# What is the name of rule: $a^2\cdot b^2 = (ab)^2$ and how to justify it mathematically?

It is easy to illustrate the rule, for instance $$10^2 \cdot 5^2 = 50^2 = 2500$$

The question is, what is the name of this rule: $$a^2\cdot b^2 = (ab)^2$$

and how to justify it mathematically?

• It is due to the fact that the order we multiply (say integers) does not matter ($(ab)^2 = \color{blue}{(ab)(ab) =a\cdot a \cdot b \cdot b} = a^2 b^2$). This fact is a consequence of the associative and commutative laws of multiplication. – Minus One-Twelfth Mar 2 '19 at 6:55
• If multiplication is associative and allows for cancellation (for example non-zero integers - though it is also easy to see that both sides are equal if either $a$ or $b$ is zero) then this is equivalent to being commutative. I can't immediately think of an interesting non-commutative system which has this property. – Mark Bennet Mar 2 '19 at 7:18
• Don't know a name for that rule, but just speak it out loud: "You may calculate the power of a product factorwise, i.e. factor by factor." – Michael Hoppe Mar 2 '19 at 9:45
• @MichaelHoppe Or, more generally, products are invariant under permutations of commuting factors - see m answer. – Bill Dubuque Mar 2 '19 at 17:01

The rule is a consequence of the commutativity of multiplication, but does not have its own name. One has $$a^2*b^2=a*a*b*b=a*b*a*b=(a*b)^2$$. More generally one has for powers of products: $$(ab)^n=a^nb^n$$ for any $$n\in\Bbb N$$, whenever '$$*$$' is an (associative and) commutative operation (or even if just $$a*b=b*a$$).

• did you use $\Bbb N$ to denote Natural number {0,1,2,}? it seems that this rule is valid for $n\in\Bbb R$ – shi95 Mar 2 '19 at 8:20
• Exponentiation with real exponents is a whole different kettle of fish. They are defined for far less kinds of multiplications, and even for numeric multiplication they have to limit the base to be a positive real number to have good properties. Also proofs for the real exponent case are quite different from the natural number case. So that is why I stated it for $\Bbb N$. – Marc van Leeuwen Mar 2 '19 at 9:05
• @shi95 This instance doesn't have a name in widespread use, but its natural extension does - see my answer. – Bill Dubuque Mar 2 '19 at 16:57
• "(or even if just $a*b=b*a$)" <-- This much is not true. Associativity is needed to even make $a^n$ and $b^n$ well-defined for $n > 2$. And even for $n = 2$, I would not expect $\left(a*b\right)*\left(a*b\right) = \left(a*a\right) * \left(b*b\right)$ to follow from commutativity alone. – darij grinberg Mar 3 '19 at 19:36
• @darij Yes of course associativity is needed; I mentioned it to show I was assuming rather than ignoring it, but I put it in parentheses to indicate my sentence was not about associativity (which I'm not sure OP even knows about) but about commutativity. For the record I am also assuming a neutral element, since I allow $n=0$. – Marc van Leeuwen Mar 3 '19 at 22:36

$$a^2b^2=(aa)(bb)$$ then with associative theorem of real numbers you get

$$(aa)(bb)=a(ab)b$$ with commutative property you get

$$a(ab)b=(ab)ab$$ again with associative you finally get

$$(ab)ab=(ab)(ab)={(ab)}^2$$

I am not sure it has a name, nor does it deserve one.

It is a consequence of the fact that multiplication (of real numbers) is commutative. To be precise, we also need that multiplication is associative: \begin{align}a^2\cdot b^2&=(a\cdot a)\cdot (b\cdot b)\\&=((a\cdot a)\cdot b) \cdot b\\&=(a\cdot(a\cdot b))\cdot b\\&=(a\cdot(b\cdot a))\cdot b\\&=((a\cdot b)\cdot a)\cdot b\\&=(a\cdot b)\cdot(a\cdot b)\\&=(a\cdot b)^2\end{align}

One name for the $$n$$-ary extension is the generalized commutative law, which states that an associative product of commuting terms remains the same under any permutation $$\sigma$$ of the terms, i.e.

$$\ \ \ \ \forall\, i,j\!: \,a_{\large i} a_{\large j} = a_{\large j} a_{\large i}\,\Rightarrow\, \,a_{\large 1} a_{\large 2}\cdots a_{\large n} = a_{\large \sigma 1} a_{\large \sigma 2} \cdots a_{\large \sigma n}$$

In particular we have the corollary: $$\, ab = ba\,\Rightarrow\, (ab)^k = a^k b^k$$

It has a straightforward inductive proof (hint: by induction, in the RHS product we can commute $$a_n$$ to the end, then by induction again we can permute the (new) first $$n\!-\!1$$ terms to be $$\:a_1 a_2\cdots a_{n-1})$$

Most good abstract algebra textbooks will discuss the generalized associative and commutative laws, e.g. see section 1.4 of Jacobson's Basic Algebra 1.

• This is probably the best answer for the long-term benefit of the OP. Let me add that an alternative way to prove the generalized commutative law is by showing first that every permutation $\sigma$ of $\left\{1,2,\ldots,n\right\}$ can be written as a composition of simple transpositions $s_i$ for $i \in \left\{1,2,\ldots,n-1\right\}$ (where a simple transposition $s_i$ only swaps $i$ with $i+1$ while leaving the other numbers unchanged). Either way, the generalized commutative law relies on the generalized associative law, which ensures that products like $a_1 a_2 \cdots a_n$ make sense. – darij grinberg Mar 3 '19 at 19:38

I'm not sure if it has a name but you can prove it (the general statement) by induction for natural numbers.

To Prove: $$a^n\cdot b^n=(a\cdot b)^n, \ n\in \mathbb{N}$$

Proof:

Let $$P(k)$$ denote the statement to be proven.

1. Base Case: $$n=1$$ $$(a\cdot b)^{1}=a\cdot b$$
2. Inductive Hypothesis: Let $$P(k)$$ be true, we need to show that $$P(k)\implies P(k+1)$$ $$P(k+1): (a\cdot b)^{k+1}=(a\cdot b)^k(a\cdot b)$$ $$a^k \cdot b^k \cdot a\cdot b =a^{k+1}\cdot b^{k+1}$$

Hence by the Principle of Mathematical Induction, $$P(k)$$ is true $$\forall \ k \in \mathbb{N}$$.

To prove this statement for all numbers will require using tools from Abstract Algebra which isn't in my toolbox yet.

• In fact its generalization has a name - see my answer. – Bill Dubuque Mar 2 '19 at 18:56

More generally, $$a^n \cdot b^n = \left(ab\right)^n$$ for any nonnegative integer $$n$$.

Even more generally, the same equality holds not just for multiplication, but for any associative binary operation, provided that $$a$$ and $$b$$ commute. For example, if $$f$$ and $$g$$ are two maps from a set $$X$$ to $$X$$ such that $$f \circ g = g \circ f$$, then $$f^n \circ g^n = \left(f\circ g\right)^n$$ for any nonnegative integer $$n$$, where $$f^n$$ means $$\underbrace{f \circ f \circ \cdots \circ f}_{n \text{ times}}$$ (and similarly $$g^n$$ and $$\left(f\circ g\right)^n$$ are defined).

I show three ways to prove this fact (for maps $$f$$ and $$g$$) in the solution to Exercise 6 (b) on UMN Fall 2017 Math 4707 homework set #2. The same arguments apply to numbers and products instead of maps and compositions.