# Half-your-age-plus-seven rule

Looking for a great application of systems of linear inequalities for your Algebra 1 or 2 class? Look no further than today’s GraphJam contribution:

You might just give this picture to students and ask THEM to come up with the equations of the three lines.

There’s also a nice discussion to be had here about inverse functions, or about intersecting lines. And there might also be a good discussion about the domain of reasonableness.

Here are the three functions:

$f_{\text{blue}}(x)=x$

$f_{\text{red}}(x)=\frac{1}{2}x+7$

$f_{\text{black}}(x)=2x-14$

This is especially interesting because I never think of the rule as putting boundaries on a person’s dating age range. Usually people talk about it in the context of “how old of a person can I date?” not “how young of a person can I date?” Or rather, if you’re asking the second question, it’s usually phrased “how young of a person can date me?” (All of these questions relate to functions and their inverses!) But in fact, the half-your-age-plus-seven rule puts a lower and and upper bound on the ages of those you can date.

As far as reasonableness, is it fair to say that my daughter who is 1 can date someone who is between the age of -12 and 7.5? I don’t think so! I’m definitely going to be chasing off those -12 year-olds, I can already tell :-).

For my daughter, the domain of reasonableness might be $x\geq 18$!

# Pictures with equations

Check out this awesome blog post by Richard Clark on the Alpha Blog.

Follow the link to see lots of great pictures made with equations. These pictures are so complicated it makes you wonder, is there any picture we can’t make with equations? My first answer is NO.

Think about vector-based graphics. Vector graphics, for those who’ve never heard the term, are pictures/graphics that are stored as a set of instructions for redrawing the picture rather than as a large array of pixels. You’ve used vector graphics if you have ever used clip-art or used the drawing tools in Microsoft Office, or if you’ve ever used Adobe Illustrator, or Inkscape. The advantages of vector graphics include very small files and infinite loss-less resizeability. How can vector graphics achieve this? Well, like I said, vector graphics are stored as rules not pixels. And by rules, we could just as easily say equations.

So the answer is certainly YES we can make any picture using equations. I think the harder question is can we make any picture using ONE equation? Or one set of parametric equations? Or one implicit equation?

What constraints do we want to impose? Do fractals/iterative/recursive rules count?

I am curious to find out how the creators of these picture-equations came up with them. It seems infeasible to do this by trial and error, given the massive size of these equations.

Oh, and if you haven’t yet seen the Batman Curve, you better go check that out too.

# “Japanese” Multiplication

My brother sent me a link to this video that teaches “Japanese” multiplication (thanks Tim!):

I learned about this technique in my History of Math class, and Vi Hart talked about it in a video back in 2011:

She does a nice job showing why there’s nothing particularly special about this Japanese “visual” multiplication. Here are a few reasons why it’s not better, as far as I’m concerned:

1. It’s not faster (sometimes it is, but most of the time not). As Vi points out, counting the number of dots in a rectangle by hand is ridiculous.
2. It’s painful when the numbers are bigger than 1, 2, or 3 and when there are more than 2 digits in the numbers (just try multiplying 976 x 8937 for example).
3. Zeros make things difficult (use dashed lines?)
4. Carrying is still required.
5. It’s perhaps more error prone, since it relies on your counting all the intersections.

In the end, to multiply two numbers you still have to multiply all their digits by each other and deal with carries, no matter which method you choose. I think it’s still worth teaching various methods of multiplication to students in an effort to make the abstract more concrete.

# Why does x represent the unknown?

I think I’ll show this to my Algebra 2 class this week.

[HT: Fred Connington]

# Rationalization Rant

Every high school math student has been taught how to rationalize the denominator. We tell students not to give an answer like

$\frac{1}{\sqrt{2}}$

because it isn’t fully “simplified.” Rather, they should report it as

$\frac{\sqrt{2}}{2}.$

This is fair, even though the second answer isn’t much simpler than the first. What does it really mean to simplify an expression? It’s a pretty nebulous instruction.

We also don’t consider

$\frac{12}{1+\sqrt{5}}$

to be rationalized because of the square root in the denominator, so we multiply by the conjugate to obtain

$2-2\sqrt{5}.$

In this particular example, multiplying by the conjugate was really fruitful and the resulting expression does indeed seem much more desirable than the original expression.

But here’s where it gets a little ridiculous. Our Algebra 2 book also calls for students to rationalize the denominator when (1) a higher root is present and (2) roots containing variables are present. Let me show you an example of each situation, and explain why this is going a little too far.

## Rationalizing higher roots

First, when a higher root is present like

$\sqrt[5]{\frac{15}{2}},$

the book would have students multiply the top and bottom of the fraction inside the radical by $2^4$ so as to make a perfect fifth root in the denominator. The final answer would be

$\frac{\sqrt[5]{240}}{2}.$

Simpler? You decide.

This becomes especially problematic when we encounter sums involving higher roots. It’s certainly possible, using various tricks, to rationalize the denominator in expressions like this:

$\frac{1}{2-\sqrt[3]{5}}.$

But is that really desirable? The result here is

$\frac{1}{2-\sqrt[3]{5}}\cdot\frac{4+2\sqrt[3]{5}+\sqrt[3]{25}}{4+2\sqrt[3]{5}+\sqrt[3]{25}}=\frac{4+2\sqrt[3]{5}+\sqrt[3]{25}}{3},$

which is, arguably, more complex than the original expression. Can anyone think of a good reason to do this, except just for fun?

## Rationalizing variable expressions

Now, let’s think about variable expressions. Here is a problem, directly from our Algebra 2 book (note the directions as well):

Write the expression in simplest form. Assume all variables are positive.

$\sqrt[3]{\frac{x}{y^7}}$

The method that leads to the “correct” solution is to multiply the fraction under the radical by $\frac{y^2}{y^2}$, and to finally write

$\frac{\sqrt[3]{xy^2}}{y^3}.$

This is problematic for two reasons. (1) This isn’t really simpler than the original expression and (2) this expression isn’t even guaranteed to have a denominator that’s rational! (Suppose $y=\sqrt{2}$ or even $y=\pi$.) Once again I ask, can anyone think of a good reason to do this, except just for fun??

## So how far do we take this?

Is it reasonable to ask someone to rationalize this denominator?

$\frac{1}{2\sqrt{2}-\sqrt{2}\sqrt[3]{5}+2\sqrt{5}-5^{5/6}}$

You can rationalize the denominator, but I’ll leave that as an exercise for the reader. So how far do we take this? I had to craft the above expression very carefully so that it works out well, but in general, most expressions have denominators that can’t be rationalized (and I do mean “most expressions” in the technical, mathematical way–there are are an uncountable number of denominators of the unrationalizable type). All that being said, I think this would make a great t-shirt:

And I rest my case.

# Good discussions in the math blog world

Here are two blog posts I saw a few weeks ago. I’ve been following the comments with great interest, and the conversations have been fruitful. You should go check them out and join the conversation!

• Critical Thinking @ dy/dan — Once again, Dan gives deserved criticism to a contrived textbook problem. Hilarious problem, and fun discussion in the comments.
• Disagreement on operator precedence for 2^3^4 @ Walking Randomly — The title says it all, but it’s the first time I had ever thought about how 2^3^4 or expressions with carets should be evaluated. Note that it’s clear how $2^{3^4}$ should be evaluated. We’re just unclear on how 2^3^4 should be evaluated.