Whole exponent vs. fraction exponent

Square1

a^x means a*a*a*a.... x times

This makes sense for whole numbers to me, but I am sort of lost about transfering the definition one x becomes a fraction. I know that the denominator in a fraction in an exponent means "the denominator root"..., and I know what a root its. But it is a jump in concepts. How do you link the two cases in a more smooth/seamless way?

Simon Bridge

They mean: $$x^{1/2} = \sqrt{x}\\ x^{1/3} = \sqrt[3]{x}\\ x^{1/a}=\sqrt[a]{x}$$ The first line asks "what number do I have to multiply to itself, once, to get x?" i.e. "what is the square-root of x?"
The second line asks "what number do I have to multiply by itself twice to get x?" i.e. "what is the cube root of x?"
The last one asks "what number do I have to multiply by itself (a-1) times to get x?" i.e. "what is the a'th root of x?"

When you see ##y=x^{3/2}## you are tempted read this as asking "what number do I have to multiply by itself on-and-a-half times to get x?"
Which is hard to make sense of until you realize that it is the same as asking "what number do I have to multiply to itself to get x³?"

In general: $$x^{a/b}=\sqrt[b]{x^a}$$

The notation makes more sense when you see it graphically ... if you plot, say, y=2^x for x=0,1,2,3... you can see that there is a clear curve that the dots follow. The points on the curve in between the dots correspond to a fractional value of x.

rbj

it's not that much of a jump in concepts.

[tex] a^n = \underbrace{ a \times a \times a \cdots \times a }_{n} [/tex]

makes perfect sense for an integer [itex]n[/itex]. now what does it mean for the reciprocal of an integer?

let

[tex] b = a^{\frac{1}{n}} [/tex]

then we know that when [itex]b[/itex] is raised to the [itex]n[/itex]^th power,

[tex] b^n = \left(a^{\frac{1}{n}}\right)^n = a^{n \cdot \frac{1}{n}} = a^1 = a[/tex]

or

[tex] \underbrace{ b \times b \times b \cdots \times b }_{n} = a [/tex]

if [itex]a[/itex] is known, we can start with a guess for [itex]b[/itex], raise it to the [itex]n[/itex]^th power, run it through the operation above, and if the result is greater than [itex]a[/itex], we must reduce [itex]b[/itex] by a little and if the result is less than [itex]a[/itex], we must increase [itex]b[/itex] a little and repeat doing this until we get sufficiently close to [itex]a[/itex].

so that covers exponents that are an integer and exponents that are the reciprocal of an integer. now any exponent that is a rational number (which is a fraction of integers), then you do a little of both:

let

[tex] b = a^{\frac{m}{n}} = \left( a^m \right)^\frac{1}{n} = \left( a^\frac{1}{n} \right)^m [/tex]

for integers [itex]m, n[/itex].

if you go through the same song and dance, then it's

[tex] b^n = a^m [/tex]

or

[tex] \underbrace{ b \times b \times b \cdots \times b }_{n} = \underbrace{ a \times a \times a \cdots \times a }_{m} [/tex]

then if you know [itex]a[/itex] and the two integers [itex]m[/itex] and [itex]n[/itex], you can compute the right side of the equation straight away and go through the same guess-and-adjust iteration with [itex]b[/itex] on the left side as we have done above.

that is how you define the meaning of fractional exponents without getting into the concept of logarithms. it's much easier to understand this when logarithms are understood.

Square1

Thanks both for the contributions. RBJ, actually I think I will start to play with the method you said more often now to at least estimate an answer, instead of jumping to my calculator right away when fractions show up.

I've been thinking/writing this reply for quite some time now to be honest and explored many different ideas, and I think I the answer I am looking for is somewhat of a trick...

I think the real underlying issue is that exponent notation is really just simply what I said I understood it to be. Fraction exponents using the definition in my first line doesn't make sense, and it is not meant to, even though algebraic tricks like what rbj showed will still work it out. Exponential notation I understand is just a convenient shorthand invention (Descarte generally credited?...) to make writing easier going that "one way" - exponentiating. Writing a whole number above the base can be relatively intuitively explained. But fractions simply don't make a good direct analogue for the reverse. Fractions? Huh?? Multiplying the number to itself 1.5 times?

So I guess then thats when we go to the definition of a root, which is itself understandable, and where we have a translation for bringing in the fraction exponents.

x^(a/b) = 'b'root(x^a)

If it weren't for it's bulkiness, I guess it would be able accommodate exponentiation and rooting really, if we just let the "seat" be blank for 1 instead of 2, and then start writing a 2 on the seat for sqrt, and keep the rest the same. The notation itself directly asks to think of two underlying concepts exponentiating, and rooting, where as exponent notation has this one way-edness.

So I don’t think there is any linking that I need to burden myself with to begin with right?

Simon Bridge

All notation is a convenient shorthand invention. The only "trick" lies in understanding what it is a shorthand for.But they do - which you can see easily and quickly if you draw the graph like I suggested.

The whole power exponents are taught first because they are simple to explain.
But that is just baby steps ... when you are used to them, you can learn the whole story.

Then you will be learning even more complete math, and new notation styles to go along with it. It never stops.

Square1

I did look and I'm not sure I see on the graph what is supposed to solve this problem though... In 2^x for example. I understand what you said about rethinking about 2^1.5 as, what do I have to multiply by itself to get 2^3. But this itself is already going back to make use of

x^(a/b) := b-root (x^a)

It is the first case that you mention that I am confused about (although still, the way you word it I think is implying using the root-method). What does it mean to multiply a number, by itself 1.5 times?

2^3 = 2*2*2 = 2*(2+2) = (2+2) + (2+2) = 8

2^(3/2) = 2^1 + 2^(1/2) perhaps, but the fraction still comes back and needs to be dealt with the root method

or..

2^(3/2) = something like this comes to mind....two multiplied by its self 1.5 times....2*(2*1/2).... :S but wrong obviously

rbj

Sq1, you should use LaTeX paste up. it makes it easier to read what you intend. i shall demonstrate how easy it is:


well it is wrong.

do you know and understand that:

[tex] \left( a^b \right)^c \ = \ a^{b c} [/tex]

first settle that for the case where [itex]b[/itex] and [itex]c[/itex] are integers. once you figure that out then let

[tex] x = 2^{1.5} [/tex]

now whatever [itex]x[/itex] is, this is true:

[tex] x^2 = \left( 2^{1.5} \right)^2 = 2^{2 \times 1.5} = 2^3 = 8 [/tex]

you get that, right?

then you have to figure out an [itex]x[/itex] such that when you square it, it becomes 8. whatever legit method you use for that (including the iterative guess-and-adjust) is fine.

HallsofIvy

Yes, "a^m means a multplied by itself m times" only makes sense when m is a positive integer. To talk about a to some [b]other[b] number we have to define what we mean. We can, in fact, extend that to other number, 0, negative numbers, fractions, irrational numbers any way we like. And what we "like" is to extend it in such a way as to keep some nice properties

It is easy to see that [itex]a^ma^n= a^{m+n}[/itex]: the first gives us a multiplied by itself mtimes and then we have a multiplied by itself n times- that's a total of m+n times.

Similarly, [itex](a^m)^n= a^{mn}[/itex]- imagine the m "a"s in a row, then m different rows: that's a total of mn "a"s.

Now, we want [itex]a^na^0= a^{n+ 0}= a^n[/itex]. For a positive, we can divide both sides by [itex]a^n[/itex] to get [itex]a^0= 1[/itex]. In order to keep the "[itex]a^na^n= a^{m+n}[/itex] rule even when n= 0, we must [b]define[b] [itex]a^0= 1[/itex].

Similarly, for any positive n, we want to have [itex]a^na^{-n}= a^{n- n}= a^0= 1[/itex] so we have to define [itex]a^{-n}= 1/a^n[/itex].

For any nonzero n, we want to have [itex](a^{1/n})^n= a^{n/n}= a[/itex] which says simply that [itex]a^{1/n}= \sqrt[n]{a}[/itex]. For any rational number n/m, we want [itex]a^{n/m}= (a^n)^{1/m}= \sqrt[m]{a^n}[/itex].

We have to go a little further to define [itex]a^x[/itex] for x an irrational number, because irrational numbers (and general real numbers) because the real numbers have to be defined "analytically", not "algebraically". That means, using some kind of limit process. The simplest is this: if x is any real number, rational or irrational, then there exist some sequence pf rational numbers, [itex]\{r_i\}[/itex] that converges to x.

Now, being "continuous" is a very nice property so in order to guarantee that the function [itex]f(x)= e^x[/itex] (x here is a variable) is continuous, we define [itex]e^x[/itex] by "if [itex]\{x_i\}[/itex] is any sequence or rational numbers that converges to x ([itex]\lim_{i\to\infty} x_i= x[/itex]) then [itex]e^x[/itex] is the limit of the sequence [itex]\{e^{x_i}\}[/itex].].

Square1

Rbj yes haha I want to start using latex and actually I dl'ed it over christmas it seems pretty daunting and lots to learn. I am not actually taking math this semester and so I wouldn't really have an economical way of learning it either.

I don't see what is different in your second reply from your first though. And yes I understand it (I would use it in the future too).

I think I should rephrase my OP question though although its nothing drastically new. What does multiplying a number by itself x.y times where y does not equal 0, mean. So far we just have strategies that avoid confronting this semantic anomaly - using ideas of radical, or the method you are showing, where in both cases natural multiplication of a number by itself "whole-number of times" takes place and then guess and adjust, once again for whole number amounts. Neither approach does some sort of explicit x.y amount of multiplication (where y != 0...). They both go around it.

Simon Bridge

Did you see how the integer powers all look like they lie on the same imaginary curve? (try it for x={-2, -1, 0, 1, 2, 3, 4, 5}) Just sketch in the curve. The extension of the concept of an exponent to handle fractions is just this: filling in the spaces between the dots.

for 2^1.5 ... go to the graph, find the point on the curve that corresponds to x=1.5, trace back to the y axis, and there you have it. The rest is about getting that curve, which you just sketched in, drawn more rigorously.

The rigorous approach is being explained to you - the graph is illustrative.

So you are getting the number theory, the illustration, and the sort of questions the notation is supposed to represent. That's basically the best that anyone can do.
The bottom line is that the exponential notation does not represent, exclusively, the way you've been thinking about it: it is not just an instruction to multiply a number by itself so-many times.

The LaTeX here is not so tricky since it is not the full language. The quick way to learn, from these forums, is to use the "quote" button from a post that contains some latex and look to see how it was done.

rbj

and, for reference, the wikipedia math pasteup page is useful.

Square1

Hey sorry I had to put this down for some time, but thanks for all the replies. It seems like Ivey is getting at what I am asking.

I feel strange to say it but I still don't really see what the graph is meant to show apart from that those points in between the integers are defined, at those corresponding points, by that function. Its just this is what I was referring to as being disjoint. The method to get to the non-integer points is not congruent with the strict definition of exponent ie with integer exponents. As Ivey shows though, we define the remaining cases for n in [itex]x^{n}[/itex] to suit existing rules algebraic rules.

Simon Bridge

It's similar to any graph when you change from integers to real numbers ... you need to use your imagination to make the connection - once done, you have a motivation for finding the algebra that describes it.
That said, there's nothing wrong with an algebra-first approach.