My daughter showed me the picture below, which came from 9GAG. It shows a question on an exam asking the student to "use the repeated addition strategy to solve: 5 x 3." The student answered "5+5+5" and the teacher gave the answer a one-point deduction, stating that the correct answer is "3+3+3+3+3."

repeated addition

I have not seen any textbook that defines the notation $a\times b$ (where $a$ and $b$ are positive integers) as $b$ added $a$ times.

Personally, I prefer to use $a\times b$ to represent $a$ added $b$ times. This is to be consistent with using the notation $a^b$ (where $a$ and $b$ are positive integers) to mean $a$ multiplied $b$ times.

My question is:

What official document defines $a\times b$ (where $a$ and $b$ are positive integers) as $b$ added $a$ times?

  • 16
    $\begingroup$ I would hope that the fault lies with an over-zealous mark scheme rather than the teacher. There is no standard for these, beyond that created by whoever came up with the list of 'multiplication strategies'. The questions and marking also seem designed to teach the wrong things. Students should be encouraged to recognise that multiplication is commutative, and to NOT do it by repeated addition (I remember reading that using repeated addition hinders students when they need to use multiplication for later tasks). $\endgroup$
    – Jessica B
    Oct 27, 2015 at 10:44
  • 7
    $\begingroup$ Joel - the second question is just as, if not more, disturbing than the first. The teacher didn't like the grid orientation? $\endgroup$ Oct 27, 2015 at 19:41
  • 4
    $\begingroup$ Who thinks 9GAG is serious? $\endgroup$ Oct 27, 2015 at 23:40
  • 2
    $\begingroup$ There are surely many definitions that specify the opposite. For example, Goodstein's definition of hyperoperations (1947) includes multiplication as $H_2 (a, b)$, and resolves inductively to $a+a+a+...$, similar to exponents and higher operations: angrymath.com/2015/10/on-hyperoperations.html $\endgroup$ Oct 29, 2015 at 22:15
  • 2
    $\begingroup$ @DanielR.Collins, thanks for the very nice reference. $\endgroup$ Oct 29, 2015 at 23:46

7 Answers 7


You asked for an official document, and I can't give that. But I will try to speak for the teacher here. I don't agree with deducting points, but want to point out that the strategy of identifying axb with b+b+...+b (a of them) may be useful for students.

Mathematics generalizes, but it is helpful if it starts with something concrete. If some of the kids think 5x3 means 5+5+5 and others think it means 3+3+3+3+3, then it may be hard for them to discuss the ideas with each other. If they all start from the same place, and are led to notice that these two different sorts of problems always end up with the same answer, then they get to discover commutativity, instead of having it forced on them. (What I remember of the 'new math' of the 60's is having to write "this is true because multiplication is commutative". Gag.)

[Some kids will already have thought about this, and will already feel that both are the same. It will be hard for those kids to be put back in a box. With my bad memory, I would never remember which way was "right". That's why they shouldn't have points deducted.]

My friend wrote a great couple of blog posts referencing this. His niece thought of multiplication this way. He later saw the classwork behind that. In the comments on that post, he and I discuss the merits of this approach.

Although my friend is in New York, the school he references at his blog turns out to be in Connecticut. For New York, on this site you can find: Common Core Learning Standard 3.OA.1: Interpret products of whole numbers, e.g., interpret 5 x 7 as the total number of objects in 5 groups of 7 objects each. For Connecticut, I'm not sure.

  • $\begingroup$ Thank you for the links. It seems that the school mentioned in the blog (that taught that $3\times 5=5+5+5$) is in New York. It would be nice to know if this interpretation is in a state standard, and not just a convention agreed upon in class. $\endgroup$ Nov 8, 2015 at 1:35
  • 1
    $\begingroup$ It looks like it might be in the standards. Look at this: engageny.org/resource/grade-3-mathematics $\endgroup$
    – Sue VanHattum
    Nov 8, 2015 at 15:45
  • 1
    $\begingroup$ Yes, it's there. Common Core Learning Standard 3.OA.1: Interpret products of whole numbers, e.g., interpret 5 x 7 as the total number of objects in 5 groups of 7 objects each. This is consistent with what the teacher in my post is using. If you can edit this into your answer, I will accept it. $\endgroup$ Nov 9, 2015 at 0:56

While there may be legitimate reasons behind the convention

In $a \times b $ the $a$ denotes the number of terms and the $b$ denotes the individual terms

the larger issue is the mismatch between the teacher's enforcement of that convention and the expressly stated purpose of the formative assessment, which is written at the top of the very same page:

I can use multiplication strategies to help me multiply

It is clear that the student is using multiplication strategies to help her multiply; she is even using the exact multiplication strategy called for in the item.

I know this is not directly an answer to your question, but this teacher clearly does not understand what "Formative Assessment" means, nor what this particular formative assessment is meant to assess.


A reason why this form might be preferred is the way one says it: $5 \times 3$ is read out "five times three" so it says take $3$ five times, hence it "is" $3+ 3+ 3 + 3 + 3$.

However I doubt there is any real standard. For what it's worth Wikipedia disagrees with itself.

  • On the page on Multiplication it has $a \times b$ as $b + \dots + b$.
  • On the page on Peano axioms when it defines multiplication recursively it states $a \times S(b)= a + a\times b$, leading to $a \times b$ when unrolled being $a+ \dots + a$.

Personally, I am very used to thinking of $a \times b$ as $b+ \dots + b$ as it lines up with scalar multiplication in vector-spaces (or more generally the fact usually modules over commutative rings are written as left-modules).

  • $\begingroup$ I am not sure if this meets what you expect from an answer. If not please just let me know an I will remove it. $\endgroup$
    – quid
    Oct 27, 2015 at 10:59
  • $\begingroup$ This is a good answer, as it cites official sources. (I'm considering Wikipedia as an official source.) $\endgroup$ Oct 27, 2015 at 12:13
  • 2
    $\begingroup$ +1, the relationship with scalar multiplication is not something I had thought of. Also, now that I think of it, $3b=b+b+b$. $\endgroup$ Oct 27, 2015 at 12:20
  • 1
    $\begingroup$ +1 for the point of scalar multiplication – when one thinks of $\mathbb R$ as a $\mathbb R$ vector space, then in $a\cdot b$ the variable $a$ is the scaling factor and $b$ is the vector (usually we write $\lambda v$ with $\lambda$ being from the field and $v$ being a vector) $\endgroup$ Oct 27, 2015 at 17:05
  • 2
    $\begingroup$ @user21820 I can see some merits in mentioning the other points you raise. I invite you to write your own answer. Please also see the comment above where I forgot the at-user notification. $\endgroup$
    – quid
    Oct 30, 2015 at 12:07

One place in math where this issue actually does come up is in defining ordinal multiplication. From an ordinal perspective, the ordinal $5$ is the order type $a<b<c<d<e$, the ordinal $3$ is the order type $x<y<z$, and $5 \times 3$ is $$a_x < b_x < c_x < d_x < e_x < a_y < b_y < c_y < d_y < e_y < a_z < b_z < c_z < d_z < e_z.$$ I would definitely describe this as more like $5+5+5$ (the student's answer), than $3+3+3+3+3$ (the teacher's), although they are isomorphic.

Indeed, ordinal multiplication is distributive from the left: $a \times (b+c) = a \times b + a \times c$, but not from the right: $(a+b) \times c$ need not equal $a \times c + b \times c$.

To see that this issue can matter, consider $2 \times \omega$ and $\omega \times 2$, where $2$ is the order type $p<q$ and $\omega$ is $0 < 1 < 2< 3 < \cdots$. Indeed, $\omega \times 2 = \omega + \omega$ and $2 \times \omega \neq \omega + \omega$.

Of course (1) this is a pretty specialized topic which is very distant from elementary school math and (2) the decision about which way $\alpha \times \beta$ should be defined was arbitrary in the first place. But it is a nice example of how issues that seem like philosphical hairsplitting on a lower level can actually matter if you get far enough into math.


I would like to suggest a (English) language based approach:

"I went to the store three times." (Should be totally clear).

"Three times I went to the store." (Still clear).

"I went to the store times three." (Maybe understandable, but rather odd).

"Three I went to the store times." (Doesn't make much sense).

Upshot: The number modifying 'times' wants to go before rather than after 'times'. Thus the number before times says how many occurences; while the other element says what is repeatedly occurring.

"Three times five"...five is occurring three times: $5 + 5 + 5$.

When teaching this idea (to preservice teachers), I sometimes suggest that we can translate 'times' as 'groups of' (or 'group of'). Then "Three times five is three groups of five ($5 + 5 + 5$). This later extends to fractions with $\frac 12 \times 6$ becomes half of a group of $6$.


I am not a math educator; however I wonder if at times it might make sense to use the commutative law first before expanding the terms. Please solve

1,000,000,000 x 3

Might make sense to use the commutative law

3 x 1,000,000,000



1,000,000,000 + 1,000,000,000 + 1,000,000,000

However the Teacher may prefer the student use their method for this operation.

  • $\begingroup$ I would enjoy to see the teachers solution for this problem. It make help our students. $\endgroup$
    – A_Andy
    Jan 16, 2016 at 20:01
  • $\begingroup$ I would enjoy seeing all the teachers that feel strongly about NOT applying the cumulative property first, solve this problem by hand as a homework problem. Possibly they could do this before coming back to work. It might help our students. $\endgroup$
    – A_Andy
    Jan 16, 2016 at 20:29
  • 2
    $\begingroup$ I think you mean commutative law. $\endgroup$ Jan 16, 2016 at 20:35
  • $\begingroup$ Yes, Thank you. I did mean commutative law. $\endgroup$
    – A_Andy
    Jan 16, 2016 at 20:58

If you want an official document from the Math universe your best bet is Bourbaki. However, that probably isn't useful, since almost no educator works from Bourbaki. You may want an official document from the Math education world, but I don't know anything about that and it probably differs from place to place anyway.

And in reality, this only matters at all for a very brief time until they realize that multiplication is commutative, so I kind of feel like you can pick whichever convention you want and if it conflicts with anyone else's convention you can just argue that they're effectively equivalent.

  • 1
    $\begingroup$ Without checking, I'd be extremely surprised if Bourbaki talks about integer multiplication by repeated addition. Elementary set theory, but I doubt this student knows what the set containing the empty set is... $\endgroup$
    – Jessica B
    Oct 27, 2015 at 19:53
  • $\begingroup$ @JessicaB right, that's why I threw out that option in the same sentence that I introduce it. But I figured OP might be looking for a fairly universal and fundamental pronouncement on the topic, and therefore thought it might be something to at least mention--just in case it's helpful. But since Bourbaki defines everything, then once it gets to the definition of multiplying natural numbers I have to imagine they pick one of the two conventions. $\endgroup$
    – Addem
    Oct 27, 2015 at 19:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.