# Can I call $(ab)^n=a^nb^n$ “Distribution”?

My grade 8 students generally know how to use the distributive property: $a(b+c)=ab+bc$

However, we're now learning exponent laws, and one of them is $(ab)^n=a^nb^n$. In order to help my students remember to raise $a$ to the $n$ and not just $b$, I often draw two little arrows from the exponent to the two bases. Because this looks a bit like The Distributive Property, I sometimes call this action "distributing."

1) Is it OK for me to call this action "distributing" as long as I try to make it clear that it is different to The Distributive Property?

2) Is there a better verb I could us instead of "distribute" in the case of the exponent law?

• If you use the logarithmic version nlog(ab)=nlog(a) + nlog(b), then it is a consequence of the distributive law and other rules involving exponentiation. However, when you don't have it, it is an algebraic consequence of associativity and commutativity. Since you might need a version like a^kb^na^(n-k), I would call it rearranging. Gerhard "Warning: Not A Professional Educator" Paseman, 2015.03.16 – Gerhard Paseman Mar 16 '15 at 20:31
• To be clear, "not having it" means not using the logarithmic form. However, you could use (ab)^n =a^nb^n as an example of proof by induction, using (ab)^nab=a^nb^n= aa^nb^nb, etc. How appropriate these diversions might be highly depends on your situation; I can see many possibilities for extending this example in small but interesting directions. Gerhard "Likes To Show Simple Examples" Paseman, 2015.03.16 – Gerhard Paseman Mar 16 '15 at 20:41
• Also, because exponentiation is generally not commutative, you have only a right distributive law when you write a^n as a exp n, for then (a*b) exp n = (a exp n)*(b exp n). However, exp does not left distribute over the positive integers. Better to use the term "rearranging". Gerhard "Making For More Interesting Questions" Paseman, 2015.03.16 – Gerhard Paseman Mar 16 '15 at 20:45

Yes. It is distribution (on one side) of exponentiation over multiplication, rather than the traditional distribution (on both sides) of multiplication over addition.

For two binary operators $\oplus$ and $\otimes$, $\otimes$ is right-distributive over $\oplus$ iff

$$(a \oplus b) \otimes c = (a \otimes c) \oplus (b \otimes c)$$

In the traditional case, $\oplus$ is addition and $\otimes$ is multiplication.

In the the case you ask about, $\oplus$ is multiplication and $\otimes$ is exponentiation.

This is the right terminology, and there's no reason to call it anything else.

Don't give the generic definition with funny-looking circles. Just write something like:

$$(a + b) \cdot c = (a \cdot c) + (b \cdot c)$$

$$(a \cdot b) \hat{\text{ }} c = (a \hat{\text{ }} c) \cdot (b \hat{\text{ }} c)$$

and the pattern is is clear. (The caret will be more familar if they use calculators that have it.)

This may be likely to get students to think about it, rather than just rearrange symbols -- and higher, littler symbols -- around according to quirky rules.

I'd much prefer explaining multiplication distributing over addition and exponentiation distributing over multiplication, than trying to make up a term.

• i think that you need to be very careful with this explanation because 9 out of 10 kids will use this as a justification for why $(x + y)^2 = x^2 + y^2$ – celeriko Mar 17 '15 at 2:00
• @celeriko, absolutely. Point out this property is special, or "rare". That's what makes it so cool! – Paul Draper Mar 17 '15 at 2:37
• I also call this distribution, saying "powers distribute over multiplication". In fact if you arrange your operations like this: [^ $\sqrt$][$\times$ $\div$][+ -] (I normally put them one group on top of the other), then you can say "each distributes over the ones directly below" (though of course division only distributes on the right of + and -) – DavidButlerUofA Mar 17 '15 at 8:40
• @DavidButlerUofA, nice. (And exponentiation and roots distribute only on the right.) – Paul Draper Mar 17 '15 at 15:42

# Operator viewpoint

Yes, you are right. See Paul Draper's answer.

# Function viewpoint

Here you should use the term algebraic linearity:

• The proportional functions $f(x)=kx$ are linear concerning addition: $$f(a+b)=f(a)+f(b)$$
• The power functions $g(x)=x^n$ are linear concerning multiplication: $$g(ab)=g(a)g(b)$$

Remark: I used linear as a translation of the German verträglich. If there is a better translation, please say so in the comments or edit it straigtaway.

# Ring theory viewpoint

Here it's just a generalization of commutativity and associativity. See Chris C's answer.

• Hyman's German-English mathematical dictionary prefers "consistent" or "compatible". Though I would probably call it "structure preserving". In the case you mentioned, however, one can probably call them homomorphisms. // None of these, however, are words that should be forced on 8th graders. – Willie Wong Mar 17 '15 at 15:43
• @WillieWong Thanks for the translations. You're right, that these terms seem rather over the top. Maybe a new word like "swappable" or "interchangeable" might be appropriate in secondary. – Toscho Mar 18 '15 at 18:28

I would be strongly hesitant in calling $(ab)^n = a^n b^n$ distribution for students at that level. You are correct in that distribution is $a(b+c) = ab+ac$, so I would stick with that. Continuing down that road would lead to questions whether $a^{n+m} = a^n a^m$ is distribution or not (it isn't). We want to caution them that distribution is distinct due to errors such as $\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}$ where a more colloquial usage of distribution leads to a more colloquial understanding.

The more proper terms here are associativity and commutativity or just that they are associative and commutative, that is, $(ab)c = a(bc)$ for associativity and $abab = aabb$ is due to commutativity, but that might be too much for 8th graders.

I would add, I recall focusing more on this material in 10th grade geometry than in 8th grade, but times may have changed.

• By that argument, there isn't a need for distribution at all, since it can be explained with community of addition. The associativity explanation only holds water as long as $n$ is an integer, i.e. you can't use it to show $a^{\frac{1}{2}}b^{\frac{1}{2}} = (ab)^{\frac{1}{2}}$. – Paul Draper Mar 17 '15 at 0:47
• I did state a more ring theoretic viewpoint, but you can justify radicals and other algebraic objects as certain algebraic extensions of the rationals. – Chris C Mar 17 '15 at 2:07