How Notation Obscures Patterns

This is another stab for me at what continues to prove to be a complicated topic: How our mess of mathematical notation obfuscates key patterns. This is also a rough draft of thoughts, not meant to be a polished product.

In Algebra II recently, the unit is radical notation and rational exponents. That means that logarithms are around the corner. This is the kernel of where our traditional notation goes from somewhat annoying to utterly slapdash. The underlying relationships are admittedly difficult, but the notation makes them so much worse.

But instead, I’ll keep it simpler. Today in class I reviewed how to add, multiply, and divide fractions, so I’ll reflect on my thoughts there.

First, let’s multiply fractions. Groovy. \(\frac{4}{5} \cdot \frac{3}{7}=\frac{12}{35}\). Multiply across the top, multiply across the bottom. Simple enough rule, but… why? Why does it work?

Set that aside. Let’s divide fractions. \(\frac{4}{5} \div \frac{3}{7}=\frac{4}{5} \cdot \frac{7}{3}=\frac{28}{15}\). In high school, we say, “Dividing by a fraction is equivalent to multiplying by its reciprocal.” In middle school, that’s given as “Keep-Change-Flip” (okay, stop screaming, Nix The Tricksers).

It is less obvious why this works.

Next, let’s add some fractions. \(\frac{4}{5} + \frac{3}{7}=\frac{28}{35}+\frac{15}{35}=\frac{43}{35}\). It’s not at all clear why this works. It’s a confusing bunch of steps that can be condensed to the dreaded “Butterfly Method” (or “Bowtie Method”).

Next up: Converting from mixed numbers to improper fractions. \(6\frac{3}{7}=\frac{6}{1}+\frac{3}{7}=\frac{42+3}{7}=\frac{45}{7}\). This has a trick, too: Multiply the integer by the denominator, add the product to the numerator, then put it over the denominator. But if we see a mixed number as a sum of an integer and a fraction, this is just another example of adding fractions (with unlike denominators).

Now… imagine our notation were different. A small tweak for now: Allow / to be used like -, that is, as a unary operator. /4 is 1/4, just as -4 is 0-4.

No more denominators. I’ll do the multiplication we just did, but no fractions and no binary division.

\((4 \times /5) \times (3 \times /7) = 4 \times /5 \times 3 \times /7\). Compare this to \((4 + -5) + (3 + -7) = 4 + -5 + 3 + -7\). Here’s a key feature: When dealing with real numbers, order doesn’t matter within addition, and it doesn’t matter within multiplication. We only need parentheses when subtraction or division are involved.

So \(4 + -5 + 3 + -7 = 4 + 3 + -5 + -7 = (4 + 3) + -(5 + 7)\) and \(4 \times /5 \times 3 \times /7 = 4 \times 3 \times /5 \times /7 = (4 \times 3) \times /(5 \times 7)\). Entirely parallel. No need for a special rule on how to multiply fractions: If you know how to deal with \((4 – 5) + (3 – 7)\), you know how to deal with \((4 \div 5) \times (3 \div 7)\).

Because I do think it’s difficult to read this new notation (I have suggestions for that, but that’s a separate item), I’ll tweak it to make it closer to what we’re used to: I’ll use \(\div\) instead of \(\times /\) for division (as I did at the end of the last paragraph). The parallel pattern persists: \(4 – 5 + 3 – 7 = 4 + 3 – 5 – 7 = (4 + 3) – (5 + 7)\) and \(4 \div 5 \times 3 \div 7 = 4 \times 3 \div 5 \div 7 = (4 \times 3) \div (5 \times 7)\).

Consider dividing fractions: \((4 \div 5) \div (3 \div 7)\). Its parallel within addition is: \((4 – 5) – (3 – 7) = 4 – 5 – 3 + 7\). When we get rid of the parentheses, we “distribute” the negative sign. What we’re really doing is distributing a coefficient of -1, but … notation!

So what happens when we distribute the division sign? If negating a negative results in a positive, what does dividing by a division do? In standard mathematics terms, that doesn’t even make any sense, but let’s just pretend. If \(-(-x) = x\), then perhaps \(/(/x) = x\) as well?

Indeed, the multiplicative inverse of the multiplicative inverse of a number is the original number, just as the additive inverse of the additive inverse of a number is the original number.

So… hold on, this is where things got weird for me when I noticed them. But it works! \((4 \div 5) \div (3 \div 7) = 4 \div 5 \div 3 \times 7\). And since each division sign only applies to its neighbor to the right (that is, because this is shorthand for \(4 \times /5 \times /3 \times 7\)), we can rearrange these items as we please. This is why \((4 \div 5) \div (3 \div 7) = (4 \times 7) \div (5 \times 3)\).

No need for “multiply by the reciprocal”: If someone can understand why \((4 – 5) – (3 – 7) = (4 + 7) – (5 + 3)\), they can understand why \((4 \div 5) \div (3 \div 7) = (4 \times 7) \div (5 \times 3)\). Completely parallel structures.

Next up is the monster: Adding fractions with unlike denominators. Note that I don’t mention subtraction. Subtraction is fairly trivial. We could get rid of subtraction today (from high school mathematics) without otherwise changing our notation just by always writing negatives instead.

Rewrite \(\frac{4}{5} + \frac{3}{7}\) to \((4 \div 5) + (3 \div 7)\) and compare it to \((4 \times 5) + (3 \times 7)\). What are we doing when we “add” fractions? We’re creating a single expression involving division and two numbers.

Step back one more pace to \((4 \times 5) + (3 \times 5) = (4 + 3) \times 5\). This works because \(4 \times 5\) and \(3 \times 5\) have a common factor. We can’t do anything similar with \((4 \times 5) + (3 \times 7)\) because there’s no common factor. If we wanted to write something like \((a + b) \times c\) for this case, we would need to introduce a common factor.

I don’t think I’ve ever seen anyone deliberately introduce a common factor for that purpose, but maybe I’ve just missed it. Regardless, though, that’s what we do when adding fractions: We effectively introduce a common factor (of sorts). Here’s the whole path: \((4 \div 5) + (3 \div 7) = \) \((4 \div 5 \div 7 \times 7) + (3 \div 7 \div 5 \times 5) = \) \((4 \times 7 \div (5 \times 7)) + (3 \times 5 \div (7 \times 5)) = \) \((28 \div 35) + (15 \div 35) = \) \((28 + 15) \div 35 = 43 \div 35\).

At first blush, this might look like more work than the traditional method, but if so, that might be more a factor of its alienness. I feel like the underlying mathematics is more consistent and logical and, more to the point, that this allows us to see ways in which division works like multiplication, as well as how the multiplication/division layer work like the addition/subtraction layer.

A more radical approach would be to get rid of division entirely. Just as we can do all of our subtraction using negation (e.g., \(5 – 3 = 5 + -3)\), we can do all of our division using reciprocals (e.g., \(5/3 = 5 \times /3\). Just as \(-(-4) = 4\), \(/(/4) = 4\). Just as we can distribute negation across a polynomial (e.g. \(-(4 + 3) = -4 + -3\)), we can distribute reciprocals: \(/(4 \times 3) = /4 \times /3\). The common symbols even lend themselves to this: Just as a + sign consists of two overlapping – signs, a \(\times\) sign consists of two overlapping / signs.

I’m not sure where I sit on such a radical approach. I personally would be eager for it, but there’s a serious argument to be made about not just throwing current notation out entirely. Plus, as a teacher, I’m fully aware of how entrenched many of my colleagues are, and how my eagerness would be generally met by their lack of it.

My bigger point here, though, is that until I started digging into alternative notation, I didn’t see these parallels myself. I knew that multiplication generally acted like addition, but I didn’t notice how completely parallel the processes were.

And throwing in the exponential layer? I’ve started to dig here, but I’m still a long way off from something I find satisfying. I do know, though, that our current notation is a total mess. And messy notation isn’t conducive for clear understanding.

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