# Decimal Approximations of Logarithms (Part 5)

While some common (i.e., base-10) logarithms work out evenly, like $\log_{10} 10,000$, most do not. Here is the typical output when a scientific calculator computes a logarithm:

In today’s post, I’ll summarize the past few posts to describe how talented Algebra II students, who have just been introduced to logarithms, can develop proficiency with the Laws of Logarithms while also understanding that the above answer is not just a meaningless jumble of digits. The only tools students will need are

To estimate $\log_{10} 5.1264$, Algebra II students can try to find a power of 5.1264 that is close to a power of 10. In principle, this can be done by just multiplying by $5.1264$ until an answer decently close to $5.1264$ arises. For the teacher who’s guiding students through this exploration, it might be helpful to know the answer ahead of time.

One way to do this is to use Wolfram Alpha to find the convergents of $\log_{10} 5.1264$. If you click this link, you’ll see that I entered

Convergents[Log[10,5.1264],15]

A little explanation is in order:

• Convergents, predictably, is the Mathematica command for finding the convergents of a given number.
• Log[10,5.1264] is the base-10 logarithm of 5.1264. By contrast, Log[5.1264] is the natural logarithm of 5.1264. Mathematica employs the convention of that $\log$ should be used for natural logarithms instead of $\ln$, as base-10 logarithms are next to useless for mathematical researchers. That said, I freely concede that this convention is confusing to students who grew up thinking that $\log$ should be used for base-10 logarithms and $\ln$ for natural logarithms. (See also my standard joke about using natural logarithms.) Naturally, the $5.1264$ can be changed for other logarithms.
• The 15 means that I want Wolfram Alpha to give me the first 15 convergents of $\log_{10} 5.1264$. In most cases, that’s enough terms to provide a convergent whose denominator is at least six digits long. In the rare instance when this doesn’t happen, a number larger than 15 can be entered.

From Wolfram Alpha, I see that $\displaystyle \frac{22}{31}$ is the last convergent with a numerator less than 100. For the purposes of this exploration, I interpret these fractions as follows:

• The best suitable power of $5.1264$ for an easy approximation on a scientific calculation will be $(5.1264)^{31}$. In this context, “best” means something that’s close to a power of 10 but less than $10^{100}$. Students entering $(5.1264)^{31}$ into a calculator will find

$(5.1264)^{31} \approx 1.009687994 \times 10^{22}$

$(5.1264)^{31} \approx 10^{22}$

In other words, the denominator of the convergent $\displaystyle \frac{22}{31}$ gives the exponent for $5.1264$, while the numerator gives the exponent for the approximated power of 10. Continuing with the Laws of Logarithms,

$\log_{10} (5.1264)^{31} \approx \log_{10} 10^{22}$

$31 \log_{10} 5.1264 \approx 22$

$\log_{10} 5.1264 \approx \displaystyle \frac{22}{31}$

$\log_{10} 5.1264 \approx 0.709677\dots$

A quick check with a calculator shows that this approximation is accurate to three decimal places. This alone should convince many students that the above apparently random jumble of digits is not so random after all.

While the above discussion should be enough for many students, some students may want to know how to find the rest of the decimal places with this technique. To answer this question, we again turn to the convergents of $\log_{10} 5.1264$ from Wolfram Alpha. From this list, we see that $\displaystyle \frac{89,337}{125,860}$ is the first convergent with a denominator at least six digits long. The student therefore has two options:

Option #1. Ask the student to use Wolfram Alpha to raise $5.1264$ to the denominator of this convergent. Surprisingly to the student, but not surprisingly to the teacher who knows about this convergent, the answer is very close to a power of 10: $10^{89,337}$. The student can then use the Laws of Logarithms as before:

$\log_{10} (5.1264)^{125,860} \approx \log_{10} 10^{89,337}$

$125,860 \log_{10} 5.1264 \approx 89,337$

$\log_{10} 5.1264 \approx \displaystyle \frac{89,337}{125,860}$

$\log_{10} 5.1264 \approx 0.70981249006\dots$,

which matches the output of the calculator.

Option #2. Ask the student to “trick” a hand-held calculator into finding $(5.1264)^{125,860}$. This option requires the use of the convergent with the largest numerator less than 100, which was $\displaystyle \frac{22}{31}$.

• Option #2A: Use the Microsoft Excel spreadsheet that I’ve written to perform the calculations that follow.
• Option #2B: The student divides the smaller denominators into the larger denominator and finds the quotient and remainder. It turns out that $125,860 = 31 \times 4060 + 0$. (This is a rare case where there happens to be no remainder.) Next, the student uses a hand-held calculate to compute

$\displaystyle \left( \frac{(5.1264)^{31}}{10^{22}} \right)^{4060} \times (5.1264)^0$

In this example, the $\times (5.1264)^0$ is of course superfluous, but I include it here to show where the remainder should be placed. Entering this in a calculator yields a result that is close to $10^{17}$. (The teacher should be aware that some of the last few digits may differ from the more precise result given by Wolfram Alpha due to round-off error, but this discrepancy won’t matter for the purposes of the student’s explorations.) In other words,

$\displaystyle \left( \frac{(5.1264)^{31}}{10^{22}} \right)^{4060} \times (5.1264)^0 \approx 10^{17}$,

which may be rearranged as

$(5.1264)^{125,860} \approx 10^{89,337}$

after using the Laws of Exponents. From this point, the derivation follows the steps in Option #1.

# Decimal Approximations of Logarithms (Part 4)

While some common (i.e., base-10) logarithms work out evenly, like $\log_{10} 10,000$, most do not. Here is the typical output when a scientific calculator computes a logarithm:

To a student first learning logarithms, the answer is just an apparently random jumble of digits; indeed, it can proven that the answer is irrational. With a little prompting, a teacher can get his/her students wondering about how people 50 years ago could have figured this out without a calculator. This leads to a natural pedagogical question:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Here’s a trial-and-error technique — an exploration activity — that is within the grasp of Algebra II students. It’s simple to understand; it’s just a lot of work. The only tools that are needed are

• The Laws of Logarithms
• A hand-held scientific calculator
• A lot of patience multiplying $x$ by itself repeatedly in a quest to find integer powers of $x$ that are close to powers of $10$.

In the previous post in this series, we found that

$3^{153} \approx 10^{73}$

and

$3^{323,641} \approx 10^{154,416}$.

Using the Laws of Logarithms on the latter provides an approximation of $\log_{10} 3$ that is accurate to an astounding ten decimal places:

$\log_{10} 3^{323,641} \approx \log_{10} 10^{154,416}$

$323,641 \log_{10} 3 \approx 154,416$

$\log_{10} 3 \approx \displaystyle \frac{154,416}{323,641} \approx 0.477121254723598\dots$.

Compare with:

$\log_{10} 3 \approx 0.47712125471966\dots$

Since hand-held calculators will generate identical outputs for these two expressions (up to the display capabilities of the calculator), this may lead to the misconception that the irrational number $\log_{10} 3$ is actually equal to the rational number $\displaystyle \frac{154,416}{323,641}$, so I’ll emphasize again that these two numbers are not equal but are instead really, really close to each other.

We now turn to a question that was deferred in the previous post.

Student: How did you know to raise 3 to the 323,641st power?

Teacher: I just multiplied 3 by itself a few hundred thousand times.

Student: C’mon, really. How did you know?

While I don’t doubt that some of our ancestors used this technique to find logarithms — at least before the discovery of calculus — today’s students are not going to be that patient. Instead, to find suitable powers quickly, we will use ideas from the mathematical theory of continued fractions: see Wikipedia, Mathworld, or this excellent self-contained book for more details.

To approximate $\log_{10} x$, the technique outlined in this series suggests finding integers $m$ and $n$ so that

$x^n \approx 10^m$,

or, equivalently,

$\log_{10} x^n \approx \log_{10} 10^m$

$n \log_{10} x \approx m$

$\log_{10} x \approx \displaystyle \frac{m}{n}$.

In other words, we’re looking for rational numbers that are reasonable close to $\log_{10} x$. Terrific candidates for such rational numbers are the convergents to the continued fraction expansion of $\log_{10} x$. I’ll defer to the references above for how these convergents can be computed, so let me cut to the chase. One way these can be quickly obtained is the free website Wolfram Alpha. For example, the first few convergents of $\log_{10} 3$ are

$\displaystyle \frac{1}{2}, \frac{10}{21}, \frac{21}{44}, \frac{52}{109}, and \frac{73}{153}$.

A larger convergent is $\frac{154,416}{323,641}$, our familiar friend from the previous post in this series.

As more terms are taken, these convergents get closer and closer to $\log_{10} 3$. In fact:

• Each convergent is the best possible rational approximation to $\log_{10} 3$ using a denominator that’s less than the denominator of the next convergent. For example, the second convergent $\displaystyle \frac{10}{21}$ is the closest rational number to $\log_{10} 3$ that has a denominator less than $44$, the denominator of the third convergent.
• The convergents alternate between slightly greater than $\log_{10} 3$ and slightly less than $\log_{10} 3$.
• Each convergent $\displaystyle \frac{m}{n}$ is guaranteed to be within $\displaystyle \frac{1}{n^2}$ of $\log_{10} 3$. (In fact, if $\displaystyle \frac{m}{n}$ and $\displaystyle \frac{p}{q}$ are consecutive convergents, then $\displaystyle \frac{m}{n}$ is guaranteed to be within $\displaystyle \frac{1}{nq}$ of $\log_{10} 3$.)
• As a practical upshot of the previous point: if the denominator of the convergent $\displaystyle \frac{m}{n}$ is at least six digits long (that is, greater than $10^5$), then $\displaystyle \frac{m}{n}$ must be within $\displaystyle \frac{1}{(10^5)^2} = 10^{-10}$ of $\log_{10} 3$… and it’ll probably be significantly closer than that.

So convergents provide a way for teachers to maintain the illusion that they found a power like $3^{323,641}$ by laborious calculation, when in fact they were quickly found through modern computing.

# Decimal Approximations of Logarithms (Part 3)

While some common (i.e., base-10) logarithms work out evenly, like $\log_{10} 10,000$, most do not. Here is the typical output when a scientific calculator computes a logarithm:

To a student first learning logarithms, the answer is just an apparently random jumble of digits; indeed, it can proven that the answer is irrational. With a little prompting, a teacher can get his/her students wondering about how people 50 years ago could have figured this out without a calculator. This leads to a natural pedagogical question:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Here’s a trial-and-error technique — an exploration activity — that is within the grasp of Algebra II students. It’s simple to understand; it’s just a lot of work.

To approximate $\log_{10} x$, look for integer powers of $x$ that are close to powers of 10.

In the previous post in this series, we essentially used trial and error to find such powers of 3. We found

$3^{153} \approx 9.989689095 \times 10^{72} \approx 10^{73}$,

from which we can conclude

$\log_{10} 3^{153} \approx \log_{10} 10^{73}$

$153 \log_{10} 3 \approx 73$

$\log_{10} 3 \approx \displaystyle \frac{73}{153} \approx 0.477124$.

This approximation is accurate to five decimal places.

By now, I’d imagine that our student would be convinced that logarithms aren’t just a random jumble of digits… there’s a process (albeit a complicated process) for obtaining these decimal expansions. Of course, this process isn’t the best process, but it works and it only uses techniques at the level of an Algebra II student who’s learning about logarithms for the first time.

If nothing else, hopefully this lesson will give students a little more appreciation for their ancestors who had to perform these kinds of calculations without the benefit of modern computing.

We also saw in the previous post that larger powers can result in better and better approximation. Finding suitable powers gets harder and harder as the exponent gets larger. However, when a better approximation is found, the improvement can be dramatic. Indeed, the decimal expansion of a logarithm can be obtained up to the accuracy of a hand-held calculator with a little patience. For example, let’s compute

$3^{323,641}$

Predictably, the complaint will arise: “How did you know to try $323,641$?” The flippant and awe-inspiring answer is, “I just kept multiplying by 3.”

I’ll give the real answer that question later in this series.

Postponing the answer to that question for now, there are a couple ways for students to compute this using readily available technology. Perhaps the most user-friendly is the free resource Wolfram Alpha:

$3^{323,641} \approx 9.999970671 \times 10^{154,415} \approx 10^{154,416}$.

That said, students can also perform this computation by creatively using their handheld calculators. Most calculators will return an overflow error if a direct computation of $3^{323,641}$ is attempted; the number is simply too big. A way around this is by using the above approximation $3^{153} \approx 10^{73}$, so that $3^{153}/10^{73} \approx 1$. Therefore, we can take large powers of $3^{153}/10^{73}$ without worrying about an overflow error.

In particular, let’s divide $323,641$ by $153$. A little work shows that

$\displaystyle \frac{323,641}{153} = \displaystyle 2115 \frac{46}{153}$,

or

$323,641 = 153 \times 2115 + 46$.

This suggests that we try to compute

$\displaystyle \left( \frac{3^{153}}{10^{73}} \right)^{2115} \times 3^{46}$,

and a hand-held calculator can be used to show that this expression is approximately equal to $10^{21}$. Some of the last few digits will be incorrect because of unavoidable round-off errors, but the approximation of $10^{21}$ — all that’s needed for the present exercise — will still be evident.

By the Laws of Exponents, we see that

$\displaystyle \left( \frac{3^{153}}{10^{73}} \right)^{2115} \times 3^{46} \approx 10^{21}$

$\displaystyle \frac{3^{153 \times 2115 + 46}}{10^{73 \times 2115}} \approx 10^{21}$

$\displaystyle \frac{3^{323,641}}{10^{154,395}} \approx 10^{21}$

$3^{323,641} \approx 10^{154,395} \times 10^{21}$

$3^{323,641} \approx 10^{154,395+21}$

$3^{323,641} \approx 10^{154,416}$.

Whichever technique is used, we can now use the Laws of Logarithms to approximate $\log_{10} 3$:

$\log_{10} 3^{323,641} \approx \log_{10} 10^{154,416}$

$323,641 \log_{10} 3 \approx 154,416$

$\log_{10} 3 \approx \displaystyle \frac{154,416}{323,641} \approx 0.477121254723598\dots$.

This approximation matches the decimal expansion of $\log_{10} 3$  to an astounding ten decimal places:

$\log_{10} 3 \approx 0.47712125471966\dots$

Since hand-held calculators will generate identical outputs for these two expressions (up to the display capabilities of the calculator), this may lead to the misconception that the irrational number $\log_{10} 3$ is actually equal to the rational number $\displaystyle \frac{154,416}{323,641}$, so I’ll emphasize again that these two numbers are not equal but are instead really, really close to each other.

Summarizing, Algebra II students can find the decimal expansion of $\log_{10} x$ can be found up to the accuracy of a hand-held scientific calculator. The only tools that are needed are

• The Laws of Logarithms
• A hand-held scientific calculator
• A lot of patience multiply $x$ by itself repeatedly in a quest to find integer powers of $x$ that are close to powers of $10$.

While I don’t have a specific reference, I’d be stunned if none of our ancestors tried something along these lines in the years between the discovery of logarithms (1614) and calculus (1666 or 1684).

# Decimal Approximations of Logarithms (Part 2)

While some common (i.e., base-10) logarithms work out evenly, like $\log_{10} 10,000$, most do not. Here is the typical output when a scientific calculator computes a logarithm:

To a student first learning logarithms, the answer is just an apparently random jumble of digits; indeed, it can proven that the answer is irrational. With a little prompting, a teacher can get his/her students wondering about how people 50 years ago could have figured this out without a calculator. This leads to a natural pedagogical question:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Here’s a trial-and-error technique — an exploration activity — that is within the grasp of Algebra II students. It’s simple to understand; it’s just a lot of work. While I don’t have a specific reference, I’d be stunned if none of our ancestors tried something along these lines in the years between the discovery of logarithms (1614) and calculus (1666 or 1684).

To approximate $\log_{10} x$, look for integer powers of $x$ that are close to powers of 10.

I’ll illustrate this idea with $\log_{10} 3$.

$3^1 = 3$

$3^2 = 9$

Not bad… already, we’ve come across a power of 3 that’s decently close to a power of 10. We see that

$3^2 = 9 < 10^1$

and therefore

$\log_{10} 3^2 < 1$

$2 \log_{10} 3< 1$

$\log_{10} 3< \displaystyle \frac{1}{2} = 0.5$

Let’s keep going. We just keep multiplying by 3 until we find something close to a power of 10. In principle, these calculations could be done by hand, but Algebra II students can speed things up a bit by using their scientific calculators.

$3^3 = 27$

$3^4 = 81$

$3^5 = 243$

$3^6 = 729$

$3^7 = 2,187$

$3^8 = 6,561$

$3^9 = 19,683$

$3^{10} = 59,049$

$3^{11} = 177,147$

$3^{12} = 531,441$

$3^{13} = 1,594,323$

$3^{14} = 4,782,969$

$3^{15} = 14,348,907$

$3^{16} = 43,046,721$

$3^{17} = 129,140,163$

$3^{18} = 387,420,489$

$3^{19} = 1,162,261,467$

$3^{20} = 3,486,784,401$

$3^{21} = 10,460,353,203$

This looks pretty good too. (Students using a standard ten-digit scientific calculator, of course, won’t be able to see all 11 digits.) We see that

$3^{21} > 10^{10}$

and therefore

$\log_{10} 3^{21} > \log_{10} 10^{10}$

$21 \log_{10} 3 > 10$

$\log_{10} 3 > \displaystyle \frac{10}{21} = 0.476190\dots$

Summarizing our work so far, we have

$0.476190\dots < \log_{10} 3 < 0.5$.

We also note that this latest approximation actually gives the first two digits in the decimal expansion of $\log_{10} 3$.

To get a better approximation of $\log_{10} 3$, we keep going. I wouldn’t blame Algebra II students a bit if they use their scientific calculators for these computations — but, ideally, they should realize that these calculations could be done by hand by someone very persistent.

$3^{22} = 31,381,059,609$

$3^{23} = 94,143,178,827$

$3^{24} = 282,429,536,481$

$3^{25} = 847,288,609,443$

$3^{26} = 2,541,865,828,329$

$3^{27} = 7,625,597,484,987$

$3^{28} = 22,876,792,454,961$

$3^{29} = 68,630,377,364,883$

$3^{30} = 205,891,132,094,649$

$3^{31} = 617,673,396,283,947$

$3^{32} = 1,853,020,188,851,841$

$3^{33} = 5,559,060,566,555,523$

$3^{34} = 16,677,181,699,666,569$

$3^{35} = 50,031,545,098,999,707$

$3^{36} = 150,094,635,296,999,121$

$3^{37} = 450,283,905,890,997,363$

$3^{38} = 1,350,851,717,672,992,089$

$3^{39} = 4,052,555,153,018,976,267$

$3^{40} = 12,157,665,459,056,928,801$

$3^{41} = 36,472,996,377,170,786,403$

$3^{42} = 109,418,989,131,512,359,209$

$3^{43} = 328,256,967,394,537,077,627$

$3^{44} = 984,770,902,183,611,232,881$

Using this last line, we obtain

$3^{44} < 10^{21}$

and therefore

$\log_{10} 3^{44} < \log_{10} 10^{21}$

$44 \log_{10} 3 < 21$

$\log_{10} 3 < \displaystyle \frac{21}{44} = 0.477273\dots$

Summarizing our work so far, we have

$0.476190\dots < \log_{10} 3 < 0.477273\dots$.

A quick check with a calculator shows that $\log_{10} 3 = 0.477121\dots$. In other words,

• This technique actually works!
• This last approximation of $0.477273\dots$ actually produced the first three decimal places of the correct answer!

With a little more work, the approximations

$3^{109} \approx 1.014417574 \times 10^{52} > 10^{52}$

$3^{153} \approx 9.989689095 \times 10^{72} < 10^{73}$

can be found, yielding the tighter inequalities

$\displaystyle \frac{52}{109} < \log_{10} 3 < \displaystyle \frac{73}{153}$,

or

$0.477064\dots < \log_{10} 3 < 0.477124$.

Now we’re really getting close… the last approximation is accurate to five decimal places.

# Decimal Approximations of Logarithms (Part 1)

My latest article on mathematics education, titled “Developing Intuition for Logarithms,” was published this month in the “My Favorite Lesson” section of the September 2018 issue of the journal Mathematics Teacher. This is a lesson that I taught for years to my Precalculus students, and I teach it currently to math majors who are aspiring high school teachers. Per copyright law, I can’t reproduce the article here, though the gist of the article appeared in an earlier blog post from five years ago.

Rather than repeat the article here, I thought I would write about some extra thoughts on developing intuition for logarithms that, due to space limitations, I was not able to include in the published article.

While some common (i.e., base-10) logarithms work out evenly, like $\log_{10} 10,000$, most do not. Here is the typical output when a scientific calculator computes a logarithm:

To a student first learning logarithms, the answer is just an apparently random jumble of digits; indeed, it can proven that the answer is irrational. With a little prompting, a teacher can get his/her students wondering about how people 50 years ago could have figured this out without a calculator. This leads to a natural pedagogical question:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Students who know calculus, of course, can do these computations since

$\log_{10} x = \displaystyle \frac{\ln x}{\ln 10}$,

$\ln (1+t) = t - \displaystyle \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \dots$,

a standard topic in second-semester calculus, can be used to calculate $\ln x$ for values of $x$ close to 1. However, a calculation using a power series is probably inaccessible to bright Algebra II students, no matter how precocious they are. (Besides, in real life, calculators don’t actually use Taylor series to perform these calculations; see the article CORDIC: How Hand Calculators Calculate, which appeared in College Mathematics Journal, for more details.)

In this series, I’ll discuss a technique that Algebra II students can use to find the decimal expansions of base-10 logarithms to surprisingly high precision using only tools that they’ve learned in Algebra II. This technique won’t be very efficient, but it should be completely accessible to students who are learning about base-10 logarithms for the first time. All that will be required are the Laws of Logarithms and a standard scientific calculator. A little bit of patience can yield the first few decimal places. And either a lot of patience, a teacher who knows how to use Wolfram Alpha appropriately, or a spreadsheet that I wrote can be used to obtain the decimal approximations of logarithms up to the digits displayed on a scientific calculator.

I’ll start this discussion in my next post.

# Engaging students: The area of a circle

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Deetria Bowser. Her topic, from Geometry: the area of a circle.

How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic? Note: It’s not enough to say “such-and-such is a great website”; you need to explain in some detail why it’s a great website.

An example of a helpful and engaging website for students is aaamath.com. On the left side of the webpage, there are a list of subjects. To find the Area of a circle lesson, select geometry and then area of a circle. The lesson is color coded with green being the “learn” part of the lesson, and blue being the “practice.”In the “learn” part of the lesson it explains briefly how to find the area of a circle. While I believe that and actually lesson should be taught before using this website, I think that the “learn” part provided by this lesson would be a great way to quickly review how to find the area of a circle. The next section (“practice”) gives a radius and the student is expected to calculate the area of the circle using said radius. I think this aspect of the lesson will help students gain speed and accuracy in computing the area of a circle. Although I do not think that this website can be used as a complete lesson on finding the area of a circle, on its own, I do believe that it could serve as a great review tool for students.

How could you as a teacher create an activity or project that involves your topic?

Hands on activities are easier to find for geometry topics, and finding the area of a circle is no exception. An example activity can be found in the YouTube video “Proof Without Words: The Circle.” In this video, the area of a circle is proved using beads and a ruler. The demonstrator creates a circle with silver beads, and shows that the radius of the circle can be measured using the ruler, and the circumference of the circle can be measured by unraveling the outermost part of the circle and measuring it (or by plugging the radius into the equation 2πr). The demonstrator then deconstructs the circle and traces the triangle created by it. From this he shows that $A=0.5bh = 0.5(2\pir)r = \pi r^2$. Instead of just using symbols to show this idea, I would create a guided explore activity where the students need to actually measure the radius and circumference of the circle they created as well at the base and height of the triangle created by deconstructing the circle they created. I would ask how the circumference and radius of the circle relate to the base and the height of the triangle. Once students recognize that the base of the triangle correlates with the circumference of the circle, and the radius correlates with the height, it will be easier to see why the area of a circle is calculated using the formula $A=\pi r^2$

What interesting (i.e., uncontrived) word problems using this topic can your students do now? (You may find resources such as http://www.spacemath.nasa.gov to be very helpful in this regard; feel free to suggest others.)

Practical uses for finding the area of a circle proved to be quite difficult. For example, most questions contain unrealistic examples such as “making a card with three semi-circles” (Glencoe). Although, many of these impractical exist, I found two example problems that could actually be used in the real world. The first example states “The Cole family owns an above-ground circular
swimming pool that has walls made of aluminum. Find the length of aluminum surrounding the pool as shown if the radius is 15 feet. Round to the nearest tenth” (Glencoe). This example is practical because when constructing a pool, one needs to know the surface area which can be found by using $\pi r^2$. The final example states “A rug is made up of a quadrant and two semicircles. Find the area of the rug. Use 3.14 for $\pi$and round to the nearest tenth!” (Glencoe). Although this seems less practical than the pool example, it is still related to real life because finding the area of a rug will help when deciding which rug to choose for a room.

References
M. (2012, May 29). Proof Without Words: The Circle. Retrieved October 06, 2017, from

(n.d.). Retrieved October 06, 2017, from http://www.aaamath.com/geo612x2.htm#pgtp
(n.d.). Retrieved October 06, 2017, from

# Engaging students: Finding the area of a right triangle

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Deanna Cravens. Her topic, from Geometry: finding the area of a right triangle.

How could you as a teacher create an activity or project that involves your topic?
One of the most common questions students ask when working with the area of a triangle is: “Why do I multiply by ½ in the formula?” It is a rather simple explanation for working with right triangles. Students could either do an explore activity where they discover the formula for the area of a right triangle, or a teacher could show this short two minute video in class.

So why do we multiply by ½? If we look at the formula ignoring the ½, you will see that it is the same formula for the area of a rectangle. Each angle in a rectangle forms 90 degrees and if we cut the rectangle along one of the diagonals, we will see that it creates a right triangle. Not only that, but it is exactly one half of the area of the rectangle since it was cut along the diagonal. Another way of showing this is doing the opposite by taking two congruent right triangles and rearranging them to create a rectangle. Either way shows how the ½ in the formula for the area of a right triangle appears and would be a great conceptual explore for students to complete.

How can this topic be used in your students’ future courses in mathematics or science?

Students are first introduced to finding the area of right triangle in their sixth grade mathematics class. One way that the topic is advanced in a high school geometry class is by throwing the Pythagorean Theorem into the mix. Students will know that formula for the area of a right triangle is A=½ bh. The way the topic is advanced is by giving the students the length of the hypotenuse and either the length of the base or the height, but not both. Students must use a^2 +b^2=c^2 in order to solve for the missing side length. The side lengths will not always be an integer, so students should be comfortable with working with square roots. Once students utilize the Pythagorean Theorem, they can then continue to solve for the area of the right triangle as they previously learned in sixth grade.

How has this topic appeared in pop culture (movies, TV, current music, video games, etc.)?

In this short music video young students at Builth Wells High School did a parody of Meghan Trainor’s “All About that Bass.” They take the chorus and put the lyrics in “multiply the base, by the height, then half it.” This music video can help several different types of learners in the classroom. Some need a visual aid which is done by specific dance movements by the students in the video. Others will remember it by having the catchy chorus stuck in their head. The parody lyrics are also put on the video to help students who might struggle with English, such as ELL students. Plus, it is a good visual cue to have the lyrics on the screen so it makes it easier to learn. No doubt with this catchy song, students will leave the classroom humming the song to themselves and have connected it to finding the area of a triangle.

# Engaging students: Midpoint

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Danielle Pope. Her topic, from Geometry: deriving the term midpoint.

How could you as a teacher create an activity or project that involves your topic?

Introducing the definition of a midpoint in the classroom will take using class time to let students explore for themselves. The activity that I would make my students do is have the entire class stand up and have 2 students stand at opposite sides of the room. I would then ask my students to line up shoulder to shoulder. Once they were in a straight line I would ask “who is perfectly in the middle of this line?” This is where I would give my students 10 minutes initially to come up with various ways of how they would prove a student was in the middle of the line. Various “proofs” that they could tell me would be that there is exactly the same number of people on each side of the middle person. If that answer was given I would make an odd number of students stand in line and ask the same question of “Who is in the middle”? They would have to reconsider this answer because they couldn’t cut the student in half but I would hope that they would come to the conclusion that they would have to half the person in order to find the perfect center. Another “proof” that they may give me is measuring the distance from one end to the other and half that distance to find the person in the middle. This can also start that same conversation of how we would find the exact “midpoint” without cutting the person into pieces.

How has this topic appeared in pop culture (movies, TV, current music, video games, etc.)?

To get just a basic definition of the midpoint, we can look at the lingo used in all sporting events. All sports have some form of a season that lasts for a certain amount of time. For this example specifically, I will be looking at the football season. Towards the middle of the season teams will know what to expect by the end. Most of the stats and predictions for teams are made already by the middle or midpoint of a season. In this article about football it relates to what changes various teams needed to make by the middle of their season. Just in the article itself, it says that “we’re now at the midpoint of the NFL season, and while some things are beginning to take shape, there’s still plenty of football left to be played.” In this context, students can understand that midpoint is being used to describe the middle of a football season. With this knowledge, they can use those context clues and just add the numbers given to them.

One of the most important people in mathematics to date would have to be Euclid. Euclid’s book, The Elements, is still the backbone of all mathematics taught from kindergarten to college. One artist took this book or manual to mathematics and put it in the form of artwork. Crockett Johnson is an artist who bases his work off of mathematics. He takes the complicated proofs, lemmas, and theorem that have been proved and puts those in a form that we see as beautiful. One piece that uses mostly all midpoints titled “Bouquet of Triangle Theorems”. This piece is based off of the many of Euclid’s propositions about triangle just used together in one piece of art. For example “the midpoints of the sides of the large triangle in the painting are joined to form a smaller one.” Giving students a copy of this picture they can find various characteristics given a ruler and other tools that can help them possibly come to this conclusion that Euclid already proved. Crockett’s pieces can also be seen at the Smithsonian so that could show kids that math really does show up everywhere in our world even in unexpected places.

http://www.foxsports.com/nfl/gallery/every-nfl-teams-biggest-weakness-at-the-midpoint-of-the-2016-season-110116

# Engaging students: Defining the term segment bisector

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Caroline Wick. Her topic, from Geometry: defining the term segment bisector.

How has this topic appeared in high culture (art, classical music, theatre, etc.)?

A segment bisector is a point, segment, line, or plane that divides a line segment into two equal parts, according to the math dictionary “intermath” (A1). This geometric term has been used throughout history to create art, even before the term was eloquently defined. The people of ancient Greece would use all sorts of geometric ideas to build their vast architectural structures and sculptures, and almost all of these structures would require segment bisectors. According to the Metropolitan Museum of Art page on Greek Architecture, “the vertical structure of [their] temple conformed to an order, a fixed arrangement of forms unified by principles of symmetry and harmony” (A2). The Ancient Greeks prided themselves on their beautiful structures that were pleasing to the eyes because of their symmetry and balance.

Take this picture above, for instance. The columns on the right are perfectly symmetrical to the top beam, and the middle column perfectly divides the top beam. It would be considered a Segment bisector.

Other examples of segment bisectors in high art can be seen in renowned artists work like Picasso who used geometry to paint/express the world in a way that one might not normally see, and other painters have used this geometrical interpretation in their works as well.

A. Applications: How could you as a teacher create an activity or project that involves your topic?

Segment bisectors could be used in a number of projects or activities. One activity could be showing the use of segment bisectors in origami, or the art of paper folding. Origami requires multiple strategic folds of paper that must be perfect if the shape is to come to fruition. One often has to fold the paper in half perfectly many times, which is the definition of a segment bisector. Students could learn how to use geometric concepts in a concrete and fun way that is applicable in the real world.

The picture above shows just how much segment bisectors are used in the art of paper folding.

Another project could be using the information above on ancient Greek architecture to create their own little architectural temples or structures. The students would use basic materials found around the house, and their knowledge of geometric definitions to create these structures. Not only would this project apply to geometry, but it would also help students see how geometry plays a role in architecture; another real-world application of school knowledge.

How can technology be used to effectively engage students with this topic?

Segment bisectors do not really sound like the most exciting topic for students to cover. Sure, they can be used in a lot of different applications, but when a student hears that they will be working with the definition of a segment bisector, they likely will not get terribly enthusiastic. However, if students learn these ancient concepts in the context of new technology, it might stick in their brains as a more interesting topic. Geogebra is a website that allows you to construct geometrical shapes and objects as if you were using a ruler and compass. Students could very easily spend hours on the site just finding different ways to construct a geometric shape. They could use the site to create and define multiple geometric concepts, including segment bisectors, so that they discover the words’ meanings for themselves.

The picture above was taken from a youtube video that shows you how to construct perpendicular segment bisectors using a ruler and compass. And though it may seem like it a more advanced subject, students will be able to see the reasoning behind the definition, and might be able to use this website and knowledge for later geometric use.

References:

# Engaging students: Perimeters of polygons

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Brittnee Lein. Her topic, from Geometry: perimeters of polygons.

How have different cultures throughout time used this topic in their society?

Finding the perimeter of a polygon has been a necessity since the implementation of architecture and engineering in society. Every advanced society has used this topic to their benefit. For example, if a person wanted to build a fence around their rectangular garden, but wanted to use the least amount of building materials possible, they could find the perimeter around the garden and then calculate how many planks of wood they would need to use before buying that wood. This is a much more effective method than buying the wood and then finding out how much one would need to use.

2. What interesting (i.e., uncontrived) word problems using this topic can your students do now?

A few interesting word problems I found online involve real world applications of decorating. Both of the problems I found are off of this website: http://www.bbc.co.uk/skillswise/worksheet/ma31peri-e3-w-perimeter-problems

Problem 1 is good for enforcing student understanding because it challenges them to think more abstractly than if the problem were merely stated as “what is the perimeter of the cake?”. This problem tests student understanding of both the meaning of a square and the meaning of perimeter.

Problem 5 is beneficial to students because it includes both a closed and an open question. The closed question allows students to practice what they have learned about finding the perimeter of a polygon and the open-ended question is worded in a way that challenges the student’s conceptual understanding. The student must not only compute the perimeter but also must explain his/her thinking. This problem also forces students to visualize the problem in their head.

3. How could you as a teacher create an activity or project that involves your topic?

An engaging activity that a teacher could create to reinforce the topic of finding the perimeter of a polygon is a game where students set out to stop a criminal from entering an a given area/robbing a bank. The students would have to find the perimeter of a building from a simple blueprint mapping the building’s structure (in this case it would just be the outline of the shape of the building with given dimensions). The student would be informed of how much area each officer can cover and they would then be expected to “secure the perimeter” by allotting a certain amount of police officers to the building and placing them along the perimeter (denoted by a given symbol). To ramp up the difficulty of the game, you could set a time limit for each building and have students compete against the clock to stop the robber and you could also increase the variety of officers in the game where each type has a specialty and can cover a different amount of area.

The idea for this activity is found on the website: https://www.teacherspayteachers.com/Product/Secure-the-Perimeter-Cover-the-Area-Hands-on-police-trainee-activity-2770696

References

“Secure the Perimeter! Cover the Area! Hands-on ‘Police Trainee’ Activity.” Teachers Pay Teachers, http://www.teacherspayteachers.com/Product/Secure-the-Perimeter-Cover-the-Area-Hands-on-police-trainee-activity-2770696.

Perimeter Problems.” BBC News, BBC, http://www.bbc.co.uk/skillswise/worksheet/ma31peri-e3-w-perimeter-problems.