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Let us assume that the students haven't been exposed to these two rules: $a^{x+y} = a^{x}a^{y}$ and $\frac{a^x}{a^y} = a^{x-y}$. They have just been introduced to the generalization: $a^{-x} = \frac{1}{a^x}$ from the pattern method: $2^2 = 4, 2^1 = 2, 2^0 = 1, 2^{-1} = \frac{1}{2}$ etc. However, some students confuse $2^{-3}$ to be $(-2)(-2)(-2)$ since they are familiar with $2^{3} = 2 \cdot 2 \cdot 2$. This is a low-income urban school and most kids in this algebra class struggle with math dealing with exponents, fractions and decimals. What would be the best approach to reach all 32 students?

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$\dfrac{a^x}{a^x}$ is? –  Amzoti Jan 7 at 2:33
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$a^{x+y} = a^{x}a^{y}$ for positive $x$ and $y$ should come first. Then start making $y$ zero and then negative. $a^{-x}$ then follows naturally –  Henry Jan 7 at 2:46
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Your requirements are unreasonable. Teach your students the "two rules" you claim they do not know. You have plenty of material from this answer to do so. –  trb456 Jan 7 at 3:29
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I wonder what the socio-economic condition of the school has to do with how to teach these concepts. Pretty much every 9th grader I've known from a variety of backgrounds has found these concepts confusing at first. –  Chan-Ho Suh Jan 7 at 5:38
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First, disabuse yourself of the assumption that there exists a single best approach. In mathematics, the more ways you can explain something, the more likely something will stick. I would start with powers of 10, as suggested by Gina. Review the meaning of positive integer exponents as repeated multiplications of the base. Working backwards, negative exponents are repeated divisions by the base. So $10^3 = 1000$, $10^2 = 100$, $10^1 = 10$, $10^0 = 10^1/10 = 1$, $10^{-1} = 10^0/10 = 1/10$. Show them the pattern in decimal, too. –  heropup Jan 7 at 7:19

24 Answers 24

You could use the rules that they already know.

Since $-n=0-n$, $$ x^{-n}=x^{0-n}=\frac{x^0}{x^n}=\frac{1}{x^n}. $$

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"the students haven't been exposed to these two rules" –  Henry Jan 7 at 2:53
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@Henry As trb456 commented below, "Then teach them these rules. You're kinda screwed without them. Just because you specify a requirement does not mean that requirement is reasonable. In this case, it's not." alo, To try to teach this "rule" before the other two seems to be teaching the "rules" in an strange order. It will make the topic more understandable to the students to teach these "rules" first. –  Daryl Jan 7 at 8:35

Tell them that they need to follow the $\frac{a^x}{a^y} = a^{x-y}$ rule, where $x=0$. This forces $\frac{1}{a^y} = a^{-y}$. This is a great opportunity to point out that their intuition is wrong, and that carefully following the existing rules is the way to enlightenment.

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"the students haven't been exposed to these two rules" –  Henry Jan 7 at 2:46
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Then teach them these rules. You're kinda screwed without them. Just because you specify a requirement does not mean that requirement is reasonable. In this case, it's not. –  trb456 Jan 7 at 2:59

You could say that the minus sign means an opposite; for instance, subtraction is the opposite of addition, and it uses the - symbol. Similarly division is the opposite of multiplication, and the division symbol has a minus sign in it (this isn't where it comes from, but is only good for memorizing purposes).

So a minus in the exponent is the opposite of multiplying over and over again, namely dividing over and over again.

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I like this explanation. If $2^{2} = (1*2)*2$ then it could follow that $2^{-2} = (1/2)/2$. And if students are still confused, maybe explicitly compare side-by-side $2^{-2}$ and $-2^{2}$ –  Dan Jan 7 at 7:54
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I don't see how it could get any more intuitive than this. –  G. Bach Jan 7 at 16:08
    
what do you mean "this isn't where it comes from"? This is exactly where it comes from: division is subtraction, just on the opposite side of the exp map. –  Vectornaut Jan 8 at 20:10
    
I meant that the 'obelus' ÷ does not come from the subtraction sign, although looking it up on wikipedia just now, they actually do seem related. –  Brian Rushton Jan 8 at 20:15
    
Ah—when you said "division symbol," I thought you meant $x^{-1}$, not the obelus. It's cool that the bar in the obelus is maybe related to the subtraction bar! –  Vectornaut Jan 8 at 22:25

How about motivate them by using the familiar example of scientific number? For example, $1.23\times 10^{2}=123$ where the dot is moved to right by 2 place, and so it make sense that the other direction apply too, that is $1.23\times 10^{-2}=0.0123$. And the only way that is true is if $10^{-2}=\frac{1}{100}$. And once you got that worked out for 10 power it's easy to just generalize it.

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Is scientific notation familiar to these students? –  KRyan Jan 7 at 15:12
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You don't necessarily need the 1.23. You can use 10^{2} and work from there. –  SPRBRN Jan 7 at 16:21
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If time permits, the movie "Powers of Ten" is useful. –  Tobias Hagge Jan 7 at 22:19
    
@rxt: the "$\times 10^{2}$" part might have been treated as an unit of measurement by people, so it would looks weird if it come without a number. Just like you don't say the length of something is $m$, you say it's $1m$. Beside, that number is part of the standard notation. –  Gina Jan 8 at 5:13
    
@KRyan: I'm pretty sure it should be familiar. The speed of light being $2.998\times 10^{8}ms^{-1}$ is just so ubiqous, and also so easy to remember in that form, that I am very sure that an average person remember more digit of it than $\pi$. And since standard scientific number is something so basic to all science, which 9th grader have been exposed to for years, the chance of someone not being familiar with it is around as low as someone made it through elementary school without being able to perform simple arithmetic. –  Gina Jan 8 at 5:16

Of course it's confusing to them.

"They have just been introduced to the generalization: $a^{-x} = \frac{1}{a^x}$ from the pattern method: $2^2 = 4, 2^1 = 2, 2^0 = 1, 2^{-1} = \frac{1}{2}$ etc"

In other words, you've made them memorize this thing that doesn't make any sense to them. If I didn't already know about exponents, I'd be confused too. A pattern is only useful if they can anticipate it. How is a normal kid supposed to go from positive exponents to a zero exponent, let alone a negative exponent, when you haven't even explained the basic rule of exponents?

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this doesn't really answer the question –  Malachi Jan 7 at 20:20

You could make a table like this:

$$ \begin{array}{c | c} n & 2^n \\ \hline \cdot & \cdot \\ \cdot & \cdot \\ \cdot & \cdot \\ 1 & 2 \\ 2 & 4 \\ 3 & 8 \\ 4 & 16 \end{array} $$

on a blackboard and ask how the "natural" continuation upwards of each column looks like. So if you read from the bottom $4,3,2,1,\ldots$ what numbers follow? And in $16,8,4,2,\ldots$ what numbers come next in that sequence (the "rule" is to half each term, clearly).

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The obvious answer would be 0, -2, -4, -8, -16... –  Alex J Jan 7 at 14:55
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@IliaJerebtsov Why? Do you think the sequence $16,8,4,2,\ldots$ would continue like that? Maybe you could say, each time we eat precisely half our cakes. The number of cakes left could become fractional, but not negative. –  Jeppe Stig Nielsen Jan 7 at 15:01
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@JeppeStigNielsen, Good idea to use a comparison to physical objects. I always had trouble in physics E&M because I had difficulty visualizing what I was learning, while I got straight A's in physics mechanics, because I could easily visualize what would happen when, for example, the ball bounced against the wall. –  Brian S Jan 7 at 15:05
    
Similarly, you could also graph the progression starting at powers of 0 and going up. Then show how powers < 0 just keep the graph going as expected. E.g. fooplot.com/… –  3noch Jan 7 at 19:14
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Graphs are problematic, because you can't explain how the "intermediate" values are calculated –  miniBill Jan 7 at 20:45

However, some students confuse $2^{−3}$ to be $(−2)(−2)(−2)$ since they are familiar with $2^3=2⋅2⋅2$.

They are familiar with the concept that an integer exponent ($a^b$) can be represented by multiplying $a$ a $b$ number of times. You can try to explain in the following way:

$$ 2^{-3} = \frac{1}{2^3} = \frac{1}{2\times2\times2} = \frac{1}{8} $$

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some students confuse $2^{-3}$ to be $(-2)(-2)(-2)$

Simple, $(-2)(-2)(-2)$ means ${-2}^{3}$, that is the first thing you have to show when they make that mistake.

Edit: That is the number minus 2, not a 2 combined with the unarry (-) and a exponent. Or it probably is, you might want to write ${(-2)}^{3}$ if it confuses you, i'm not even sure which bind stronger.

Next, show the relation between $2^3$ and $2^2$. One equals $2\cdot2\cdot2$ and the other $2\cdot2$. So $2^3$ = $2^2\cdot2$ and $\frac{2^3}{2} = 2^2$. Also, ${2\cdot2\cdot2\over2}=2\cdot2$

We can then move forwards (well, backwards),

$\frac{2^2}{2} = 2^1$,

$\frac{2^1}{2} = 2^0$,

$\frac{2^0}{2} = 2^{-1}$

etc.

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$-2^3$ means $-(2^3)$, not $(-2)^3$. –  MJD Jan 7 at 14:44
    
@MJD, but Dorus said 'equals', not 'means'. and both forms you cite are equal ;) –  Nevik Rehnel Jan 7 at 15:20
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And don't even get me started on those who think that $-2^2 = 4$ because when they enter - 2 x ENTER on their calculator it answers back 4 (or they cite Excel for giving that same answer). –  kahen Jan 7 at 15:37
    
@NevikRehnel I thought very carefully before posting that comment, and I decided that although Dorus said "equals", she meant "means". –  MJD Jan 7 at 18:18
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This comment thread made me consider that there may be an "order of operations" explanation that could help ... I think that the base of the exponent (being 2, not -2) would need to be well understood before this answer could help (and serve as a self-checking mechanism). –  opello Jan 7 at 19:14

(First of all, I’d echo @heropup’s comment: there not one best way to explain things. Different explanations will work better for different students. That said, this is my personal favourite explanation.)

Start by asking (either rhetorically or Socratically): what is exponentiation meant to mean? It means multiplying by a number multiple times. 10^5 means: multiply 10 together 5 times. 10^9 means: multiply 10 together 9 times. And so on.

So, $10^{-2}$ (for instance) ought to mean: multiply 10 together $-2$ times. But how can we make sense of that? Well, doing something negative-many times should be the opposite of doing it that many times. Taking (-1) step forward means taking one step back. And the opposite of multiplying by something is dividing by it. So, $10^{-2}$ should be the same as dividing by 10 twice — but that’s the same as multiplying by $\frac{1}{10}$ twice, so it’s $\left( \frac{1}{10} \right)^2$, or $\frac{1}{100}$.

The key point which I find is helpful for many students,is emphasising that it’s not just about trying to satisfy some identities, and it’s certainly not an arbitrary convention (no matter what some teachers say). It’s about defining it so that it means something useful, some coherent concept. The identities it satisfies express the ways in which the concept is more meaningful and coherent than (say) defining $a^{-b} := (-a)^b$ would be. But the meaning, the concept, is what underlies it all.

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  1. Help for confusing $$ 2^{-3}\neq (-2)(-2)(-2)$$

    Advise them to watch out for the exact position of the - sign,

    and tell them since it isn't next to 2, it simply does not belong to the 2, instead it belongs to the 3.

    Example would be

    $$2^{-3} = \frac{1}{2^3} = \frac{1}{2\times2\times2} = \frac{1}{8}$$ as already proposed by @DoktoroReichhard

  2. Learning the rule:

    $$a^{x+y} = a^{x}a^{y}$$ First example (with x = 2, y = 3) would be $$ 10^{2+3} = 10^{5} = 100 ~000= 100 \cdot 1000 = (10\cdot10) ~ \cdot ~ (10\cdot10\cdot10) = 10^{2}\cdot10^{3} $$ Now substitute 10 with a and you get $$a^{2+3}= a^{5} = aaaaa =aa\cdot aaa =a^{2}\cdot a^{3}=a^{2+3} $$ Then use 2 instead of 10 and calculate each single step on the board together with class

  3. Learning the rule: $$a^{x-y} = \frac{a^{x}}{a^{y}}$$ First example (with x = 2, y = 3) would be $$ 10^{2-3} = 10^{-1} =\frac{1}{10}= \frac {100}{1000} = \frac {10^{2}}{10^{3}} $$ Now substitute 10 with a and you get $$a^{2-3}=a^{-1} = \frac{1}{a}= \frac {aa}{aaa} = \frac {a^{2}}{a^{3}} $$ Then use 2 instead of 10 and calculate each single step on the board together with class

In the end feel free to point out (for the ambitious students) that these rules will also work with crazy numbers like $$ 2^{1.45 - 2.45} =2^{-1} =\frac{1}{2} $$

It's great that you are so passionate about teaching your class.

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Teaching counter-examples always seems like a bad method to me. Students will mess up enough on their own without help. –  ebyrob Jan 7 at 20:45
    
@ebyrob: To what do you refer as counterexample? (Using 2 instead of 10, or telling that these rules will also work with crazy numbers?) –  AddingColor Jan 8 at 12:29
    
all of item 1) above. It may only be one of your problem sets, but it is the first one (first and last are always most prominent). –  ebyrob Jan 8 at 14:17

The rule a ^(x+y) = a^x * a^y can easily be illustrated and explained for positive integers x,y (use several examples).

Then, e.g.

2^(3-3)= 2^0 = 1 =(rule) 2^3 * 2^(-3) = 8 * 2^(-3)

leads to

1 = 8 * 2^(-3) and therefore

2^(-3) = 1/8 as inverse to 8 = 2^3.

Of course you have to use several examples, but I think this makes it very easy.

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I would teach them the general rules but starting with increments and decrements of only 1. I would motivate them something like this:

If I tell you 2^10 is 1024, and then ask you for 2^11, you don't need to multiply 2x2x2.... you can just take the 1024 and multiply it by another 2. Everyone agree?

Once that's clear, you can try getting 2^9 by starting at 1024 and dividing by 2.

With that in place, you can go from 2^1 is 2 to 2^0 is 1 (and if you like anything^0 is 1) pretty easily. You still haven't taught the general rule about 2^(x+y) at this point. And now you can ask them to guess (figure out) what 2^-1 is. And from that, 2^-2.

General rules can follow, but if they get it moving one extra power up or down at a time, they get it.

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I'd tell the students if you want to make the negative "go away", put it under a 1. Maybe that sounds redundant, but putting it in a context where they can take something confusing and convert it to something manageable sometimes makes the rule stick better.

In the context of the particular example, you could point out the addition of the exponents when the problem is broken out.

$2^{3}$ = $2^{1}$ x $2^{1}$ x $2^{1}$

$-2^{3}$ = $-2^{1}$ x $-2^{1}$ x $-2^{1}$

$2^{-3}$ = $2^{-1}$ x $2^{-1}$ x $2^{-1}$

$\frac{1}{2^3}$ = $\frac{1}{2^1}$ x $\frac{1}{2^1}$ x $\frac{1}{2^1}$

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To a 9th grader, I would say "whenever you see a minus sign in the exponent, you always flip the number."

$$ 2^{-3} = \frac{1}{2^{3}} = \frac{1}{8} $$

I would simply do 10-20 examples on the board, and hammer the point until they start to get it.

You may have to review fractions with them here too.

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This is what a 9th grader needs. I don't know if you agree with the term simplify, but they need to reflexively understand that a negative exponent 'flips' the number. OP says that they have been introduced to this concept, but from the nature of the question, it's obvious that they don't understand it. Once they understand this, students seem to grasp the idea of cancelling factors in a ratio. From there, you can show that cancelling factors is really adding and subtracting exponents. –  MEH Jan 7 at 18:22

If they understand that $2^3=(2)\times(2)\times(2)$, then explain to them that this means $2^3 = (2^1)\times(2^1)\times(2^1) = 2^{(1+1+1)}$. Then tell them that this means that if $a = b + c +d$, then $X^a = (X^b)\times(X^c)\times(X^d)$. Simply tell them that if we if we multiply 2 to the power (a) by 2 to the power (b), we get $2^{(a+b)}$. Then tell them that if we divide instead of multiply, we add a minus sign instead. For example, $\frac{(2^5)}{(2^3)}=2^{(5-3)}$. This is a simple case of the general rule $\frac{(a^x)}{(a^y)}=a^{(x−y)}$. So $2^{-3}=2^{(0-3)}=\frac{(2^0)}{(2^3)}=\frac{1}{(2^3)}=\frac{1}{(2\times2\times2)}$

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Based on heropup's comment, draw the following pictures:

2:    --|--.---   
10^2:   100.000

1:    ---|-.---
10^1:    10.000

0:    ----|.---
10^0:     1.000

-1:   -----.|--
10^-1:     .100

-2:   -----.-|-
10^-2:     .010

and so on. The upper ones are number lines; the lower ones are decimals. Of course, one usually writes the number line with positives on the right, but I had to reverse them because we also write numbers with the most significant digit on the left.

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Multiplication of exponents:

$4^2 * 4^3 = 4^5$

The powers are added: $2 + 3 = 5$

Division of exponents:

$\dfrac{4^2}{4^3}\ = 4^{-1}$

The powers are subtracted: $2 - 3 = -1$

Now tell them to try and find the value of $4^{-1}$

Since $4^2 = 16$, and $4^3 = 64$

$\dfrac{16}{64}\ = 1/4 = 4^{-1}$

This is the way I would have wanted to have been taught it.

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I disagree with others saying that you need to teach the rule of 'flipping the number'. It may amount to this, but they need to know why first. Blindly learning rules puts a barrier in the place of deeper learning. Teach why then work through deducing the various rules with the students.

I think the answer from Jeppe was good - it builds on your existing answer, and using a physical example will help too. Just because the kids don't know some other 'rules' doesn't mean you can't teach this intuitively. In fact I think that this would be a good base for learning the two rules you quote.

(NOTE: this answer could probably have better been left as a couple of comments, but I am unfortunately low on rep on this stack-exchange site)

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I think that there have been a couple of answers that touch on this concept, I am hoping that I can state it clearly enough.

the Negative symbol ($-$) specifies direction and so does the Positive Symbol ($+$)

negative means left on the number line and positive means right on the number line, so from that we can place the numbers on the line.

$$ \begin{array}{c c c c c} 2^{-2} & 2^{-1} & 2^0 & 2^1 & 2^2 \\ \hline \frac 1 4 & \frac 1 2 & 1 & 2 & 4 \end{array} $$

as you go right you multiply by 2 (positive direction)

as you go left you divide by 2 (negative direction)

that is how I remember negative exponents, along with $$ 2^{-2} = \frac 1 {2^2} \\ 2^{-1} = \frac 1 {2^1} $$

that the Negative sign means the reciprocal of the power.

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thank you @miniBill –  Malachi Jan 7 at 20:40
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$\LaTeX$ isn't hard, you should learn it ;) –  miniBill Jan 7 at 20:42
    
my cousin told me that I should learn it –  Malachi Jan 7 at 20:44
    
It's a markup language, if you already know one you will pick it up fast –  miniBill Jan 7 at 20:46

Negative exponents are similar to but different than positive exponents, but they fit together well because of their algebraic properties. In explaining exponents, you need to explain what they are ($2^{-3}=1/2^3$ by definition) but alo why they are defined the way they are.

If $a>0$ is the base and $m,n>0$ are integers, then

  • $a^n=\underbrace{a*a*\cdots *a}_{\text{n times}}$
  • $a^{m+n}=a^m a^n$ just by counting the number of terms you are multiplying together
  • If $m<n$, then $a^{n-m}=\frac{a^n}{a^m}.$ There are two wys to explain this. First, write out factors and use the fact that, in fractions, you can cancel factors. Second, $a^{n-m}a^m=a^{(n-m)+m}=a^n$ by our previous property, so we can divide both sides by $a^m$.
  • Assuming $a\neq 0$, Write out the positive powers of $a$ as a sequence. We can move left or right by multiplying or dividing by $a$. Why? Because of the previous two properties.
  • If we wanted to continue the pattern further to the left, then we would want $a^0=1$. Define $a^0=1$ and show that all the previous properties we had still work.
  • Can we continue the pattern further to the left? If we could, we would have to define $a^{-1}=1/a, a^{-2}=1/a^2$, etc., and so we make a definition: $a^{-n}=1/a^n$.
  • Show that with this definition, the properties still work: $a^{m+n}=a^m a^n$ even if $m$ and $n$ aren't necessarily positive. This turns our addition and subtraction formulas into a single formula.
  • Recap: We had a definition, it had certain nice properties/patterns. If we tried to extend the patterns, it forced a new definition on us. Surprisingly, the new definition actually works and simplifies things, turning two patterns into one pattern.

If you had already discovered the formula $(a^m)^n=a^{mn}$ for $m, n>0$, you can simplify the search for the definition $a^{-n}=1/a^n$ to extending the pattern from $a^{-1}=1/a$. However, I don't have strong feelings on when this formula should be presented.

Negetive exponents are different than positive exponents, and if you're coming from only having positive exponents, you need to make it clear that you're making a new definition and not just stating an obvious property, or else things will seem unnecessarily mysterious. The definition isn't just pulled from a hat, however, and they should know there are good reasons to have the new definition. Remember, even if the students are just going to memorize formulas, it's important that they know there are simple ideas behind the formulas or else they will view mathematics as either a dry manipulation of symbols or an arcane and impenetrable fortress not worth trying to enter.

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Doing it with numbers may not reach all students because some people think better in other ways.

In addition to some of the stellar explanations here with numbers, I would do it visually for other students.

Plot 2^1, 2^2, 2^3 ... 2^5 on a graph for them:

A

Add on 2^0:

B

Then extend in to the negatives, and ask them where they think it will go:

C

I think there are two options for what they'll guess. They may guess that the pattern flips over:

D

The issue then though is that it flips at 1, so you'd have to think that 2^0 and 2^-0 would be different values (1 and -1 respectively). And that can't be. So that should lead them to guessing that it doesn't suddenly hop negative, but rather, it continues getting smaller and smaller:

E

As to what those numbers are, I suggest using one of the other explanations for the math, but to not confuse as to the direction, or the pattern, seeing it visually may help some students.

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I'm a 9th grader myself and I learned this in 7th grade and 8th grade but wasn't satisfied with the explanation until 9th grade. I first understood $x^{a+b}$ and $x^{a-b}$ because they're obvious once you write out the exponentiation as a bunch of individual multiplications and then see the terms cancel or whatever. Then you prove $x^0=1$ by showing $x^0 = x^{n-n}=\frac {x^n} {x^n} = 1$. Then you do negatives with $x^{-n} = x^{0-n} = \frac{x^0}{x^n} = \frac{1}{x^n}$. Then you demonstrate rational exponents by showing that since ${(x^{\frac{1}{n}})}^n = x^1$, $x^{1/n} = \sqrt[n]{x}$.

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Ninth-graders are not naive. Avoid metaphors and demonstrate what happens when you decrease positive exponents by one, then suggest that the rule should continue to apply as we get to zero and negative exponents.

$2^4 = 16$

$2^3 = 8$

$2^2 = 4 = 8/2$

$2^1 = 2 = 4/2$

(Same with other bases if necessary, to see the general pattern.) So, now that we have the pattern we can continue applying it to smaller exponents:

$2^0 = 2/2 = 1$,

$2^{-1} = 1/2$,

$2^{-2} = 1/4 = 1/2^2$, etc.

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During my middle school and high school education there were a couple things from mathematics that I just accepted 'as is'. This was my recourse when the teacher couldn't explain it in terms I could understand, they said that my questions would be answered in a higher grade or in college, or they themselves didn't understand it (although they never worded it that way).

However, once I got to those levels of mathematics to understand the rules of operations with the natural numbers, the integers, the rational numbers, the reals, and the complex numbers, I learned it was, essentially, all a sham. Quite literally, these number systems and associated binary operations are constructed to satisfy these nice properties. These systems and operations are supposed to represent intuitive, real-world results, but axiomatic mathematics has not (and perhaps cannot) capture this. So instead, they become definitions.

The reason that addition and multiplication on the natural numbers is associative and commutative is (almost) immediate from their definition. The reason that $0$ is the identity for $+$ and $1$ for $\times$ is by their definition. The reason that $a^{n+m}=a^n\cdot a^m$ is a quick corollary of the definition of exponentiation on the natural numbers.

The reason that $-n$ is the additive inverse of $n$ in the integers is because of the definition of $-$. The reason that $(-n)(-m)=nm$ is because of the definition of multiplication on the integers.

This pattern goes right on up to the complex numbers where $i^2=-1$ is true because of the definition of complex multiplication.

Again, these definitions are made to capture something we see to be intuitive and useful in real life. But to tell yourself that there are deep, lucid explanations for why things are, if you just wait, is a lie.

This also applies to your problem. How are you going to teach a 9th grader that $3^{-2}=\frac{1}{9}$? You have to remind them that $3^{-2}$ is the number $x$ such that $x\cdot 3^{2}=1$. This is because of the way exponentiation is defined on the rational numbers. Exponentation is defined on the rational numbers so as to satisfy these rules:

  • for nonzero $a$ we have $a^0=1$
  • for $n>1$ we have $a^{n}=a\cdot a^{n-1}$
  • for nonzero $a$ we have that $a^{-1}$ is the unique rational number such that $a\cdot a^{-1}=1$
  • for $n<-1$ and nonzero $a$ we have that $a^{n}=a^{-1}\cdot a^{n+1}$

Recall that $\frac{1}{a}$ is also defined as the unique number such that $a\cdot\frac{1}{a}=1$. So, it is immediate from definitions that $a^{-1}=\frac{1}{a}$.

These laws determine exponentiation and give rise to the nice properties:

  • $a^{x+y}=a^xa^y$
  • $(a^x)^y=a^{xy}$

And the fact that $a^{-x}=\frac{1}{a^x}$ follows from the observation above that $a^{-1}=\frac{1}{a}$.

So in my opinion, there are two ways to teach exponentiation to high schoolers.

Option 1: just tell them this is how things are and they just need to memorize the rules. You can lighten this approach by showing them patterns, like the ones already mentioned, which hint at why this should be the way things are. But I know I always appreciated bluntness from the teacher, so that I could shift my brain into 'as is' mode.

Option 2: give them the 'real' definition at the risk of confusing most of them, intriguing a couple, but still telling them that this is how it is because a lot of people wanted those nice properties. In fact, instead of doing this in class, you could make this a hand-out for the interested student (this might help the most amount of students).

This may be somewhat disappointing as it is your job to teach these things and hopefully imbue understanding. And it also goes often unsaid that it is also your job to get them ready for 'the test' which will be one of the largest factors that determine their path in life. And that's a lot of pressure (especially for that demographic). However, there isn't much you can do for a student who does not see how this will be applicable anywhere else in life. And when so many kids are criticizing the same stuff the same way almost everywhere, we need to step back and consider that maybe the math curriculum in secondary schools is so flawed that even adolescents can point out the asinine.

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protected by Alexander Gruber Jan 7 at 21:13

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