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I have a friend with dyscalculia and was teaching her some some mathematics (namely, solving a linear equation, simplifying certain expressions, and what (affine linear) functions are).

She understood solving equations of the form $ax + b = 0$ by first adding $-b$ to both sides and then diving by $a$. Dealing with negative $a$ and with expression $b - b$ was something of a problem, but I hope she figured it out, also.

Adding slightly more complexity created more problems. For example: $2x + 3 = -7$. We subtract 3 from both sides, getting $2x = -7 - 3$. She has great trouble seeing that $-7-3 = -10$.

How to communicate and teach the concepts here? I tried using the thermometer analogy, explaining how $a - a = a + (-a) = 0$ and, somewhat poorly, that $-7 - 3 = - (7 + 3) = -10$. How to justify the last attempt in a useful way? What other models or intuitions are there for understanding the negative numbers and particularly summing them?

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    $\begingroup$ What's going on here? This is a barely-appropriate question on mathoverflow (I'd vote to close if I weren't a moderator). But people are voting up the answers like crazy! I think this is a great example of how easy material gets lots of votes, but the hard and interesting stuff that we're actually after wallows in obscurity. $\endgroup$
    – Kim Morrison
    Commented Dec 5, 2009 at 20:32
  • $\begingroup$ Also, a few of the explanations below seem not to have seriously considered what what be useful to someone who has trouble parsing mathematical symbols. Jose's would probably confuse most high schoolers. $\endgroup$
    – Ben Webster ♦
    Commented Dec 5, 2009 at 20:53
  • $\begingroup$ @Ben: Quite possibly if explained as written. However the answer is meant for the person doing the teaching. This is why I edited it to state explicitly that my answer should be translated into a suitable narrative, perhaps via analogies as in the answers inspired by "finance". $\endgroup$
    – José Figueroa-O'Farrill
    Commented Dec 5, 2009 at 21:02
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    $\begingroup$ I don't think this question belongs on Math Overflow. Let's keep to questions so that you should be a mathematician to give a competent answer, rather than a neuroscientist or developmental psychologist. $\endgroup$
    – Anton Geraschenko
    Commented Dec 5, 2009 at 21:09
  • $\begingroup$ @Scott/Ben: Speaking for myself, I voted up Jose's answer precisely because as written it would be confusing. That and because it's always lots of fun to sneak in some general nonsense when teaching a general audience. (Agree that it's off-topic in general, though.) $\endgroup$
    – Harrison Brown
    Commented Dec 6, 2009 at 0:35

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It would help to know a little more about the nature of your friend's dyscalclia. The sense of the integer line comes from the inferior parietal cortex (and causes difficulty with problems like "what number is halfway between 7 and 11?") while rote memory-type problems ("what is 4 times 7?") are associated with the basal ganglia. It is possible to have trouble with one but not the other.

If your friend has difficulty with rote memory, you need to invoke their visual intuition (that is, rely heavily on a number line). Because the inferior parietal cortex is associative (that is, it's located in the brain at a convergence of senses) I would recommend also including their kinesthetic sense by making a physical number line and having them practice by walking along it to do number problems (I use playing cards, black for positive and red for negative).

If your friend instead has difficulty with mathematical intuition, I would present a systemized set of rules that can be incorporated into rote memory. I wouldn't stop there though, if they can master the rules that way -- I would then go back to associating those rules with mathematical intutition. It is possible to teach someone mathematical intuition who is normally impaired this way; this is a technique using neuroplasticity which involves "rewiring" around problematic parts in the brain.

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    $\begingroup$ Wow! I think this is the first time that my wife (a neuroscientist) has understood anything on MO. $\endgroup$
    – Alicia Garcia-Raboso
    Commented Dec 5, 2009 at 17:07
  • $\begingroup$ She does not know which sort of dyscalclia she has. She does not have multiplications memorised, so I'd guess that basal ganglia is where there are problems, at least. She has problems remembering which hand is the right and the left one and, when given a fraction, is denominator above or below the line, and similar issues. (So do I, for that matter.) Does this help with the classification? $\endgroup$
    – Tommi
    Commented Dec 5, 2009 at 18:07
  • $\begingroup$ Right hand / left hand issues are in the inferior parietal cortex. Math Overflow is not the appropriate place to continue this, although she is free to contact me (follow my name to my website). $\endgroup$
    – Jason Dyer
    Commented Dec 5, 2009 at 18:21
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You can think of them as numbers with directions. Draw a line with the integers on it, adding a positive number $b$ to a given one $a$ is counting $b$ positions to the right starting from $a$ (no matter if it is positive or negative), if $b$ is negative, you count towards the left.

This point of view will also help when dealing with multiplication of negative numbers, thinking as -1 as "turning around" makes easy to understand $(-1)\times (-1) = 1 $. In my experience the interpretation of negative numbers as "debts" never works well.

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  • $\begingroup$ Actually, there is a good debt model that works. If you are interested, please see my comment below. $\endgroup$
    – Anna Varvak
    Commented Dec 5, 2009 at 14:44
  • $\begingroup$ From a pedagogical point of view, I think javier´s explanation is the best one one can find. $\endgroup$
    – M.G.
    Commented Dec 5, 2009 at 18:29
  • $\begingroup$ I agree that this is the best explanation. Arnol'd also gives a good exposition of this in his memoirs "Yesterday and Long Ago". $\endgroup$
    – Kevin H. Lin
    Commented Dec 5, 2009 at 19:38
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You could think of integers as pairs of natural numbers $(a,b)$ modulo the equivalence $$(a,b) \sim (c,d)\quad \mathrm{if}\quad a+d = b+c.$$ (In other words, you think of $(a,b)$ secretely as $a-b$.) Addition is just entrywise addition of natural numbers: $$(a,b)+(c,d) = (a+c,b+d).$$ Negative numbers are numbers of the form $(0,a)$. Now the equation $-7 -3 = -10$ is $$(0,7) + (0,3) = (0,10).$$ In summary, you can restrict yourself to adding natural numbers, provided that you consider pairs.

I'm not sure this helps in your friend's case, though.

Edit

Negative numbers do not arise in nature, of course. It is the result of comparison. The above set up with pairs is precisely that. $(a,b)$ is positive if $a>b$, negative if $a<b$ and zero if $a=b$. This could perhaps be woven into some sort of narrative for your friend.

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    $\begingroup$ I'm not sure if this answer was serious, but I'm giving a +1 for boldness. $\endgroup$
    – Harry Gindi
    Commented Dec 5, 2009 at 13:27
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    $\begingroup$ I just realized that the chips model that I provided in my comment is a physical model of exactly what Jose describes: operations with a pair of numbers. It also happens to be the classic accounting model of debits and credits, so it is all kinds of useful. $\endgroup$
    – Anna Varvak
    Commented Dec 5, 2009 at 14:46
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    $\begingroup$ It's actually extremely useful. It's Grothendieck's construction of the K-group of a commutative monoid, which forms the basis of K-theory :) $\endgroup$
    – José Figueroa-O'Farrill
    Commented Dec 5, 2009 at 15:37
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    $\begingroup$ @Jose, whence came the +1 for boldness. I just don't know how much this would help someone who doesn't already understand negative numbers. $\endgroup$
    – Harry Gindi
    Commented Dec 5, 2009 at 20:21
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    $\begingroup$ Sorry, by nature I mean Physics. One never really measures negative numbers. Sure, apparata often read negative numbers, but that is only because internally they are doing a comparison between two positive quantities. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Dec 5, 2009 at 20:24
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There are some nice physical models of integers that I learned about when I was working with elementary school teachers. One that I find particularly useful is the following:

You have two kinds of chips (say, red and black, to correspond with the standard accounting practice). The rule is that a red chip cancels the black chip. If after that, if you have k black chips, that represents the positive integer k; if you have k red chips, that represents the negative integer -k, and if you got nothing, that's 0.

The first thing that a student needs to figure out is that there are a lot of ways to represent any integer, cause you can always add a pair of red and black chips. This eventually gets the point across that adding $(k-k)$ doesn't change the value.

Addition of two integers is easy: put both piles of chips together and figure out what integer you got.

Subtraction is not much more difficult: it's a take-away operation, analogous to what people are familiar with in counting numbers. To do $a-b$ as a take-away operation, but you may need to modify $a$ by adding more pairs of chips to have enough to take away $b$ from it.

For example: 2-5 2 is represented five black and three red chips. Take away five black chips that represent 5, and you are left with three red chips which represent -3.

The other advantage of this model is that it's easy to demonstrate that $a-b=a+(-b)$. For example, consider 2-5 as 2+(-5): two black chips, put together with five red ones, which after cancellation yield three red chips representing -3.

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Although anything to do with finance tends to turn me off, for adding negative numbers I find debts to be the easiest model to think about. E.g., if I owe you \$a and I owe \$b to someone else, how much money do I owe altogether? And if I then get \$c, what's my total amount of debt?

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  • $\begingroup$ This answer feels like it's begging the question. $\endgroup$
    – Harry Gindi
    Commented Dec 5, 2009 at 20:43
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    $\begingroup$ Well, that depends. For some things an abstract issue is completely clarified by thinking about a well-chosen concrete example, for which the answer appears obvious. For me, the answer appears obvious in the case of this example, and abstracting from this example is one way to understand the general situation. For many other people this example may be no clearer than the general question, in which case you're absolutely right. $\endgroup$
    – Mark Meckes
    Commented Dec 7, 2009 at 16:32

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