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

Question

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?

Answer 1

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 $a*b=b*a$).

Answer 2

$a^2b^2=(aa)(bb)$ 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.

Answer 3

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

Answer 4

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

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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 $\forall \ k \in \mathbb{N}$.

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

Answer 5

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.

Answer 6

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.