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Hows and Whys of Exponents (and other tidbits)

  1. Sep 1, 2006 #1
    I'm new here, and totally green to "real" Maths, to be nice :)

    But really simple question: I understand why x^1 == x, because the Identity Element of the multiplication operator is 1, so x^1 == 1*x == x, so x^0 must be just 1 on its own. (I was never told this in GCSE Maths, so I don't know how I got an A in it)

    But why do negative exponents produce reciprocals? I see it does division instead of multiplication, but why?

    Finally, there's non-integer exponents, I haven't yet read the Wikipedia article on those yet (seeming as it is 2AM right now) but can anyone explain those in an easy-to-digest format?

    ...that's all I can think of for now, I'll reply to this thread with other things I don't understand.

  2. jcsd
  3. Sep 1, 2006 #2


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    Hi, I'm not a mathematician, but I've got a theory about how the history of the exponents unfolded!

    Consider a a real number and n a positive integers. When a is multiplied by itself n times, we write that as [itex]a^n[/itex] instead of [itex]a\cdot a\cdot \ ... \ \cdot a[/itex]. An immediate consequence of this notation is the rule [itex]a^na^m = a^{n+m}[/itex]. It is on the "principle of conservation of that rule" that each of the generalization we are to make to the notion of exponent is based.

    The first step in generalizing the notion of exponent to any number is to consider a number raised to the 0: What is [itex]a^{0}[/itex]? Well, if the rule [itex]a^na^m = a^{n+m}[/itex] is to hold for any positive or null integer n,m we'd have to define [itex]a^{0}[/itex] as the number 1, because we want [itex]a^0a^n=a^{0+n}=a^{n}[/itex], which automatically defines [itex]a^{0}[/itex] as the multiplicative identity elements of the real, i.e. the number 1.

    The next step is to consider negative integers exponents. What is [itex]a^{-n}[/itex]? Well, if the rule [itex]a^na^m = a^{n+m}[/itex] is to hold for any integer n,m we'd have to define [itex]a^{-n}[/itex] as [itex]1/a[/itex] because in particular, we want [itex]a^{-n}a^n=a^{-n+n}=a^0=1[/itex], which automatically defines [itex]a^{-n}[/itex] as the multiplicative inverse of [itex]a^{n}[/itex], which is 1/a.

    The rest of the story will come after dinner.
    Last edited: Sep 1, 2006
  4. Sep 1, 2006 #3
    Definitions, definitions and definitions.
  5. Sep 1, 2006 #4


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    On to rational numbers. What is [itex]a^{p/q}[/itex] ([itex]p,q \in \mathbb{Z}, \ q\neq 0[/itex])? It follows from the notation [itex]a^n[/itex] that [itex](a^n)^m=a^{nm}[/itex] ([itex]n,m\in \mathbb{N}[/itex]). It is easily seens that this holds also for exponents in [itex]\mathbb{Z}[/itex]. We want to extend this propery also to exponents in [itex]\mathbb{Q}[/itex]. So we define [itex]a^{p/q}[/itex] by "its solution" [itex]a^{p/q}=x[/itex] plus the requirement that our [itex](a^n)^m=a^{nm}[/itex] law holds for rational integers as well. That is to say, given that our law holds, we have [itex](a^{p/q})^q=a^p=x^q[/itex]. So we say that [itex]a^{p/q}[/itex] represents the number(s) x such that [itex]x^q=a^p[/itex]. It could be that there are no such x, one such x or 2 such x in [itex]\mathbb{R}[/itex], but we don't care; at least rational exponents, when they exist, obey the same laws as integer ones.

    Finally, the real exponents. What's [itex]a^b[/itex] ([itex]b\in\mathbb{R}[/itex])? The exponential function [itex]\exp (x):\mathbb{R}\rightarrow (0,+\infty)[/itex] is defined by

    [tex]\exp(x)=\lim_{n\rightarrow +\infty}\sum_{k=0}^n\frac{x^k}{k!}[/tex]

    The property exp(a)exp(b)=exp(a+b) is readily proved. We call e the value of [itex]\exp(1)[/itex]. Since it can be shown that [itex]\forall x\in\mathbb{Q}[/itex], [itex]\exp(x)=e^x[/itex], it seems only natural to define [itex]e^x[/itex] [itex]\forall x\in\mathbb{R}[/itex] as [itex]\exp(x)[/itex]. We also define [itex](e^x)^y=e^{xy}[/itex] ([itex]x,y \in \mathbb{R}[/itex]) in order to fit the [itex](a^n)^m=a^{nm}[/itex] property. Now we have a precise definition of [itex]a^b[/itex] ([itex]b\in\mathbb{R}[/itex]) in the very special case [itex]a=\exp(1)[/itex]. But with the help of the inverse of exp(x), we'll see that this special definition automatically defines [itex]a^b \ \ \ \forall a>0[/itex]).

    It can be shown also that exp(x) is bijective, which assures the existence of an inverse, which we call [itex]\ln (x) :(0,+\infty)\rightarrow \mathbb{R}[/itex]. Being exp(x)'s inverse, ln(x) has the property that exp(ln(x))=x. With the chain rule and the fond. them of calculus, we find an analytic form of ln(x) in the person of


    The conclusion tomorrow. :zzz:
    Last edited: Sep 2, 2006
  6. Sep 2, 2006 #5


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    There are reasons that definitions are what they are. That is what w3bbo asked, that is what quasar987 is explaining.
  7. Sep 2, 2006 #6
    quasar987: I'm afraid to say you've gone completely over my head. I was able to comprehend you up until you mentioned the "principle of conservation". Can you explain it in (even) simpler terms? Thanks :)
  8. Sep 2, 2006 #7


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    Dearly Missed

    You said you agreed to that if we only consider NATURAL numbers (i.e, the counting numbers) in the exponents, then we observe the rule:
    [itex]a^{n}a^{m}=a^{n+m}[/itex] where n and m are natural numbers (say, for example, 2 and 3).

    Now, consider the folllowing:
    The rule [itex]a^{n}a^{m}=a^{n+m}[/itex] is very simple to remember; whenever we raise some number "a" to some power "n", and then multiply that answer with "a" raised to some power "m", then the rule states that the answer you get out of this is exactly the same as raising "a" to the power n+m.

    Now, wouldn't it be nice if this rule could be said to work also for OTHER types of numbers than just the positive integers?
    I hope you agree that that would have been great!

    But, there is nothing in maths that exists independently of how we choose to DEFINE it!

    Thus, we really ought to reformulate our question to the following one:
    HOW should we define non-natural powers so that the nice rule governing natural exponents also can be shown to hold whenever we deal with other types of exponents?

    Are you following so far?
  9. Sep 2, 2006 #8


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    The promised conclusion:

    Given our definition [itex](e^x)^y=e^{xy}[/itex], we have, in the case x=ln(a) and y=b: [itex](e^{\ln(a)})^b=a^b=e^{b\ln(a)}[/itex].
  10. Sep 2, 2006 #9
    I am not sure how can you prove exp(a)exp(b) = exp(a+b) by using the definition for e. I mean, you would have to prove the following right?

    [tex]\Big( \sum_{k=0}^{\infty}\frac{a^k}{k!} \Big)\Big( \sum_{k=0}^{\infty}\frac{b^k}{k!} \Big) = \sum_{k=0}^{\infty}\frac{{(a+b)}^k}{k!}[/tex]

    which seems hard to do...
  11. Sep 2, 2006 #10


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    Seems is the right word, because in truth, it's not. :smile:

    Just substitute (a+b)^k by its binomial sum and form the cauchy product of the left hand side. You'll see that they are equal.

    (Edit: convergence of the Cauchy product is assured by the absolute convergence of the exp(x) series.)
    Last edited: Sep 2, 2006
  12. Sep 2, 2006 #11
    It comes from group theory. The definition of whole exponents in a group is

    e=a^0, a=a^1, a^n = a(a^[n-1])

    so first we prove if b+c = n then (a^b)(a^c) = a^n

    This relationship holds for n = 1
    (a^1)(a^0) =(a^1)e=a^1 = e(a^1) =(a^0)(a^1)

    This relationship holds for n = 2
    (a^2)(a^0) =(a^2)e=a^2 = e(a^2) =(a^0)(a^2)
    a^2=a(a^1)=(a^1) (a^1)

    Suppose this relation holds for numbers from 1 to n-1, and let b and c be any numbers that satisfy b+c=n. Claim: if this is the case, the relationship also holds for n.

    a^(n) = a(a^[n-1]) = a[a^(b+c-1)]= a[a^([b-1]+[c])]

    now [b-1]+[c] = n-1<n, so [a^([b-1]+[c])] satisfies our inductive hypotheses

    so a[a^([b-1]+[c])] = a([a^(b-1)])a^c = (a[a^(b-1)])a^c = (a^b)(a^c)

    But since we are talking about a group, any a^n has to have an inverse. I.e. some element p that satisfies p(a^n) = e. Let a! denote a’s inverse. Notice P must = (a!)^n (this can also easily be proved with induction). So we define a’s inverse to be a^(-1) and a^n’s inverse to be a^(-n). Handily this satisfies our “if b+c = n then (a^b)(a^c) = a^n” relationship.
  13. Sep 3, 2006 #12


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    Yes, we define an for n a positive integer from analogy with multiplication :an= a "multiplied by itself n times" and from that get the very nice properties that [itex]a^x a^y= a^{x+y}[/itex] and [itex](a^x)^y= a^{xy}[/itex]. We then define [itex]a^0= 1[/itex] and [itex]a^{-n}= 1/a^n[/itex] as we do in order to keep the first formula true and then define [itex]a^{1/n}= ^n/sqrt{a}[/itex] and [itex]a^{m/n}= ^n\sqrt{a^m}[/itex] as we do in order to keep the second formula true.
    However, we do not define ax, for x irrational, in order to keep either of those true- we define them by "continuity". That is, if x is an irrational number and {ri} is a sequence of rational numbers converging to x (and such a sequence always exists) then ex is defined as the limit of the sequence [itex]a^{r_i}[/itex]. Then the funciton f(x)= ax is continuous for all x.
    Of course, if r is a rational number close to x, then er must be close to ex because of that continuity. That is what allows us to enter 3.141592 into a calculator instead of [itex]\pi[/itex] and know that we will get an accurate answer.
  14. Sep 3, 2006 #13
    i didnt learn calculus in any institution, but i read a book once, and im pretty sure that the exponent integer was first used in the log derivation...

    after doing some playing with the general derivation of log, you meet the expression (1+x)^(1/x) when x--->0.
    it can be canged to the expression: (1+1/x)^x when x-->infinity
    and then you may use newton's binum, and get the number 2.71 which is "e".

    just read the full development of the log. there youll see where the ecxponent came from...
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