Category: Algebra

Posted in Algebra

sorting pi, e, and root 2

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Mathematicians, always economical,   love to categorize numbers according to their properties. This is because numbers belonging to the same category behave in the same way. You don’t have to deal with each one! That’s an economical way of preserving the energy demand of brain cells.  In the grades we give pupils tasks that involve sorting numbers. Whole numbers  can be sorted out as odd or even, prime or composite, for example. This is a very good way of giving the students a sense of how strict definitions are in mathematics and in understanding the nature of numbers. In the higher grades they meet other numbers which they can categorize as imaginary or real, transcendental or algebraic. The same mathematical thinking is used.

\pi is one of the most widely known irrational number. Ask a student or a teacher to give an example of an irrational number, the chances are they will give \pi as the first example or the second one, after square root of 2.  And of course at a distance third is the number e. Now, although they belong to the same set of numbers, the irrationals, they don’t really belong to the same category. For example, \pi and e are both irrationals but pi is transcendental and square root of 2 is algebraic.  The number e is also transcendental. Here’s a short and simple explanation.

 

 

A  transcendental number is one that cannot be expressed as a solution of ax^n+bx^(n-1)+…+cx^0=0 where all coefficients are integers and n is finite. For example, x=sqrt(2), which is irrational, can be expressed as x^2-2=0. This shows that the square root of 2 is nontranscendental, or algebraic.

It is very easy to prove that a number is not transcendental, but it is extremely difficult to prove that it is transcendental. This feat was finally accomplished for ? by Ferdinand von Lindemann in 1882. He based his proof on the works of two other mathematicians: Charles Hermite and Euler.

In 1873, Hermite proved that the constant e was transcendental. Combining this with Euler’s famous equation e^(i*?)+1=0, Lindemann proved that since e^x+1=0, x is required to be transcendental. Since it was accepted that i was algebraic, ? had to be transcendental in order to make i*? transcendental. Click here for source.

Of course understanding the proof of pi as a transcendental number is beyond the level of basic mathematics and hey, we don’t even talk about transcendental numbers before Grade 10. But students at this level can understand the expression ax^n+bx^(n-1)+…+cx^0=0 where all coefficients are integers and n is finite. With proper scaffolding or if they have been exposed to similar task of sorting numbers before  students can make sense of the logic and reasoning shown above which characterizes most of the thinking in mathematics.

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Image from http://studenthacks.org/wp-content/uploads/2007/10/pumpkin-pi.jpg

Posted in Algebra, Geometry, Math blogs

Math and Multimedia Carnival #7

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Welcome to the 7th edition of Math and Multimedia blog carnival.

Before we begin Carnival 7, let’s look at some of the trivias about the number seven:

Now, lets start with posts that involve mathematics sans technology.

Guillermo P. Bautista Jr., the organizer of Mathematics and Multimedia Carnival, presents Generating Pythagorean Triples posted at Mathematics and Multimedia, saying, “A simple strategy in generating Pythagorean Triples.”

Mike Dimond presents Squares ending in 5 – Two Digit Numbers posted at Education For All, saying, “Learn how to quickly calculate the square for two digit numbers ending in five. The post goes over how to quickly calculate 75 * 75.”

I also grab the post Numbers and Variables, the first in the series of post on teaching algebra to students in their first year of High School from the blog Learning and Teaching Math.

John Golden presents Math Hombre: Variable and a Problem posted at Math Hombre, saying, “This post tries to give a couple of contexts for middle school or Algebra I development of the concept of variable.”

Let me include on this list my latest post titled  Counting Smileys which shows several solutions to counting problems that are used to introduce variables and algebraic expressions.

click link to view source

Now, for mathematics with technology:

David Wees presents Is Interactivity in Mathematics Important posted at Professional blog | 21st Century Educator, saying, “This blog post is a discussion of the importance of using interactive tools when teaching mathematics.” This is one way indeed to involve students in the learning.

Alexander Bogomolny presents Fascination with Tessellations posted at CTK Insights. The post presents several Java applets that illustrate various hinged tessellations and ways of inserting hinges into an existing tessellation.

Terrance Banks presents Treasure Hunt Activity posted at So I Teach Math and Coach?, saying, “Review Activity – Treasure Hunt for Algebra”

Gianluigi Filippelli presents Gravity vs height posted at Science Backstage, saying, “The dependance of gravity by height plotted with Scilab”

Tamarah Buckley presents Instant Feedback posted at Infinitely Many Solutions, saying, “My blog focuses on using iPads in a secondary math classroom.”

Pat Ballew presents Microsoft Mathematics is FREE! posted at Pat’sBlog, saying, “Software for every kid, at just the right price…”

Finally, let me share my post on Squares and Square Roots which presents a series of activities for teaching these concepts meaningfully using the free software, GeoGebra.

That concludes this edition. Submit your blog article to the next edition of mathematics and multimedia blog carnival using our carnival submission form. Past posts and future hosts can be found on our blog carnival index page

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Posted in Algebra

Teaching algebraic expressions – Counting smileys

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This is an introductory lesson for teaching the concept of variable and algebraic expressions through problem solving. The problem solving task combines numerical, geometric, and algebraic thinking.  The figure below shows the standard version of the task. Of course some easier versions would ask for the 5th figure, then perhaps 10th figure, then the 100th figure, and then finally for the nth figure. This actually depends on the mathematical maturity of the students.

An alternative version which I strongly encourage that teachers should try is to simply show first the diagrams only (see below).

Study the figures from left to right. How is it growing? Can you think of systematic ways of counting the number of smileys for a particular “Y” that belongs to the group? This way it will be the students who will think of which quantity (maybe the number of smileys in the trunk of the Y or the position of the figure) they could represent with n.The students are also given chance to study the figures, what is common among them, and how they are related to one another. These are important mathematical thinking experiences. They teach the students to be analytical and to be always on the lookout for patterns and relationships. These are important mathematical habits of mind.

Here are possible ways of counting the number of smileys: The n represents the figure number or the number of smiley at the trunk.

1. Comparing the smileys at the trunk and those at the branches.

In this solution, the smileys at the branches is one less than those at the trunk. But there are two branches so to count the number of smileys, add the smileys at the trunk which is n to those at the two branches, each with (n-1) smileys. Hence, the algebraic expression representing the number of smileys at the nth figure is n+2(n-1).

2. Identifying the common feature of the Y’s.

The Y’s have a smiley at the center and has three branches with equal number of smileys. In Fig 1, there are no smiley. In Fig 2, there is one smiley at each branch. In fact in a particular figure, the number of smileys at the branches is (n-1), where n is the figure number. Hence the algebraic expression representing the number of smileys is 1+ 3(n-1).

3. Completing the Y’s.

This is one of my favorite strategy for counting and for solving problems about area. This kind of thinking of completing something into a figure that makes calculation easier and then removing what were added is applicable to many problems in mathematics. By adding one smiley at each of the branches, the number of smileys becomes equal to that at the trunk. If n represents the smileys at the trunk (it could also be the figure number) then the algebraic representation for counting the number of smilesy needed to build the Y figure with n smiley at each branches and trunk is 3n-2, 2 being the number of smileys added.

4. Who says you’re stuck with Y”s?

This is why I love mathematics. It makes you think outside the box. The task is to count smileys. It didn’t say you can not change or transform the figure. So in this solution the smileys are arrange into an array. With a rectangular array (note that two smileys were added to make a rectangle), it would be easy to count the smileys. The base is kept at 3 smileys and the height corresponds to the figure number. Hence the algebraic expression is (3xn)-2 or 3n-2.

The solutions show different visualization of the diagram, different but equivalent algebraic expressions, and all yielding the same solution. Of course there are other solutions like making a table of values but if the objective is to give meaning to algebraic symbols, operations, and processes, it’s best to use the visuals.

A more challenging activity involved Counting Hexagons. Click the link if you want to try it with your class.

Posted in Algebra

Attention Math Bloggers

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This blog will be hosting the 7th edition of Math and Multimedia Blog Carnival. It will be posted on 31st January 2011. This special edition carnival will focus on teaching and learning algebra at all levels. Click the link to send your posts/ articles Math and Multimedia Blog Carnival.