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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?

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  • $\begingroup$ 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. $\endgroup$ – Minus One-Twelfth Mar 2 '19 at 6:55
  • $\begingroup$ 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. $\endgroup$ – Mark Bennet Mar 2 '19 at 7:18
  • $\begingroup$ 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." $\endgroup$ – Michael Hoppe Mar 2 '19 at 9:45
  • $\begingroup$ @MichaelHoppe Or, more generally, products are invariant under permutations of commuting factors - see m answer. $\endgroup$ – Bill Dubuque Mar 2 '19 at 17:01
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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 $a*b=b*a$).

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  • $\begingroup$ did you use $\Bbb N$ to denote Natural number {0,1,2,}? it seems that this rule is valid for $n\in\Bbb R$ $\endgroup$ – shi95 Mar 2 '19 at 8:20
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    $\begingroup$ 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 $\Bbb N$. $\endgroup$ – Marc van Leeuwen Mar 2 '19 at 9:05
  • $\begingroup$ @shi95 This instance doesn't have a name in widespread use, but its natural extension does - see my answer. $\endgroup$ – Bill Dubuque Mar 2 '19 at 16:57
  • $\begingroup$ "(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. $\endgroup$ – darij grinberg Mar 3 '19 at 19:36
  • $\begingroup$ @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 $n=0$. $\endgroup$ – Marc van Leeuwen Mar 3 '19 at 22:36
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$a^2b^2=(aa)(bb)$ 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.

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

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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.

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  • $\begingroup$ 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. $\endgroup$ – darij grinberg Mar 3 '19 at 19:38
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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.

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

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