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There are long debates among Indonesian netizens about this http://www.globalindonesianvoices.com/15785/is-4x6-the-same-as-6x4-this-primary-school-math-made-controversy-in-social-media/

u100899
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    Yes. ${}{}{}{}{}{}{}{}$ – amWhy Feb 09 '15 at 15:56
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    Is this still an ongoing debate in Indonesia? – Frank Vel Feb 09 '15 at 15:59
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    It doesn't matter. You define $4\times 6$ one way, I define it the other way, we prove multiplication is communicative so our definitions are equivalent, everything is dandy. –  Feb 09 '15 at 16:00
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    Is this serious? – Gahawar Feb 09 '15 at 16:15
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    This is why I only use square numbers. – Milo Brandt Feb 09 '15 at 16:15
  • Actually such teachers as those described in the link exist virtually everywhere. Some time ago I read similar news about a teacher in Russia. That's of course the teacher who is bad, not the student. – Ruslan Feb 09 '15 at 16:22
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    I suppose there is a point to answer in the question. It is all very well to say that multiplication is commutative and it doesn't matter which way round you do it (true, of course), but you can't prove that if if you don't understand the definition of multiplication in the first place. – Geoff Robinson Feb 09 '15 at 17:17
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    If you read it as "four times six", that literally means you have a six, but you have it four times over, so $4\times6$ is indeed better represented as $6+6+6+6$. That is, there is a difference (not a numerical difference, but still a difference) between six groups of four and four groups of six. I think it's definitely important that a person be able to understand the distinction (if a person honestly can't see the distinction, that would make me worry about their capacity for abstract thought). Whether taking marks off without comment is the best way to teach this is debatable. – Jack M Feb 09 '15 at 17:39
  • I think the answer will depend on exactly what you have in mind by multiplication. Do you mean the intuitive, common-sense version of repeated addition? In that case, common sense would suggest that it doesn't matter which, although as Jack M pointed out, you could get very strict about how the English reads to get a unique answer. Do you mean the rigorously, axiomatically defined mathematical symbol? In that case, you would use the Piano axioms, which Ross Millikan explained in the answers. – Addem Feb 09 '15 at 18:09

8 Answers8

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Are you aware, the multiplication operation is commutative over $\mathbb{N}$.

As such, $6 \times 4 = 4 \times 6$.

And we have also $4+4+4+4+4+4=24=6+6+6+6= 4 \times 6 = 6 \times 4$

EDIT: I forgot the following:

$6+6+6+6=(4+2)+(4+2)+(4+2)+(4+2)=(4+4+4+4)+(2+2+2+2)=(4+4+4+4)+(4+4)=4+4+4+4+4+4$

Thanks to other properties of the additive operation, such as commutativity and associativity...

EDIT2: Changed to $\mathbb{N}$ from $\mathbb{R}$ per suggestion.

Martigan
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If you refer to the Peano axiom for multiplication, it says $a \cdot S(b)=a+a\cdot b$, so expanding this we would have $4 \cdot 6=4+4+4+4+4+4$. What definition of multiplication are you using?

Ross Millikan
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Surely, multiplication is commutative in R and I think the teacher was a bit narrow-minded when correction it.

Based on my native language (Portuguese), I would opt for 4x6 = 6+6+6+6, just because this is how I read it.

I can also say 4 objects = (object + object + object +object) and 4 m = 4 * meter, but it would be weird to say (objects x 4) or note a distance as m 4.

toliveira
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When written $4 \times 6$ this is often, in English read as four sixes, which implies $6+6+6+6$ rather than $6 \times 4$ which would be read as six fours which would imply $4+4+4+4+4+4$., but this is a pure linguistic convention and not how we define these things in maths.

Conrad Turner
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Let $A \times B$ mean the addition of $A$ repeated $B$ times.

Since multiplication is commutative in $\mathbb{R}$, that tells us that $A \times B = B \times A$. In other words, the addition of $A$ repeated $B$ times is the same as the addition of $B$ repeated $A$ times.

Thus, $4 \times 6 = 4+4+4+4+4+4 = 6 \times 4 = 6+6+6+6$

MT_
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    So... what's $\pi\times\sqrt2$? Adding $\pi$ to itself $\sqrt2$ times? – Asaf Karagila Feb 09 '15 at 17:10
  • @AsafKaragila: No, that would be $\pi\times(1+\sqrt2)$. –  Feb 09 '15 at 17:22
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    @Yves: Makes no sense. – Asaf Karagila Feb 09 '15 at 17:23
  • @Yves: I just saw that you edited the comment; fine, $\pi\cdot\sqrt2$ is "adding $\pi$ repeated $\sqrt2$ times". How do you repeat something $\sqrt2$ times? How is that different from repeating $\sqrt3$ times, or $\frac14$ times? What does it even mean to repeat something a non-ordinal number of times? – Asaf Karagila Feb 09 '15 at 18:45
  • Just joking. But actually, this raises no real difficulty: $\pi+\frac\pi{10}+\frac\pi{10}+\frac\pi{10}+\frac\pi{10}+\frac\pi{100}+\frac\pi{1000}+\frac\pi{1000}+\frac\pi{1000}+\frac\pi{1000}+\frac\pi{10000}+\frac\pi{10000}+\frac\pi{100000}+\frac\pi{1000000}+\frac\pi{1000000}+\frac\pi{1000000}\cdots$ –  Feb 09 '15 at 20:25
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Let $a\cdot b$ denote the cardinality of the set $A\times B$, where $A$ has cardinality $a$ and $B$ has cardinality $b$. There is no addition involved in this definition. But of course he have to introduce the notion of ordered pair to speak about $A\times B$, and you may have fun starting your controversy again: Is $\langle x,y\rangle$ a language primitive or defined as $\{\{\{x\},\emptyset\},\{\{b\}\}\}$ or $\{\{a,1\},\{b,2\}\}$ or $\{\{a\},\{a,b\}\}$ or $\{a,\{a,b\}\}$ or ... ?

Hagen von Eitzen
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If this teacher explicitly defined the notation $a\times b$ to be short for the sum of $b$ terms $a$ and insisted that this convention be enforced, then he was right to count errors for not following the instructions.

But nobody that I know uses this strict and somewhat silly/misleading notation.

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Another approach to it, less "mathematical" than my other answer, but linked to the "intuition" of the multiplication.

We are dealing with small children, and multiplication is, for them, repeated addition.

The question is $6+6+6+6=6 \times 4$ or is it $4 \times6$ ?

Some people tend to prefer (they "see") the following reasoning: they count the number of times they will multiply, and then put the number (or the elements, items etc.) they want to "add multiple times". In that case, $6+6+6+6=4 \times 6$.

However, you might want to see that writing $6+6+6+6$, you are dealing with sixes, and that you deal with them four times. So you start by identifying the number at hand ($6$), and then write the number of times you "saw" the number, that is, the number of items added together. In that case, you have $6+6+6+6=6 \times 4$.

Of course, we all know this is the same thing, but intuitively this is different for someone starting with multiplication. Both ways are valid to my eyes...

Martigan
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