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Student was asked to convert the following statement into multiplication format $$7+7+7+7+7+7$$ She wrote the answer as $7\times 6=42$ and was marked wrong as the teacher expected $6\times 7=42$.

Is there any rule that can clarify the answer format?

The same with converting a multiplication sum into adding $6\times 3$

she wrote $3+3+3+3+3+3$ and once again was marked as wrong. Teacher expected $6+6+6=18$

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This teacher should not be teaching mathematics to children. – user22805 May 11 '12 at 6:22
No wonder most teenagers hate math – Andrea Mori May 11 '12 at 6:53

I'd say both variants are fine.

The variant $7\cdot6=7+7+7+7+7+7$ seems pretty much consistent with how ordinal multiplication (which generalizes multiplication of natural numbers) is usually defined. And there, when it comes to transfinite ordinals, this is the only way that makes sense: $\omega\cdot2$ means $\omega$ followed by another $\omega$, i.e. $\omega+\omega$. On the other hand $2\cdot\omega$ means $2$ followed by $2$ followed by $2$ etc.

So this way of looking at things is consistent with how ordinal multiplication is usually written.

And since we already have some answers telling us how people tend to prefer $7\cdot 6=6+6+6+6+6+6+6$, I'd have say both interpretations are indeed fine.

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I suspect there is no commonly accepted convention on whether $2+2+2$ is $2\times 3$ or $3\times 2$. But note that $2\times 2\times 2$ is $2^3$, so I personally prefer to say that $2+2+2=2\times 3$.

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I asked a question about this at… and I got a few answers. – Joel Reyes Noche Sep 24 '14 at 13:36
See also my question at Mathematics Educators Stack Exchange. – Joel Reyes Noche Dec 18 '15 at 2:35

I favor Austin Mohr, for scalar multiplication, the commutativity law may hold, but in other cases (say vector multiplication), the commutativity law may not hold.

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Just for the sake of concreteness, suppose we are counting apples. Many educators prefer to consistently interpret $m \times n$ as "$m$ groups of $n$ apples". In your example, $$7 + 7 + 7 + 7 + 7 + 7 = 6 \times 7$$ because it is 6 groups with 7 apples in each group. Similarly, $$6 + 6 + 6 + 6 + 6 + 6 + 6 = 7 \times 6$$ because it is 7 groups with 6 apples in each group. The fact that these both count the same number of apples overall is known as the commutative property of multiplication and should not be taken for granted (many other mathematical operations do not commute).

If this is what the teacher had in mind, then indeed $$ 6 \times 3 = 3 + 3 + 3 + 3 + 3 + 3, $$ so perhaps the he/she is mistakenly inconsistent in grading the second example.

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I disagree: at the grade level indicated by the assigned problems a student who understands and takes for granted the commutativity of multiplication is ahead of the game. I do not think that the teacher's behavior would be defensible even if it were consistent. – Brian M. Scott May 11 '12 at 8:16
@BrianM.Scott My intent wasn't to defend or condemn the teacher, but rather to explain why s/he might consider $m \times n$ different from $n \times m$. – Austin Mohr May 11 '12 at 20:10
I understand that, but should not be taken for granted still bothers me, because I think that at this level it's simply wrong. I also think that it does have the effect of defending the teacher. – Brian M. Scott May 11 '12 at 20:14
The fact that students have to face such a teacher year after year, I would give a warning (if I come across such grading) and the second such mistake would cause termination. – Kirthi Raman May 13 '12 at 21:06

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