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Imagination for Math

June 23, 2017

I once told my students that simply having an imagination isn’t enough to be a good mathematician— foremost, solid skills are needed. I wish I could take my words back. Even Einstein, who had a great mathematical skill, is noted as saying:

I am enough of the artist to draw freely upon my imagination. Imagination is more important than knowledge. Knowledge is limited. Imagination encircles the world.

What does imagination have to do with math? Well, Imagination is critical in aiding visualization of a problem, especially when there are multiple interrelated unknowns. Did you know that before Einstein mathematically developed his famous General Theory of Relativity, also popularly associated with the formula E = mc^2 , he first envisioned the problem as a mental experiment.

Einstein is definitely an exceptional human being; however, the power of visualization is not only limited to exceptional individuals. Research has shown that students who are able to employ visualization strategies are more successful in both math and science than their peers who do not. In the book, Mathematical Mindsets, Jo Boaler of Stanford University provides great examples and benefits of visual thinking along with many activities to help motivate visual thinking about math. You may also consider visiting her web initiative, youcubed, which is full of strategies for teachers, parents, and students.  In a nutshell:

The good news is that visualizing

complex problems can be learned.

But… only if visualizing could be done by talking about it. Let’s consider the following problem:

Kevin, Jill, and Elise are three cousins. Together they are 12 feet tall. Jill is one-half as tall as Kevin, and Elise and Jill together are five-sixths as tall as Kevin. Then, how tall is Elise?

If you feel comfortable with elementary Algebra, your instinct might be to set up this problem as a system of equations. Instead, entertain this problem using your basic number sense and visual representations. How would you solve it without algebraic procedures?

The key challenge in solving any problem

is first understanding the question.

The problem is asking to find out Elise’s height. The challenge is that the problem doesn’t specifically say anything about Elise’s height. In fact, upon closer inspection, you may realize that to decipher Elise’s height you would have to also find out Jill’s height, whose height is then only solvable with reference to Kevin’s height.  Essentially, this problem has three parts: finding out Kevin, Jill, and Elise’s heights.

To visualize this problem, you may use a tape-diagram. This is often a great tool for solving problems involving interrelated parts. To start, draw a simple tape for Kevin. Start with his height because his cousins' height is based on his height.  He is your 'target.'

Now that you have created a tape for Kevin, you need a method to compare Kevin, Jill and Elise. You know that:

Jill is one-half as tall as Kevin.

Elise and Jill together as five-sixths of Kevin.

You can compare all three by converting their ratios into a common denominator, which in this case would be 6. Since Kevin's tape is the original whole, you may split his tape into six-sixths or one whole.

But if Kevin is six-sixths, how would you draw the tapes for 'Jill,' and 'Elise and Jill?'

Before you read on any further, attempt to complete the diagram on your own.

Compare your diagram with the one below. Does your diagram look similar to this?

By converting all three cousins’ heights into sixths and representing them visually, you can now see that if Kevin represented six-sixths and Jill represented three-sixths of Kevin, then Elise must be two-sixths of Kevin. Together, they would represent eleven-sixths of Kevin's height. You can also check this by adding the first line (Kevin's) and the third line (Elise and Jill's) of the tape diagram together. You should get eleven-sixths.

To find out Elise's height, you would have to first divide 12 feet into 11 equal parts, which would gives approximately 1.09 feet for each one-sixth part of Kevin's height. Visually, this can be represented as:

Now, all you have to do is add up the respective parts for each cousin to find their total height?  What do you get?

Kevin is approximately 6.55 feet.

Jill is approximately 3.27 feet.

Elise is approximately 2.18 feet.

If you add the individual height of the three cousins together, what should it equal to?  Also, above I mentioned that "eleven-sixths of Kevin's height equals the whole." Now that we know Kevin's height, we can check if this statement is true: 11/6 x 6.55 feet = 12 feet.  Isn't that the total height of all three tapes?

While many countries use centimeter and meter units for height measurement, some countries also use the foot (ft) and the inch (in) units. The challenge with measuring in feet and inches is not to mistakenly evaluate the decimal as inches. To further simplify the decimals into inches, a unit conversion is needed. Specifically, 1ft = 12in. For example, for Kevin's height, you can calculate the remaining inches by: 6ft + (0.55 x 12in) = 6ft and 6.6in.  By following the same rule for all of the cousins, we get:

Kevin is approximately 6 feet 7 inches.

Jill is approximately 3 feet 3 inches.

Elise is approximately 2 feet 2 inches.

Any creative exercise requires a fair deal of imagination. Before solving a problem, it is very important to understand the problem. Once you are comfortable with the question, you may let your imagination and visualization fly; then, you can employ your numerical calculations and manipulations to decipher the answer.

If you still need some motivation to use your math imagination, you may enjoy James Tanton's Exploding Dots. In this video below he provides a great visual explanation of long division both very imaginative and inspiring. Endless Imagination!

The global math week, taking place October 10-17, 2017, will celebrate mathematical joy, imagination and visualization, focusing on exploding dots. Register with the global math project to find out more.

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