {x^n}}=x^{m-n}
$$

{\bf Careful}: Don't do things in your head! Do it step-by-step.
It is very easy to make mistakes by doing ``mental math'',
remember ``minus minus is plus''.

{\bf Examples:}

$$
{ { x^{-1} } \over {x^9} }=
x^{-1-9}=x^{-10}={{1} \over {x^{10} }} \quad .
$$


$$
{ { t^{-10} } \over {t^{-4}}} =t^{-10--4}=t^{-6}={{1} \over {t^{6}}} \quad .
$$


$$
\left ( {{8x^2y} \over { 4x^4y^{-3} } } \right )^4=
\left ( {{2x^2y} \over { x^4y^{-3} } } \right )^4=
$$
$$
2^4 (x^{-2} y^{4})^4=
16(x^{-8} y^{16})=  {{16y^{16}} \over {x^8}} \quad .
$$


{\bf Do Right Now!}
$$
{ {z^{-10} } \over {z^{-5}}}=
$$
$$
{{w^{-3} } \over {w^{8}} }=
$$
$$
{{x^{13} } \over {x^{18}} }=
$$

$$
\left ( {{6x^3y^2} \over { 3xy^{-2} } } \right )^2=
$$
$$
\left ( {{16x^{-3}y^{-2}z^3} \over { 4x^{-1} y^{2}z^2 } } \right )^4=
$$
$$
\left ( {{9x^{3}y^{3}z^3} \over {3x^{-1} y^{-2}z^{-2} } } \right )^{-1}=
$$

\end


