| | |
In the context of both braids and permutations, we have seen notations that mirror that of traditional powers of real numbers. In this section, we remind ourselves of what we know about these (multiplicative) powers by exploring their dependencies on the associativity, identities, inverses and commutativity.
Doing so justifies the traditional exponent rules for real numbers - but also explains why braid concatenation and permutation composition can be treated so similarly (since these closed operations also are associative, have identities and inverses, and are -- in certain circumstances -- commutative).
In all three contexts, we use exponents to more efficiently denote a repeated application of some operation. With braids, we were concatenating a braid to itself some number of times. With permutations, we were composing a permutation with itself over and over. With the more traditional (read "familiar") use of exponents that students have likely already seen, we are multiplying some value by itself several times.
There are practical reasons why we developed this notation in the first place. One need only look to some of the values that naturally pop up in the sciences or elsewhere for motivation:
As was the case with "powers" of braids and permutations, working with the values shown above can be tedious if they are left in these long forms -- especially if the calculations are done by hand. However, with exponents we can of course write each of these in a much better (shorter) form...
Note, in the first two examples the long strings of zeros...
Recall, we can add zeros to the end of any number by multiplying by some number of $10$'s. So, for example, we could turn $3$ into $300,000,000$ by multiplying by ten exactly eight times. Similarly, we could multiply $6.022$ by ten exactly twenty three times to arrive at the number used in the second example.
Of course, there is nothing inherently special about the value $10$, as the third example demonstrates. Here we have a number made solely from a product of two's -- forty three of them, to be exact.
Given what we have seen with braids and permutations (and likely what you have seen in previous math courses) it should be no surprise that we adopt the following shorthand: We denote by $x^n$ the product resulting from multiplying $n$ values of $x$ together: $$x^n = \underbrace{x \cdot x \cdot x \cdots x}_{n\textrm{ times}}$$
The following will undoubtedly be familiar to students, but as a matter of verbiage, $x^n$ is read as "$x$ raised to the $n^{th}$ power". We refer to $n$ as the exponent on $x$, while $x$ itself is called the base. While the choice of the word base probably makes sense (i.e., the number "on the bottom"), one might wonder from where the term exponent came. The first recorded use of this word appears in the book Arithemetica Integra written by Michael Stifel in 1544. In Latin, the prefix ex- means "out (of)", while the word pono is the verb "place". So exponent corresponds directly to the little number "placed outside of" of the base. As a natural extension of this, the act of raising a value to some power is called exponentiation.
We call $x^2$ the square of $x$, as it equals the area of a square with side length $x$. Similarly, we call $x^3$ the cube of $x$ as $x^3$ equals the volume of a cube with that same side length.
The act of finding squares and cubes actually significantly predates the calculation of expressions involving more general exponents. As evidence of this, consider the ancient Babylonian tablet below (known as "Plimpton 322", and dating back to 1800 BC). The depressions made into this clay tablet can be deciphered in a systematic way to numbers, where for any given row the numbers in the second and third columns are part of a Pythagorean triplet (i.e., integers $a$, $b$, and $c$, such that $a^2+b^2=c^2$). An example demonstrating this for one row of the tablet is highlighted in red and blue below.
The term power, which formally represents the resulting value of an exponentiation, connects to these older ideas of square and cube. Specifically, the relationship between different measurable magnitudes (e.g., length, area, volume) in segments, squares, cubes, shows up in the writing of Euclid, an ancient greek mathematician (somewhere around 300 BC). In his writings, he uses the equivalent of the phrase "in power" when he writes (in Greek, of course) that "magnitudes are commensurable in power when their squares are commensurable". Henry Billingsley, the first English translator of Euclid in 1570 cemented this verbiage in his translation of Euclid's Second book, where he writes "The power of a line is the square of the same line."
By the way, two values are "commensurable" when their quotient is a rational value (i.e., equivalent to a quotient of two integers). The values of $\sqrt{2}$ and $1$ are provably not commensurable, while their squares $2$ and $1$ certainly are. But we are getting ahead of ourselves. Wait until we have talked about square roots and why some are irrational, and then come back here and read this again.
At the moment, let us assume that $n \ge 2$ is an integer, although we will quickly see that more general values of $n$ can be considered.
Using exponents, we can write all of the numbers in our earlier examples much more succinctly:
Of course, it would be silly to develop this shorthand for writing such products if we needed to go back to the original form every time we needed to use them in a calculation. Just like we saw with braids and permutations, there are some very useful (and very similar) rules that govern how we can combine powers in the context of repeated multiplication.
One can combine powers in several ways, of course. We can add two powers together, subtract them, consider their product, or quotient -- we could even raise one power to another. Considering sums and differences of powers at this point in time would likely lead us into a discussion of polynomials. However, we choose to motivate polynomials in a slightly different way in the future -- so for now, let us just consider only products, quotients, and powers...
Consider the following product:
$$\begin{array}{rcl} x^3 \cdot x^5 &=& (x \cdot x \cdot x) \cdot (x \cdot x \cdot x \cdot x \cdot x) \\ &=& x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \\ &=& x^8 \end{array}$$Was there really a need to expand the powers above to find the overall product? Absolutely not! We know $x^3$ contributes three factors of $x$ to the overall product, and $x^5$ contributes five more, so $x^3 \cdot x^5$ must be the product of $3+5=8$ factors of $x$ (that is to say, $x^8$). This suggests the following rule: $x^m \cdot x^n = x^{m+n}$ when $m$ and $n$ are positive integers. $\require{cancel}$
There is some potential ambiguity in the above argument though, often neglected by most textbooks. When we write something like $x \cdot x \cdot x \cdot x \cdot x$, in what order do those multiplications occur?
Most will recall the so-called "order of operations" from grade school that insists when any parentheses are absent, a string of multiplications (and/or divisions) should be evaluated "left-to-right".
As an example to clarify what is meant by this "left-to-right" multiplication, consider $2 \cdot 3 \cdot 4 \cdot 5$. First, we multiply $2$ and $3$ to get $6$. Then we multiply that by $4$ to get $24$. Finally, we multiply that result by $5$ to get $120$. In other words, "left-to-right" multiplication means we are implicitly grouping things in a "leaning-left" way, as shown below: $$2 \cdot 3 \cdot 4 \cdot 5 = (((2 \cdot 3) \cdot 4) \cdot 5)$$ This means that $x^m$ is really the following: $$x^m = \underbrace{x \cdot x \cdot x \cdots x}_{m \textrm{ times}} = \underbrace{((((x \cdot x) \cdot x) \cdots) \cdot x)}_{m \textrm{ occurrences of } x}$$ So to accept what was done to $x^3 \cdot x^5$ above as a general argument for dealing with $x^m \cdot x^n$, we would need to accept the following: $$\begin{array}{rcl} x^m \cdot x^n &=& \underbrace{(x \cdot x \cdot x \cdots x)}_{m \textrm{ times}} \cdot \underbrace{(x \cdot x \cdot x \cdots x)}_{n \textrm{ times}}\\ &=& \underbrace{((((x \cdot x) \cdot x) \cdots) \cdot x)}_{m \textrm{ occurrences of } x} \cdot \underbrace{((((x \cdot x) \cdot x) \cdots ) \cdot x)}_{n \textrm{ occurrences of } x}\\ &=& \underbrace{(((((((x \cdot x) \cdot x) \cdots) \cdot x) \cdot x) \cdot x) \cdots ) \cdot x}_{m+n \textrm{ occurrences of } x}\\ &=& \underbrace{x \cdot x \cdot x \cdots x}_{m+n \textrm{ times}}\\ &=& x^{m+n} \end{array}$$ But how do we get from the second step above to the third?
The key is associativity!
Remember, if an operation (like multiplication) is associative, it means $(xy)z = x(yz)$. This in turn allows us to regroup larger products (i.e., products of more than 3 things) and evaluate them in many different ways, without changing the final result.
As an example, notice how we can use the associative property as shown below to re-express $(((ab)c)d)e$ as $(ab)((cd)e)$: $$ \begin{array}{rcl} (((ab)c)d)e &=& ((ab)c)(de)\\ &=& (ab)(c(de))\\ &=& (ab)((cd)e)\\ \end{array}$$
In each of the steps above, we apply the associative property $(xy)z = x(yz)$. The reader will benefit from asking what plays the role of $x$, $y$, and $z$ in each step.
In this light, consider the following argument (that uses only the associative property in each step) for re-expressing $x^4 \cdot x^4$ as $x^5 \cdot x^3$: $$\begin{array}{rcl} x^4 \cdot x^4 &=& (((x \cdot x) \cdot x) \cdot x) \cdot (((x \cdot x) \cdot x ) \cdot x)\\ &=& (((x \cdot x) \cdot x) \cdot x) \cdot ((x \cdot (x \cdot x)) \cdot x)\\ &=& (((x \cdot x) \cdot x) \cdot x) \cdot (x \cdot ((x \cdot x) \cdot x))\\ &=& ((((x \cdot x) \cdot x) \cdot x) \cdot x) \cdot ((x \cdot x) \cdot x)\\ &=& x^5 \cdot x^3 \end{array}$$
This is an incremental step in the right direction, as we have essentially moved one $x$ from the power on the right to the power on the left.
However, we can apply the exact same strategy again to move another $x$ from the right to the left. (Can you figure out how?) Consequently, we can rewrite $x^5 \cdot x^3$ as $x^6 \cdot x^2$, and then (upon doing it again) as $x^7 \cdot x$, until finally we exhaust all of the $x$ factors on the right, leaving only $$(((((((x \cdot x) \cdot x) \cdot x) \cdot x) \cdot x) \cdot x) \cdot x) = x^8$$
This same strategy can be applied more generally, to turn $x^m \cdot x^n$ into $x^{m+1} \cdot x^{n-1}$, and then that into $x^{m+2} \cdot x^{n-2}$, and so on -- until we finally get to $x^{m+n}$ form.
In this way, any time we have an associative "multiplication", whether that means normal real number multiplication, or braid concatenation, or permutation composition -- the same argument can be applied. As such, in all these situations one adds exponents when multiplying powers (presuming the exponents are integers): $$\boxed{x^m \cdot x^n = x^{m+n}}$$
Taking this a bit further, consider a power that is in turn raised to some other power. As long as the "multiplication" in question is associative, the above rule holds for any integers $m$ and $n$. This then allows us to also argue:
$$\begin{array}{rcl} (x^m)^n &=& \underbrace{x^m \cdot x^m \cdot x^m \cdots \cdot x^m}_{n \textrm{ times}}\\\\ &=& \underbrace{(((x^m \cdot x^m) \cdot x^m) \cdots) \cdot x^m}_{n \textrm{ occurrences of }x^m}\\\\ &=& \underbrace{((x^{m+m} \cdot x^m) \cdots) \cdot x^m}_{\textrm{product of } n-1 \textrm{ powers}}\\\\ &=& \underbrace{(x^{m+m+m+\cdots}) \cdot x^m}_{\textrm{product of } n-2 \textrm{ powers}}\\\\ &=& x^{\overbrace{m + m + \cdots + m}^{\textrm{sum of } n \textrm{ terms}}}\\\\ &=& x^{mn} \end{array}$$ which suggests that raising a power to a power can be dealt with by multiplying the exponents: $$\boxed{\displaystyle{(x^m)^n = x^{m\,n}}}$$The above rules for products of powers, and powers of powers, both hold when the exponents involved are positive integers -- but what about exponents that are zero or negative integers?
First, recall that for any real number $x$, we have $$x \cdot 1 = 1 \cdot x = x$$ This means that the value $1$ plays the role of the multiplicative identity. We have denoted identities for other operations (e.g., braid concatenation, permutation composition) by $I$ in the past -- partly because $I$ is the first letter of "identity" and partly because it looks a bit like a $1$ too.
Additionally, every non-zero real number $x$ has a multiplicative inverse where $$x \cdot x^{-1} = x^{-1} \cdot x = 1$$ As examples, $5^{-1} = 0.2$ and $0.1^{-1} = 10$.
A similar result holds for braids, where $1$ is replaced by the identity braid. The same can be said of permutations when $1$ is replaced by the permutation that leaves the order untouched.
Consider what this means for $x^3$, as we appeal to associativity and the properties of the identity and inverses: $$\begin{array}{rcl} x^3 \cdot (x^{-1})^3 &=& (x \cdot x \cdot x) \cdot (x^{-1} \cdot x^{-1} \cdot x^{-1})\\ &=& (x \cdot x) \cdot (x \cdot x^{-1}) \cdot (x^{-1} \cdot x^{-1}) \quad {\scriptstyle \textrm{after using associativity}}\\ &=& (x \cdot x) \cdot 1 \cdot (x^{-1} \cdot x^{-1}) \quad {\scriptstyle \textrm{taking advantage of inverses}}\\ &=& (x \cdot x) \cdot (x^{-1} \cdot x^{-1}) \quad {\scriptstyle \textrm{using the property of the identity}}\\ &=& x \cdot (x \cdot x^{-1}) \cdot x^{-1} \quad {\scriptstyle \textrm{then we repeat the process, using associativity}}\\ &=& x \cdot 1 \cdot x^{-1} \quad {\scriptstyle \textrm{again, as products of inverses yields the identity}}\\ &=& x \cdot x^{-1} \quad {\scriptstyle \textrm{again remembering multiplying by the inverse leaves things unchanged}}\\ &=& 1 \quad {\scriptstyle \textrm{we finish by taking advantage of inverses one last time}} \end{array}$$
Similar calculations reveal $(x^{-1})^3 \cdot x^3 = 1$
More generally, for any integer $n \ge 2$ we have $$(x^{-1})^n \cdot x^n = x^n \cdot (x^{-1})^n = 1$$
As we have seen, something similar holds for braids, for permutations, and -- of course -- for real numbers. Anytime we have an operation that is closed on some set, with associativity, an identity, and guaranteed inverses, it must be the case that $(x^{-1})^n$ is the inverse of $x^n$ for any integer $n \gt 1$.
Given this, we abbreviate $(x^{-1})^n$ by $x^{-n}$, as this then allows us to extend our previous rules of $(x^m)^n = x^{mn}$ and $x^m \cdot x^n = x^{m+n}$ to work with many more pairs of integer exponents, $m$ and $n$. As a quick example, we can easily show $x^5 \cdot x^{-3} = x^2$: $$x^5 \cdot x^{-3} = x^{2+3} \cdot x^{-3} = (x^2 \cdot x^3) \cdot x^{-3} = x^2 \cdot (x^3 \cdot x^{-3}) = x^2 \cdot (x^3 \cdot (x^{-1})^3) = x^2 \cdot 1 = x^2$$
However, we need to make two more definitions so that these two rules work with ALL integer exponents $m$ and $n$:
Consider $x^2 \cdot x^{-2}$.
The properties of associativity, the identity, and inverses tell us the above must equal the multiplicative identity, $1$.
However, the rule $x^m x^n = x^{m+n}$ tells us this must be $x^0$.
We said before to assume $x^n$ meant the product resulting from multiplying $n$ values of $x$ together -- but what we mean by "multiplying $0$ values of $x$ together" is not exactly clear. Thus, our current definition of $x^n$ can't really be used to tell us what $x^0$ is. However, to make everything consistent, we define:
In a like manner, when considering what $x^1$ should mean, we note that "multiplying $1$ value of $x$ together" is also somewhat suspicous (although possibly less so).
Perhaps we could tweak the language we use to avoid this little hickup. That said, note if $x^m x^n = x^{m+n}$ is to work for all integers, then $x^1 = x^{2+(-1)} = x^2 \cdot x^{-1} = (x \cdot x) \cdot x^{-1} = x \cdot (x \cdot x^{-1}) = x \cdot 1 = x$, encouraging us to agree that $x^1 = x$ must be true for all $x$.
†Note: we leave $0^0$ undefined here, as the topics we address are meant not only to expose students to the greater universe of mathematics, but also to prepare one for calculus -- where $0^0$ is considered an undefinable "indeterminant form". However, in other contexts this is not always the best course of action. As Ronald L. Graham, Donald Knuth, and Oren Patashnik argue in their classic text Concrete Mathematics:
"Some textbooks leave the quantity $0^0$ undefined, because the functions $x^0$ and $0^x$ have different limiting values when $x$ decreases to $0$. But this is a mistake. We must define $x^0=1$, for all $x$, if the binomial theorem is to be valid when $x=0,y=0$, and/or $x=-y$. The theorem is too important to be arbitrarily restricted! By constrast, the function $0^x$ is quite unimportant." -- Graham, Knuth, & Patashnik
We define the division of one real number by another as the product of the first and the multiplicative inverse of the second, writing the result (called the quotient) either using the $\div$ operator, or as a fraction, as seen below:
$$x \div y = \frac{x}{y} = x \cdot y^{-1}$$It should hopefully be obvious that if two real values agree, then their products with some other number should also agree. That is to say, for any real values $a,b$, and $c$, if $a = b$, then $ac = bc$.
But then, for any non-zero real value $x$ we have: $$\begin{array}{rcl} x^{-n} \cdot x^n &=& 1 \quad \quad \quad {\scriptstyle \textrm{which, if we consider $c=x^{-n}$ implies}}\\ (x^{-n} \cdot x^n) \cdot x^{-n} &=& 1 \cdot x^{-n}\\ 1 \cdot x^{-n} &=& \cfrac{1}{x^{n}}\\ x^{-n} &=& \cfrac{1}{x^{n}}\\ \end{array}$$
As the quotient of $1$ and any non-zero real value is called the reciprical of that value, we see that negative exponents are connected to recipricals of powers: $$\boxed{x^{-n} = \cfrac{1}{x^n}}$$
Even more immediate, note that if all the rules above hold, we can also show the following for integers $m$ and $n$ (and non-zero $x$): $$\cfrac{x^m}{x^n} = x^m \cdot x^{-n} = x^{m+(-n)} = x^{m-n}$$ which confirms (for integer exponents anyways) that we subtract exponents when dividing powers: $$\boxed{\cfrac{x^m}{x^n} = x^{m-n}}$$
Having considered products and quotients of powers, we now turn our attention to the reverse -- powers of products and quotients. Recall how real number multiplication "distributes over addition and subtraction": $$(a + b)c = ac + bc \quad \quad \textrm{and} \quad \quad (a - b)c = ac - bc$$ In a similar manner, we can show that because multiplication of real values is commutative (i.e., $ab = ba$ for all real values $a,b$), (integer) exponents will "distribute" over products and quotients. That is to say: $$(a \cdot b)^c = a^c \cdot b^c \quad \quad \textrm{and} \quad \quad (a \div b)^c = a^c \div b^c$$
To see how we know the first statement holds (i.e., $(a \cdot b)^c = a^c \cdot b^c$), consider a representative case where the exponent involved is $3$. That is, we will show $(xy)^3 = x^3 y^3$. Notice how the property of commutivity (along with associativity) allows us to do this by gradually move all the $y$ values to the right, where they can be collected into single power -- and how this leaves all the $x$ values on the left, where they too can be collected together into a single power:
$$\begin{array}{rcl} (xy)^3 &=& (xy)(xy)(xy)\\ &=& xyx(yx)y\\ &=& xyx(xy)y\\ &=& x(yx)xyy\\ &=& x(xy)xyy\\ &=& xx(yx)yy\\ &=& xx(xy)yy\\ &=& xxxyyy\\ &=& x^3 y^3 \end{array}$$ Interestingly, one can deduce the sequence of applications of the commutative property (like the one shown above) to accomplish any similar re-ordering by considering the permutation that would take the starting order of factors to the desired order of factors, and then considering a braid that accomplishes the same permutation of its threads to express that permutation as a series of transpositions.
To see this, consider the below diagrams -- especially the one on the right which keeps crossings from happening simultaneously -- which suggest we can turn $(xy)^3 = xyxyxy$ into $xxxyyy = x^3y^3$ by first transposing the 4th and 5th factors (orange and magenta), and then the 2nd and 3rd (green and blue), and then the 3rd and 4th (green and magenta). Notice how this matches the applications of commutativity given in the chain of equality seen above.
Demonstrating the second result (i.e., $(a \div b)^c = a^c \div b^c$) through a representative example can be done similarly -- although we'll express the quotients as fractions instead of using the "$\div$" symbol: $$\begin{array}{rcl} \left(\cfrac{x}{y}\right)^3 &=& (xy^{-1})^3\\ &=& x^3 (y^{-1})^3 \quad {\scriptstyle \textrm{(now we use $(ab)^n = a^n b^n$, which of course requires commutativity)}}\\ &=& x^3 y^{-3}\\ &=& x^3 (y^3)^{-1}\\ &=& \cfrac{x^3}{y^3} \end{array}$$
The upshot of this is that certainly for all integers $n$ and real values $x$ and $y$ (where $y \neq 0$) -- but also when $n$ is an integer and $x$ and $y$ are some elements from some set of things which under some operation (denoted in a "multiplicative way" here) is closed, associative, with an identity, inverses, and (fully) commutative -- the following are both true:
$$\boxed{\displaystyle{(xy)^n = x^n y^n \quad \quad \text{and} \quad \quad \left( \frac{x}{y} \right)^n = \frac{x^n}{y^n}}}$$That said, in other contexts (especially when commutativity is not present) we may not be able to rely on this being true -- recall, such was the case with both braids and permutations!
While there will be exceptions, we often try to simplify expressions involving powers of real numbers to their shortest written forms.
Notice in particular the rules $x^m x^n = x^{m+n}$, $x^m / x^n = x^{m-n}$, and $(x^m)^n = x^{mn}$ offer immediate reductions in how much pencil lead is required to write an expression. Consequently, one should typically seek to manipulate expressions as needed to create opportunities to apply these specific rules when simplifying.
Also -- as a useful matter of convention when dealing with powers of real numbers -- when simplifying expressions using the above rules, one should avoid the presence of negative exponents in the final form. One advantage of such a convention is that it helps reduce the number of "equivalent forms" a completely simplified expression may take.
Below an example of how an expression might be simplified is provided. Note, this is not the only way to simplify the expression given -- any legal application of the rules developed in all the discussion above that seeks to write the expression as briefly as possible should get you to the same final quotient. So feel free to follow a different path! 🙂
$$\begin{array}{rcl} \cfrac{(x^2 y^{-5})^{-3} z^{-2} w^{-7}}{x^6 z^{-2} x^2 y^4} &=& \cfrac{(x^2 y^{-5})^{-3} z^{-2} w^{-7}}{x^6 x^2 z^{-2} y^4} \quad {\scriptstyle \textrm{we used commutativity to get the bottom two powers of $x$ near one another}}\\ &=& \cfrac{(x^2 y^{-5})^{-3} z^{-2} w^{-7}}{x^8 z^{-2} y^4}\\ &=& \cfrac{x^{-6} y^{15} z^{-2} w^{-7}}{x^8 z^{-2} y^4} \quad {\scriptstyle \textrm{we used commutativity to distribute the $-3$ exponent}}\\\\ &=& x^{-6} y^{15} z^{-2} w^{-7} (x^8 z^{-2} y^4)^{-1}\\\\ &=& x^{-6} y^{15} z^{-2} w^{-7} x^{-8} z^2 y^{-4} \quad {\scriptstyle \textrm{we used commutativity to restore the original order of powers after taking the inverse}}\\\\ &=& x^{-6} x^{-8} y^{15} y^{-4} z^{-2} z^2 w^{-7} \quad {\scriptstyle \textrm{we used commutativity several times to get powers with the same base adjacent}}\\\\ &=& x^{-14} y^{11} z^{0} w^{-7}\\ &=& \cfrac{y^{11}}{x^{14}w^7} \end{array}$$
The reader will benefit by identifying each rule being applied to generate next steps in the calculation above. As a special note, be very careful about noticing when commutativity is being applied (see the several provided comments in the calculations above), recalling that can't always make similar calculations in other contexts (e.g., braids and permutations).