## Exponents Part 1

exponents part 1

Exponents Part 1

One of the biggest things to understand about math is how it is written. The spatial arrangement of characters is syntax. Syntax, in English, refers to the arrangements of words to convey meaning.

Exponents are just a way of writing repeated multiplication. If we are multiplying a number by itself repeatedly, we can use an exponent to tell how many times the number is being multiplied. That’s it. Nothing tricky exists with exponents, no new operations or concepts to tackle. If you’re familiar with multiplication and its properties, exponents should be accessible.

That said, it is not without its pitfalls. A balance between conceptual understanding and procedural short-cuts is needed to avoid those pitfalls. The only way to strike that balance is through a careful progression of exercises and examples. An answer-getting mentality will lead to big troubles with exponents. People wishing to learn how exponents work must seek understanding.

Let’s establish some facts that will come into play with this first part of exponents.

1.      Exponents are repeated multiplication

To simplify simple expressions with exponents you only need to know a few short-cuts, but to recall and understand, we need more. These facts are important.

With an exponential expression we have a base, the number being multiplied by itself, and the exponent, the small number on the top right of the base which describes how many times the base is being multiplied by itself.

${a}^{5}$

The number a is the base. We don’t know what a is other than it is a number. It’s not a big deal that we don’t know exactly what number it is, we still know things about this expression.

Five is the exponent, which means there are five a’s, all multiplying together, like this: $a\cdot a\cdot a\cdot a\cdot a.$

Something to keep in mind is that this expression equals another number. Since we don’t know what a is, we cannot find out exactly what it is, but we do know it’s a perfect 5th power number, like 32. See, 25 = 32.

What if we had another number multiplying with a5, like this:

${a}^{5}\cdot {b}^{3}$

If we write this out, without the exponents we see we have 5 a’s and 3 b’s, all multiplying together. We don’t know what a or b equals, but we do know they’re multiplying so we could change the order of multiplication (commutative property) or group them together in anyway we wish (associative property) without changing the value.

${a}^{5}\cdot {b}^{3}=a\cdot a\cdot a\cdot a\cdot a\cdot b\cdot b\cdot b$

And these would be the same:

$\left(a\cdot a\right)\left(a\cdot a\cdot a\right)\left(b\cdot b\cdot b\right)$

$\left(a\cdot a\right)\left[\left(a\cdot a\cdot a\right)\left(b\cdot b\cdot b\right)\right]$

$\left(a\cdot a\right){\left[ab\right]}^{3}$

${a}^{2}{\left[ab\right]}^{3}$

This is true because the brackets group together the a and b, making them both the base. The brackets put them together. The base is ab, and the exponent is 3. This means we have ab multiplied by itself three times.

Keep in mind, these are steps but exploring how exponents work to help you learn to read the math for the intended meaning behind the spatial arrangement of bases, parenthesis and exponents.

Now, the bracketed expression above is different than ab3, which is $a\cdot b\cdot b\cdot b$.

${\left(ab\right)}^{3}\ne a{b}^{3}$

Let’s expand these exponents and see why this is:

${\left(ab\right)}^{3}\ne a{b}^{3}$

Write out the base ab times itself three times:

$\left(ab\right)\left(ab\right)\left(ab\right)\ne a\cdot b\cdot b\cdot b$

The commutative property of multiplication allows us to rearrange the order in which we multiply the a’s and b’s.

$a\cdot a\cdot a\cdot b\cdot b\cdot b\ne a\cdot b\cdot b\cdot b$

Rewriting this repeated multiplication we get:

${a}^{3}{b}^{3}\ne a{b}^{3}$

The following, though, is true:

$\left(a{b}^{3}\right)=a{b}^{3}$

On the right, the a has only an exponent of 1. If you do not see an exponent written, it is one. If we write it out we see:

$\left(a\cdot b\cdot b\cdot b\right)=a\cdot b\cdot b\cdot b$

In summary of this first exploration, the base can be tricky to see. Parenthesis group things together. An exponent written outside the parenthesis creates all of the terms inside the parenthesis as the base. But if numbers are multiplying, but not grouped, and one has an exponent, the exponent only belongs to the number just below it on the left. For example, $4{x}^{3},$ the four has an exponent of just one, while the x is being cubed.

Consider: ${\left(x+5\right)}^{3}.$ This means the base is x + 5 and it is multiplied by itself three times.

Repeated Multiplication Allows Us Some Short-Cuts

Consider the expression:

${a}^{3}×{a}^{2}.$

If we wrote this out, we would have:

$a\cdot a\cdot a×a\cdot a$.

(Note: In math we don’t use colors to differentiate between two things. A red a and a blue a are the same. These are colored to help us keep of track of what’s happening with each part of the expression.)

This is three a’s multiplying with another two a’s. That means there are five a’s multiplying.

${a}^{3}×{a}^{2}={a}^{5}$

Before we generalize this to find the short-cut, let us see something similar, but is a potential pitfall.

${a}^{3}×{b}^{2}$

If we write this out we get:

$a\cdot a\cdot a×b\cdot b$

This would not be an exponent of 5, in anyway. An exponent of five means the base is being multiplied by itself five times. Here we have an a as a base, and three of those multiplying, and a b as a base, and two of those multiplying. Not five of anything.

The common language is that if the bases are the same we can add the exponents. This is a hand short-cut, but if you forget where it comes from and why it is true, you’ll undoubtedly confuse it with some of the other short-cuts that follow.

Short-Cut 1: If the bases are the same you can add the exponents. This is true because exponents are repeated multiplication and the associative property says that the order in which you group things does not matter (when multiplying).

${a}^{m}×{a}^{n}={a}^{m+n}$

The second short-cut comes from groups and exponents.

${\left({a}^{3}\right)}^{2}$

This means the base is a3, and it is being multiplied by itself.

${a}^{3}×{a}^{3}$

Our previous short cut said that if the bases are the same, we can add the exponents because we are just adding how many of the base is being multiplied by itself.

${a}^{3}×{a}^{3}={a}^{3+3}={a}^{6}$

But this is not much of a short cut. Let us look at the original expression and the outcome and look for a pattern.

${\left({a}^{3}\right)}^{2}={a}^{6}$

Short-Cut 2: A power raised to another is multiplied.

${\left({a}^{m}\right)}^{n}={a}^{m×n}$

Be careful here, though:

$a{\left({b}^{3}{c}^{2}\right)}^{5}$ = $a{b}^{15}{c}^{10}$

Practice Problems

 1.       ${x}^{4}\cdot {x}^{2}$ 8. ${\left(5xy\right)}^{3}$ 2.       ${y}^{9}\cdot y$ 9. ${\left(8{m}^{4}\right)}^{2}\cdot {m}^{3}$ 3.       ${z}^{2}\cdot z\cdot {z}^{3}$ 10. ${\left(3{x}^{5}\right)}^{3}{\left({3}^{2}{x}^{7}\right)}^{2}$ 4.       ${\left({x}^{5}\right)}^{2}$ 11. $7{\left({7}^{2}{x}^{4}\right)}^{5}\cdot {7}^{3}{x}^{5}$ 5.       ${\left({y}^{4}\right)}^{6}$ 12. ${5}^{3}+{5}^{3}+{5}^{3}+{5}^{3}+{5}^{3}$ 6.       ${x}^{3}+{x}^{3}+{x}^{3}+{x}^{8}+{x}^{8}$ 13. ${3}^{2}\cdot 9$ 7.       ${4}^{x}+{4}^{x}+{4}^{x}$ 14. ${4}^{x}\cdot {4}^{x}\cdot {4}^{x}$

## The Purpose of Homework and My Response

The purpose of homework is to promote learning.  That’s it.  It’s not a way to earn a grade or something to keep kids busy.  It’s also not something that just must be completed in order to stay out of trouble.  Homework is a chance to try things independently, make mistakes and explore the nature of those mistakes in order to better learn the material at hand.

If students are not learning from the homework, it is a waste of time and effort.  There are a few things that could cause students not to learn from the homework.  Even if the assignments are of high quality, without the reflection and correction piece, students will not learn much from homework.

Reflection and correction go together.  It’s not about getting right answers, but thinking about what caused mistakes, identifying misconceptions or procedural inefficiencies and replacing those.  To reflect a student should NOT erase their incorrect working but instead should write on their homework, in pen, what went wrong and what would have been better.

It is quite possible more can be learned when reviewing homework than any other time.  It is certainly a powerful experience.

Textbooks and videos, tutors and peer help offer little appropriate support to help make homework, or practice, meaningful.  Textbooks only provide correct answers, YouTube videos usually do similar treatment to topics as textbooks offer.

I wish to help students learn and believe that reviewing work that has been done is too powerful of an opportunity to pass.  The trick is, how can I provide reflection and insight when to someone I am not sitting with and talking to?  I think I can help provide this reflection piece by doing all of the practice problems myself on a document camera and discussing pitfalls and mistakes, as well as sharing my thinking about the problems as I tackle them.  Further, I can share typical mistakes I see from students as they are learning topics.

So as I develop the Algebra 1 content I will be working on adding videos and short written responses to the assignments to help students think about what they’ve done, its appropriateness, correctness and their level of understanding.

## Adding and Subtracting Algebraic Fractions

part 2

Introduction to Algebraic Fractions
Part 2

In the last section we saw both how to reduce Algebraic Fractions, which if you recall, are also called Rational Expressions, but also how the math works. Because of the relationship between division and multiplication, and multiplication’s commutative property, we can reduce Algebraic Fractions like the one below:

$\frac{3a+6}{12{a}^{2}-9}$

When dealing with Algebraic Fractions, your work is not done until you’ve reduced completely. The example above is the unfinished answer to the first problem we will do to introduce addition and subtraction of Algebraic Fractions.

$\frac{2a-5}{12{a}^{2}-9}+\frac{a+11}{12{a}^{2}-9}$

Since these have a common denominator, we just add the like terms in the numerator. Note, in math when we say add, it means combine with addition and subtraction.

$\frac{2a-5+a+11}{12{a}^{2}-9}$

Combining like terms, we end up with:

$\frac{3a+6}{12{a}^{2}-9}$

But since each term has a factor of 3, we can reduce each term by 3:

$\frac{3a+6}{12{a}^{2}-9}\to \frac{\overline{)3}a+\overline{)3}\cdot 2}{\overline{)3}\cdot 4\cdot {a}^{2}-3\cdot \overline{)3}}=\frac{a+2}{4{a}^{2}-3}$.

Our answer has four total terms. While some share a common factor, not all four terms share a common factor, so we are finished.

If the denominators are the same, you just combine the numerators.

With subtraction it is slightly trickier. Let’s simplify a problem and see how this works.

5 $–$ 4 + 3

5 $–$ (4 + 3)

The two expressions above are not the same. The first equals four, while the second is -2. The parenthesis make a group of the four and three, which is being subtracted from the five. The four and the three are both being subtracted from the five. But, great care is in order here. It is easy to mess up these signs.

5 $–$ (4 $–$ 3) = 4

Because 5 $–$ 4 - -3 is 5 $–$ 4 + 3.

Remember that fraction bars also create groups in Algebra. So, instead of parenthesis, you will see:

$\frac{5}{x}-\frac{4+3}{x}$

This is the same as:

.

An example with variables would be:

$\frac{{x}^{3}-5{x}^{4}}{9}-\frac{5{x}^{4}-{x}^{3}}{9}$

Note: Exponents are repeated multiplication, they do not change from addition. Also, in order to be like terms, the variables and exponents must be the same. x3 and x5 are not like terms.

$\frac{{x}^{3}-5{x}^{4}-\left(5{x}^{4}-{x}^{3}\right)}{9}$

Caution and care are in order when dealing with subtraction and Algebraic Fractions!

$\frac{{x}^{3}-5{x}^{4}-5{x}^{4}+{x}^{3}}{9}$

Combining like terms we get

$\frac{2{x}^{3}-10{x}^{4}}{9}$.

Since there is not a common factor between all terms, we are done.

Adding and subtracting Algebraic Fractions with unlike denominators involves finding the LCM (lowest common multiple) of each denominator. We will restrict our denominators to monomials for now, as to keep this appropriate for beginning Algebra students.

Let’s begin with the denominators a and b. All we know is that a and b are numbers that cannot be zero, but we don’t know their exact value. So, we assume they are relatively prime, making their LCM their product, a×b.

$\frac{3}{a}+\frac{2}{b}$

We arrive at common denominators through multiplication. Don’t get confused here, we are multiplying by a number that equals one if it were reduced, but we don’t want to reduce until we are finished. Also, note, that since e are multiplying by one, the expression will look different but will have the same value. It is not unlike the difference between a twenty dollar bill versus a ten and two five dollar bills.

Our denominator will be ab. To change a into ab, we multiply by b. We multiply b by a, writing the variables in alphabetical order will help to recognize that they are the same. (Sometimes students write ab and then ba. While they’re the same, the order in which they’re written can confuse you.)

$\frac{b}{b}\cdot \frac{3}{a}+\frac{2}{b}\cdot \frac{a}{a}$

In many respects, adding or subtracting Algebraic Fractions is easier because there is less calculation taking place.

Let’s walk through an uglier problem involving subtraction and negative signs.

$\frac{7{x}^{2}-2y}{5{x}^{2}y}-\frac{4x-2}{5{x}^{2}}$

We need to find the LCM of . The LCM will be 5x2y. So we need to multiply the second fraction by y over y.

$\frac{7{x}^{2}-2y}{5{x}^{2}y}-\frac{4x-2}{5{x}^{2}}\cdot \frac{y}{y}$

Now we are multiplying the entire group of 4x $–$ 2 by y, not just whatever term is written next to the y.

$\frac{7{x}^{2}-2y}{5{x}^{2}y}-\frac{4xy-2y}{5{x}^{2}y}$

Normally I would not write the step above, but did so to help make sure you understand why the fraction on the right is 4xy, not just 4x. We have to distribute the y to the entire group.

Now with care for that negative sign, let’s put it all together.

$\frac{7{x}^{2}-2y-\left(4xy-2y\right)}{5{x}^{2}y}$

which will become:

$\frac{7{x}^{2}-2y-4xy+2y}{5{x}^{2}y}$

Combining like terms, we get:

$\frac{7{x}^{2}-4xy}{5{x}^{2}y}$.

Before we can say we’re finished we need to check for a common factor (GCF) that could be divided out. These don’t always exist but if one does, and you had the answer correct up to that point, it would be a shame to mess up the last little step, so check.

$\frac{7{x}^{\overline{)2}}-4\overline{)x}y}{5{x}^{\overline{)2}}y}=\frac{7x-4y}{5xy}$.

In review, you need a common denominator which will be the LCM of the denominators. You must take care to both distribute property in the numerator, and watch for sign errors, especially with subtraction, when combining like terms. The last thing is to check for a common factor between all terms when you’ve finished combining like terms.

Practice Problems
Instructions: Add or Subtract as indicated

1. $\frac{3x}{5}+\frac{2}{a}$

2. $\frac{4}{9a}+7$

3. $\frac{4}{9a}+\frac{7}{9}$

4. $\frac{4}{9a}+\frac{7}{a}$

5. $\frac{2x-1}{3{x}^{2}}-\frac{2x-1}{{x}^{2}}$

## Combining Like Terms & The Order of Operations

Be fair-warned…this might seem pedantic, and I agree, it is.  But, with good cause. The point of this discussion is to cause students to pause and think about something they’ve always done in math without any understanding.  It’s true, you can combine like terms, but why?  To that point, and I do not say this to put anybody down, but I bet most math teachers have never wondered this on their own.

### Why Can’t We Combine Unlike Terms?

I pose this question to Algebra 1 students every year, for over a decade now.  The responses are always with an air of incredulity.  Their looks say to me:

Come on Mr. Brown, you should be smarter than to ask such a dumb question!

When I press for an answer they’re suddenly tongue-tied.  The most common answer, often spit at me with some venom is:

Because they’re not like terms.

Okay then, why can you add things that ARE like terms?

### Are We in Violation of PEMDAS

The order of operations is the set of instructions by which math is to be performed.  It’s kind of a big deal!  Mess it up and you’re in a lot of trouble.  And yet, how is it that we can combine like terms if there’s multiplication occurring first?

Consider 3x + 5x – 2a.  Even the most average of 9th grade, Algebra 1 students will tell you that obviously equals 8x – 2.  My question is this:  How can you add 3and 5x if there’s multiplication taking place.  The three and the x are multiplying, right?  And in the order of operations, doesn’t multiplication come before addition?

Student Responses

When I ask students why like terms can be combined even though there’s multiplication taking place and they add before multiplying, they just say, “cuz.”  This is a classic example of something that’s been taught and accepted without understanding.  It’s not a complicated issue, but promoting mathematical literacy is my crusade of late.

Consider the product of four and five.  I remember being a young child and counting groups of five to find the product of five and a number.  4 x 5 = 5 + 5 + 5 + 5

Good to go on that point?

Then 3x = x + x  + x

And 5x = x + x + x + x + x

Then 3x + 5x =  x + x  + x  + x + x + x + x + x8x

Because multiplication is repeated addition, and we’re adding the same thing repeatedly, we can combine the coefficients of x.

Why Can’t You Combine Unlike Terms?

Consider our original expression: 3x + 5x – 2a.  Two times a is a + a.  Three and five makes 8x.  But x and a are probably not the same thing (we don’t REALLY know because they’re unknown), so we cannot combine them with addition.

8x + 2

x + x  + x  + x + x + x + x + x + a + a

This cannot be simplified further than 8x + 2a.

To combine unlike terms would be in violation of the order of operations.  Since and a are different, we would need to multiply 8 and 2 by them respectively, before adding them.

I hope this short discussion has caused you to pause and think about what you thought you knew about Algebra.  Let me know what you think in the comments.  Thank you, again, for reading.

Philip

## Promoting Mathematical Literacy through Non-Procedural Questions

Math was okay until they threw the alphabet in it.

I have heard, Math was okay until they threw the alphabet in it, so many times from so many adults, who I wish were trying to be witty, but are in fact completely serious.  And these are reasonable, intelligent people.  Why does the abstraction of a variable or an unknown become so confusing that it creates a huge disconnect?

I’d like to share with you an assignment that I believe will help students transition from the concrete application and properties of Real Numbers to the abstractions we deal with in Algebra and mathematics beyond.

The desired outcomes of this assignment are:

• Improve mathematical literacy by encouraging students to read the mathematical meanings created by the spatial arrangement of numbers and symbols
• Improve their understanding of the Order of Operations and to help student realize that the Order of Operations is not its own topic, compartmentalized, but rather an over-arching understanding of how math is performed.
• Promote abstract thinking about numbers and their properties
• Introduce some concepts that will come into play later in Algebra, like finding the x-intercepts of a quadratic once it is factored

How to Introduce the Assignment

Students must be aware that they will be dealing with abstract ideas and that there are sometimes more than one right answer.  Also, a student can be right, but not completely right, they could also be wrong, but not always.  By fostering a healthy discussion about these problems you can introduce the idea that in order for something to be mathematically true, it must always be true.  If a single circumstance is untrue, then the statement is untrue.

Consider the problem: Given that a and b are real numbers, and the following is true, what do you know about the numbers a and b?   a×b=0

A student might say, In this case a and b are both zero.

That is correct in one case, but there are many cases where that is not true.  It is true that one of them must be zero.

The First Prompt

Given that a and b are real numbers, and the following statement is true, what can you conclude at the numbers a and b?

a – b = 0

At first have them think and write on their own.  Make sure they’re all working, not avoiding this uncomfortable notion.

After a given amount of time (short, maybe one minute), instruct them to talk with two different people.  Be clear that the expectation is that they take turns, one person shares, the other listens and responds.

After the time is up, have a whole-class discussion, but avoid being the authority until the discussion is winding down.  Only be the authority on the subject to help summarize.

The Second Prompt

With the same properties of the numbers and the statement being true, provide them with the equation:

(a + 5)(b –7) = 0

Conduct conversations as you did with the first, maybe allowing more time for them to talk together as this is a more complicated situation.

The Third Act

Switching gears from properties of variables to applying some properties of real numbers will promote their understanding of when the associative and commutative properties as well as challenge their understanding of how math is written.  We are really trying to promote their ability to read and write math and their fact that the spatial arrangement has meaning in math.

Have the students try and add parenthesis, as many as they like, to the following equation, so that it will be true.  Have them try it on their own first, then provide a short amount of time for peer discussion.

3•2 – 72 + 5 = 80

When you conduct a whole-class discussion, make sure it’s student lead, your role is as a mediator, not a disseminator of facts.

Fourth Act

Instruct students to create a similar problem by making a statement they know is true and removing the parenthesis.   For example, they might make up:

8(5 – 3) + 11 = 27

But would only write

8•5 – 3 + 11 = 27

When they’re completed their problem, have them show you.  Once everybody has a problem, hand out 3×5 cards.  On the front of the card the student will write their problem, without the parenthesis.  On the back, they’ll write their name.  Have them pass the cards forward and you can distribute them at to another class the following day.

Last Thing for the Lesson:

The idea here is the same as with the previous activity, but we are accessing their knowledge from a different angle.  They will be given an expression with parenthesis and be asked if the parenthesis can be removed without changing the value of the expression.

For example:  (5 + 4) – 2, or 11 – (4 – 9).

Introduce these problems in the same fashion, with quiet thinking first, then small group discussions, then whole-class discussion.

The Homework:

The homework is critical here because it will challenge students to think and examine how the way in which math is written changes the meaning.  It will also force them to think about numbers in a general fashion.

Timing

The last thing I’d like to mention is that this could easily be done over two separate days depending on the aptitude of the class you’re teaching.

I hope this is helpful and food for thought.

## Why Teaching Properties of Real Numbers is Important

If you are going to do a fraction review, the lesson here might be of some help.  I believe things are best reviewed in context, but this is a decent set of information that also introduces the real numbers and some other basics of math.

The PDF icon to the left has a lesson outline you can feel free to use with the PowerPoints of in any way you see fit.

The structure is all there in the lessons, but they're not over scripted.  Remember, I believe the majority of a lesson should be spontaneous.  It should be anticipated and prepared for, but how the lesson really unfolds depends on the audience.

Below you will find an overview of how and why I teach real numbers as well as two PowerPoint icons you can download and use as your own.  I only ask that you share where you found them.

Anything you purchase from Amazon.com through the banner below goes to producing more materials, and at no cost to you.

### What Good Is It?

The Real Number Line has always been one of the dullest lessons I have to teach.

Natural Numbers are the set of numbers you can count on your fingers, beginning with one.  The Whole Numbers are the Natural Numbers and Zero...Integers are ...

Blah Blah Blah

I have to teach it because it's in the curriculum.  And I always wonder, what use is it if a student knows the difference between a whole number and a natural number?

It is hypocritical of me to complain in such a fashion because I laud the virtues of education being greater than a set of skills or a body of knowledge.  Education is about learning to think, uncovering something previously unknown that ignites excitement and interest.  Education should change how you see yourself, how you think about the world.  It should enrich our lives.

Teaching the Real Number Line can be a huge first step in that direction, if done properly.

### Math is About Ideas, Not Just Computation

There are some rich, yet entirely approachable, mathematical ideas that can be introduced with the Real Number Line (RNL).  For example, a series of questions to be posed to students could be:

1.  The Natural Numbers are infinite, meaning, they cannot be counted entirely.  How do we know that?
2. The Integers are also infinite.  How do we know that?
3.  Is infinity a number?
4. Which are there more of, Natural Numbers of Integers?  How can you know, if they're both infinite?

The idea of an axiom can be introduced.  Most likely, students assume math is true, or entirely made up, but correct or incorrect, because it is written in a book and claimed to be such by a teacher.  The idea of how we know what we know and if math is an invention or a discovery can be introduced by talking about axioms.  For example:

1. Is it true that 5 + 4 = 4 + 5 ?
2. If a and are Real Numbers, would it always be true that b = b?  (What if they were negative?)
3. Is it also true that b = b - a?  How do we know that?
4. Is the following also true:  If a = b, and b = c, then a = c?  How do we know?

The idea here is not to teach students the difference between the Associative Property and the Commutative Property, but to use these properties to introduce students to math as a topic that can be discussed, and that it is not about answer getting, but instead about ideas.

For more on this topic and a few other related items, visit this page.

### Why Are Some Rational Numbers Non-Terminating Decimals?

If you had a particularly smart group of students, you could pose this question.  I mean, after all, 1/3 = 0.3333333333333333...  And yet, we are told rational numbers include decimals that can be written as a fraction (the ratio of two integers).

How it works is sometimes very clear and clean.  For example, 0.7 is said, "Seven tenths." And "Seven tenths," can also be written as the ratio of seven and ten.  And the number seven tenths is of course equal to itself, regardless of how it is written.  The number 0.27 is said, "twenty seven hundredths," which can easily be written as the ratio of twenty seven, for the numerator, and one hundred, as the denominator.  And this can continue so long as the decimal terminates.  But try the same thing with the a repeating decimal and you do not end up with things that are equal.

The algorithm to convert a repeating, but non-terminator decimal into a fraction is pretty straight forward.

But that does not address why a rational number would be a non-terminating decimal.

Click the PPT Icon to the left to download a lesson on converting repeating decimals into fractions for honors students.  It includes a proof of why the square root of two is irrational.

The simple reason that some rational numbers cannot be expressed as terminating decimals has to do with our numbering system.  We use base 10 numbers.  Our decimal system provides us an easy way to write fractions with denominators that are powers of 10.

That means that we have ten numbers, including zero, that fill up one column, like a car’s odometer.  When you travel 9 miles the odometer will read 000009.  When you travel the tenth mile the odometer will read 000010.

Not all of our number systems are base ten. While the metric system is base 10, or at least translates into powers of ten, the Imperial system is base 12 from inches to feet, but 5,280 feet to the next unit of miles, and so on.

Time is another great example of bases other than ten.  Seconds and minutes are base sixty.  You need sixty seconds before you have an hour, not ten.  But hours are base 24 because 24 hours are needed to make one of the next category, which is days.

In time, 25 minutes of an hour is the ratio:

But in base ten this is 0.4166666666666666... Our decimal system does math in base ten, not base sixty.  This is not 41 minutes!  A typical mistake would be two say 25 minutes is 0.25 of an hour.

Back to our original example of 1/3.  Not all numbers can be cleanly divided into groups of ten, like 3.  If we had a base 3 numbering system, where after the third number we moved, then 1/3 would just be 0.1.  But in our numbering system, 0.1 is one tenth.

Other numbers, like four, translate into ten more easily.  Consider the following:

The only issue remaining is that 2.5/10 is not a rational number because 2.5 is not an integer and rational numbers are ratios of two integers.  This can be resolved as follows:

Let's try the same process with 1/3.

As you can see, we will keep getting ten divided by three, forever.

This is a great example of how exploring a question can uncover many topics within the scope of the course being taught.

I hope this has caused you to pause and think of how exploring questions, relationships and properties in mathematics can lead to greater understanding than just teaching process and answer getting.