## Vestiges of the Past Making Math Confusing

Something in Math HAS to Change

Convention is a beautiful thing.  It allows us to use symbols to convey little things like direction or a sound.  We can piece those things together to make larger things, and eventually use it to create something like what you’re reading now.  There are no inherent meanings to these shapes we call letters, or the sounds we use when speaking.  It all works because we agree, somehow, upon what they mean.  Of course, over generations and cultures, and between even different languages, some things get crossed up in translation, but it’s still pretty powerful.

The structure of writing, punctuation, and the Oxford comma, they all work because we agree.  We can look back and try to see the history of how the conventions have changed and sometimes find interesting connections.  Sometimes, there are artifacts from our past that just don’t really make sense anymore.  Either the language has evolved passed the usefulness, or the language adopted other conventions that conflict.

One example of this is the difference between its and it’s.  An apostrophe can be used in a conjunction and can also be used to show ownership.  Pretty simple rule to keep straight with its and it’s, but whose and who’s.  Why is it whose, with an e at the end?

According to my friendly neighborhood English teacher there was a great vowel shift, which can be read about here, where basically, people in around the 15th century wanted to sound fancy and wanted their words to look fancy when written.  So the letters e and b were added to words like whose and thumb.

Maybe we should take this one step further, and use thumbe.  Sounds good, right?

But then, there’s the old rule, i before e except after c, except in words like neighbor and weight, and in the month of May, or on a Tuesday.  Weird, er, wierd, right?

All said, not a big deal because those tricks of language will not cause a student to be illiterate.  A student can mix those things up and still have access to symbolism and writing and higher level understanding of language.

There are some conventions in math that work this way, too.  There are things that simply are a hold-over of how things were done a long time ago.  The convention carries with it a history, that’s what makes it powerful.  But sometimes the convention needs to change because it no longer is useful at helping making clear the intentions of the author.

One of the issues with changing this convention is that the people who would be able to make such changes are so well versed in the topic, they don’t see it as an issue.  Or, maybe they do, but they believe that since they got it right, figured it out, so could anybody else.

There is one particular thing in math that stands out as particularly problematic.  The radical symbol, it must go!  There’s a much more elegant method of writing that is intuitive and makes sense because it ties into other, already established ways of writing mathematics.

But, before I get into that exactly, let me say there’s an ancillary issue at hand. It starts somewhere in 3rd or 4th grade here in the US and causes problems that are manifested all the way through Calculus.  Yup, it’s multiplication.

Let me take just a moment to reframe multiplication by whole numbers and then by fractions for you so that the connection between those things and rational exponents will be more clear.

Consider first, 3 × 5, which is of course 15.  But this means we start with a group that has three and add it to itself five times.

Much like exponents are repeated multiplication, multiplication is repeated addition.  A key idea here is that with both we are using the same number over and again, the number written first.  The second number describes how many times we are using that first number.

Now of course 3 × 5 is the same as 5 × 3, but that doesn’t change the meaning of the grouping as I described.

3 + 3 + 3 + 3 + 3 = 3 × 5

Now let’s consider how this works with a fraction.

15 × ⅕.  The denominator describes how many times a number has been added to itself to arrive at fifteen.  We know that’s three.  So 15 × ⅕ = 3.

3 + 3 + 3 + 3 + 3 = 15

Three is added to itself five times to arrive at fifteen.

Let’s consider 15 × ⅖, where the five in the denominator is saying we are looking for a number that’s been repeatedly added to get to 15, but exactly added to itself 5 times.

In other words, what number can you add to itself to arrive at 15 in five equal steps?  That’s ⅕.

The two in the numerator is asking, how far are you after the 2nd step?

3 + 3 + 3 + 3 + 3 = 15

The second step is six.

Another way to see this is shown below:

3 →6→9→12→15

Step 1: 3 → Step 2: 6 → Step 3: 9 → Step 4: 12 → Step 5: 15

Thinking of it this way we can easily see that 15 × ⅘ is 12 and 15 × 5/5 is 15.  All of this holds true and consistent with the other ways we thinking about fractions.

So we see how multiplication is repeated addition of the same number and how fractions ask questions about the number of repeats taken to arrive at an end result.

Exponents are very similar, except instead of repeated addition they are repeated multiplication.

Multiplication:  3 × 5 = 3 + 3 + 3 + 3 + 3

Exponents:  3⁵ = 3 × 3 × 3 × 3 × 3

Do you see how the trailing numbers describe how many of the previous number there exists, but the way the trailing number is written, as normal text or a superscript (tiny little number up above), informs the reader of the operation?

Pretty cool, eh?

Just FYI, 3 times itself 5 times is 243.

15 × ⅕ = 3, because 3 + 3 + 3 + 3 + 3 = 15.  That is, three plus itself five times is fifteen.

2431/5= 3 because 3 × 3 × 3 × 3 × 3 = 243.  That is, three times itself five times is two hundred and forty three.

You might be thinking, big deal... but watch how much simpler this way of thinking about rational exponents is with something like an exponent of ⅗.  Let’s look at this like steps:

3 × 3 × 3 × 3 × 3 = 243

3→9→27→81→243

Step one is three, step two is nine, step three is twenty-seven, the fourth step is eighty one, and the fifth step is 243.  So, 2433/5is asking, looking at the denominator first, what number multiplied by itself five times is 243, and the numerator says, what’s the third step?  Twenty-seven, do you see?

Connecting the notation this way makes it simple and easy to read.  The only tricky parts would be the multiplication facts.

## Why Does the Order of Operations Work?

Why does the order of operations help us arrive at the correct calculation?  How does it work, why is it PEMDAS?  Why not addition first, then multiplication then groups, or something else?

I took it upon myself to get to the bottom of this question because I realized that I am so familiar with manipulating Algebra and mathematical expressions that I know how to navigate the pitfalls.  That instills a sense of conceptual knowledge, but that was a false sense.  I did not understand why we followed PEMDAS, until now.

By fully understanding why we follow the order of operations we can gain insight into the way math is written and have better abilities to understand and explain ourselves to others.  So, if you’re a teacher, this is powerful because you’ll have a foundation that can be shared with others without them having to trip in all of the holes.  If you’re a student, this is a great piece of information because it will empower you to see mathematics more clearly.

Let’s start with the basics, addition and subtraction.  First off, subtraction is addition of negative integers.  We are taught “take-away,” but that’s not the whole story.  Addition and subtraction are the same operation.  We do them from left to right as a matter of convention, because we read from left to right.

But what is addition?  In order to unpack why the order of operations works we must understand this most basic question.  Well, addition, is repeated counting, nothing more.  Suppose you have 3 vials of zombie vaccine and someone donates another 6 to your cause.  Instead of laying them out and counting from the beginning, we can combine six and three to get the count of nine.

And what is 9?  Nine is | | | | | | | | |.

What about multiplication?  That’s just skip counting.  For example, say you now have four baskets, each with 7 vials of this zombie vaccine.  Four groups of seven is twenty-eight.  We could lay them all out and count them, we could lay them all out and add them, or we could multiply.

Now suppose someone gave you another 6 vials.  To calculate how many vials you had, you could not add the six to one of the baskets, and then multiply because not all of the baskets would have the same amount.  When we are multiplying we are skip counting by the same amount, however many times is appropriate.

For example:

4 baskets of 7 vials each

4 × 7 = 28

or

7 + 7 + 7 + 7 = 28

or

[ | | | | | | | ]   [ | | | | | | | ]   [ | | | | | | | ]   [ | | | | | | | ]

Consider the 4 × 7 method of calculation.  We are repeatedly counting by 7.  If we considered the scenario where we received an additional six vials, it would look like this:

4 × 7 + 6

If you add first, you end up with 4 baskets of 13 vials each, which is not the case.  We have four baskets of 7 vials each, plus six more vials.

Addition compacts the counting.  Multiplication compacts the addition of same sized groups of things.  If you add before you multiply, you are changing the size of the groups, or the number of the groups, when in fact, addition really only changes single pieces that belong in those groups, but not the groups themselves.

Division is multiplication by the reciprocal.  In one respect it can be thought of as skip subtracting, going backwards, but that doesn’t change how it fits with respect to the order of operations.  It is the same level of compacting counting as multiplication.

Exponents are repeated multiplication, of the same thing!

Consider:

3 + 6

4 × 7 = 7 + 7 + 7 + 7

74 = 7 × 7 × 7 × 7

This is one layer of further complexity.  Look at 7 × 7.  That is seven trucks each with seven boxes.  The next × 7 is like seven baskets per box.  The last × 7 is seven vials in each basket.

The 4 × 7 is just four baskets of seven vials.

The 3 + 9 is like having 3 vials and someone giving you six more.

So, let’s look at:

9 + 4 × 74

Remember that the 74 is seven trucks of seven boxes of seven baskets, each with seven vials!  Multiplication is skip counting and we are skip counting repeatedly by 7, four times.

Now, the 4 × 74 means that we have four groups of seven trucks, each with seven boxes, each with seven baskets containing seven vials of zombie vaccine each.

The 9 in the front means we have nine more vials of zombie vaccine.  To add the nine to the four first, would increase the number of trucks of vaccine we have!

Now parentheses don’t have a mathematical reason to go first, not any more than why we do math from left to right.  It’s convention.  We all agree that we group things with the highest priority with parentheses, so we do them first.

(9 + 4) × 74

The above expression means we have 9 and then four more groups of seven trucks with seven boxes with …  and so on.

Exponents are compacted multiplication, but the multiplication is of the same number.  The multiplication is compacting the addition.  The addition is compacting the counting.  Each layer kind of nests or layers groups of like sized things, allowing us to skip over repeated calculations that are the same.

We write mathematics with these operations because it is clean and clear.  If we tried to write out 35, we would have a page-long monstrosity.  We can perform this calculation readily, but often write the expression instead of the calculation because it is cleaner.

The order of operations respects level of nesting, or compaction, of exponents, multiplication and addition, as they related to counting individual things.  The higher the order of arithmetic we go, up to exponents, the higher the level of compaction.

Exponents are repeated multiplication, and multiplication is repeated addition, and addition is repeated, or skip counting.  We group things together with all of these operations, but how that grouping is done must be done in order when perform the calculations.

## Cube Roots and Higher Order Roots

other roots

Cube Roots
and

Square roots ask what squared is the radicand. A geometric explanation is that given the area of a square, what’s the side length? A geometric explanation of a cube root is given the volume of a cube, what’s the side length. The way you find the volume of a cube is multiply the length by itself three times (cube it).

The way we write cube root is similar to square roots, with one very big difference, the index.

$\begin{array}{l}\sqrt{a}\to \text{\hspace{0.17em}}\text{\hspace{0.17em}}square\text{\hspace{0.17em}}\text{\hspace{0.17em}}root\\ \sqrt[3]{a}\to \text{\hspace{0.17em}}\text{\hspace{0.17em}}cube\text{\hspace{0.17em}}\text{\hspace{0.17em}}root\end{array}$

There actually is an index for a square root, but we don’t write the two. It is just assumed to be there.

Warning: When writing cube roots, or other roots, be careful to write the index in the proper place. If not, what you will write will look like multiplication and you can confuse yourself. When writing by hand, this is an easy thing to do.

$3\sqrt{8}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[3]{8}$

To simplify a square root you factor the radicand and look for the largest perfect square. To simplify a cubed root you factor the radicand and find the largest perfect cube. A perfect cube is a number times itself three times. The first ten are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1,000.

Let’s see an example:

Simplify:

$\sqrt[3]{16}$

Factor the radicand, 16, find the largest perfect cube, which is 8.

$\sqrt[3]{8}×\sqrt[3]{2}$

The cube root of eight is just two.

$2\sqrt[3]{2}$

The following is true,

$\sqrt[3]{16}=2\sqrt[3]{2}$,

only if

${\left(2\sqrt[3]{2}\right)}^{3}=16$

Arithmetic with other radicals, like cube roots, work the same as they do with square roots. We will multiply the rational numbers together, then the irrational numbers together, and then see if simplification can occur.

${\left(2\sqrt[3]{2}\right)}^{3}={2}^{3}×{\left(\sqrt[3]{2}\right)}^{3}$

Two cubed is just eight and the cube root of two cubed is the cube root of eight.

${2}^{3}×{\left(\sqrt[3]{2}\right)}^{3}=8×\left(\sqrt[3]{2}\right)\left(\sqrt[3]{2}\right)\left(\sqrt[3]{2}\right)$

${2}^{3}×{\left(\sqrt[3]{2}\right)}^{3}=8×\sqrt[3]{2\cdot 2\cdot 2}$

${2}^{3}×{\left(\sqrt[3]{2}\right)}^{3}=8×\sqrt[3]{8}$

The cube root of eight is just two.

$8×\sqrt[3]{8}=8×2$

${\left(2\sqrt[3]{2}\right)}^{3}=16$

Negatives and cube roots: The square root of a negative number is imagery. There isn’t a real number times itself that is negative because, well a negative squared is positive. Cubed numbers, though, can be negative.

$-3×-3×-3=-27$

So the cube root of a negative number is, well, a negative number.

$\sqrt[3]{-27}=-3$

Other indices (plural of index): The index tells you what power of a base to look for. For example, the 6th root is looking for a perfect 6th number, like 64. Sixty four is two to the sixth power.

$\sqrt[6]{64}=2\text{​}\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}because\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{2}^{6}=64.$

A few points to make clear.

·         If the index is even and the radicand is negative, the number is irrational.

·         If the radicand does not contain a factor that is a perfect power of the index, the number is irrational

·         All operations, including rationalizing the denominator, work just as they do with square roots.

Rationalizing the Denominator:

Consider the following:

$\frac{9}{\sqrt[3]{3}}$

If we multiply by the cube root of three, we get this:

$\frac{9}{\sqrt[3]{3}}\cdot \frac{\sqrt[3]{3}}{\sqrt[3]{3}}=\frac{9\sqrt[3]{3}}{\sqrt[3]{9}}$

Since 9 is not a perfect cube, the denominator is still irrational. Instead, we need to multiply by the cube root of nine.

$\frac{9}{\sqrt[3]{3}}\cdot \frac{\sqrt[3]{9}}{\sqrt[3]{9}}=\frac{9\sqrt[3]{9}}{\sqrt[3]{27}}$

Since twenty seven is a perfect cube, this can be simplified.

$\frac{9}{\sqrt[3]{3}}=\frac{9\sqrt[3]{9}}{3}$

And always make sure to reduce if possible.

$\frac{9}{\sqrt[3]{3}}=3\sqrt[3]{9}$

This is a bit tricky, to be sure. The way the math is written does not offer us a clear insight into how to manage the situation. However, the topic we will see next, rational exponents, will make this much clearer.

Practice Problems:

Simplify or perform the indicated operations:

$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[4]{64}\\ \\ \\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[3]{9}+4\sqrt[3]{9}\\ \\ \\ \\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[3]{9}×4\sqrt[3]{9}\\ \\ \\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[5]{64}\\ \\ \\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\sqrt{7}}{\sqrt[3]{7}}\end{array}$

## Addition and Subtraction of Square Roots

Mathematical Operations and Square Roots

Part 1

In this section we will see why we can add things like $5\sqrt{2}+3\sqrt{2}$ but cannot add things like $2\sqrt{5}+2\sqrt{3}$. Later we will see how multiplication and division work when radicals (square roots and such) are involved.

Addition and Subtraction: Addition is just repeated counting. The expression $5\sqrt{2}$ means $\sqrt{2}+\sqrt{2}+\sqrt{2}+\sqrt{2}+\sqrt{2}$, and the expression So if we add those two expressions, $5\sqrt{2}+3\sqrt{2},$ we get $8\sqrt{2}$ . Subtraction works the same way.

Consider the expression $2\sqrt{5}+2\sqrt{3}$. This means $\sqrt{5}+\sqrt{5}+\sqrt{3}+\sqrt{3}.$ The square root of five and the square root of three are different things, so the simplest we can write that sum is $2\sqrt{5}+2\sqrt{3}$.

A common way to describe when square roots can or cannot be added (or subtracted) is, “If the radicands are the same you add/subtract the number in front.” This is not a bad rule of thumb, but it treats square roots as something other than numbers.

$5×3+4×3=9×3$

The above statement is true. Five groups of three and four groups of three is nine groups of three.

$5\sqrt{3}+4\sqrt{3}=9\sqrt{3}$

The above statement is also true because five groups of the numbers squared that is three, plus four more groups of the same number would be nine groups of that number.

However, the following cannot be combined in such a fashion.

$3×8+5×2$

While this can be calculated, we cannot add the two terms together because the first portion is three $–$ eights and the second is five $–$ twos.

$3\sqrt{8}+5\sqrt{2}$

The same situation is happening here.

Common Mistake: The following is obviously wrong. A student learning this level of math would be highly unlikely to make such a mistake.

$7×2+9×2=16×4$

Seven $–$ twos and nine $–$ twos makes a total of sixteen $–$ twos, not sixteen $–$ fours. You’re adding the number of twos you have together, not the twos themselves. And yet, this is a common thing done with square roots.

$7\sqrt{2}+9\sqrt{2}=16\sqrt{4}$

This is incorrect for the same reason. The thing you are counting does not change by counting it.

Explanation: Why can you add $5\sqrt{2}+3\sqrt{2}$? Is that a violation of the order of operations (PEMDAS)? Clearly, the five and square root of two are multiplying, as are the three and the square root of two. Why does this work?

Multiplication is a short-cut for repeated addition of one particular number. Since both terms are repeatedly adding the same thing, we can combine them.

But if the things we are repeatedly adding are not the same, we cannot add them together before multiplying.

What About Something Like This: $3\sqrt{40}-9\sqrt{90}$?

Before claiming that this expression cannot be simplified you must make sure the square roots are fully simplified. It turns out that both of these can be simplified.

$3\sqrt{40}-9\sqrt{90}$

$3\cdot \sqrt{4}\cdot \sqrt{10}-9\cdot \sqrt{9}\cdot \sqrt{10}$

The dot symbol for multiplication is written here to remind us that all of these numbers are being multiplied.

$3\cdot \sqrt{4}\cdot \sqrt{10}-9\cdot \sqrt{9}\cdot \sqrt{10}$

$3\cdot 2\cdot \sqrt{10}-9\cdot 3\cdot \sqrt{10}$

$6\sqrt{10}-27\sqrt{10}$

$-21\sqrt{10}$

What About Something Like This: $\sqrt{7+7}$ versus $\sqrt{7}+\sqrt{7}.$

Notice that in the first expression there is a group, the radical symbol groups the sevens together. Since the operation is adding, this becomes:

$\sqrt{7+7}=\sqrt{14}$.

Since the square root of fourteen cannot be simplified, we are done.

The other expression becomes:

$\sqrt{7}+\sqrt{7}=2\sqrt{7}.$

Summary: If the radicals are the same number, the number in front just describes how many of them there are. You can combine (add/subtract) them if they are the same number. You are finished when you have combined all of the like terms together and all square roots are simplified.

Practice Problems: Perform the indicated operation.

$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{25}-5\sqrt{5}+5\\ \\ \\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{48}+3\sqrt{3}\\ \\ \\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\sqrt{75}+8\sqrt{24}+\sqrt{75}\\ \\ \\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{200}+8\sqrt{8}-2\sqrt{32}\\ \\ \\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}-2\sqrt{98}+16\sqrt{2}\end{array}$

## What Do Grades Really Mean?

The following is highly contentious.  Many of the situations discussed here should ultimately be considered on an individual basis.  The purpose of this is not to create a rubber-stamp solution to all problems that arise with grade assignment and student ability and or performance, but is to provide a general framework so that those individual decisions can be made in fairness and with respect to what is best for the student.

In a previous post I asked about a student in summer school that obviously knew Algebra 1 (he earned 100% on his quizzes and tests), but failed during the year because he didn’t do his classwork.  The question is, Does he deserve to fail Algebra 1?

When you flip the situation around it is equally interesting.  There are many kids who work hard, but do not really understand or learn the math.  Do they deserve to pass based on the merits of effort?

The real issue with both of these situations is what grades mean, or what should they mean.  When I worked at Cochise Community College I adopted their definition of letter grades which is described below:

A – Mastery

B – Fluency

C – Proficiency

D – Lacking Proficiency

Those are clean and inoffensive definitions of grades.  A student with an A has mastered the material.  To be fluent means you can navigate the materials but not without error.  To be proficient means you can get the job done, but there are some gaps in ability, but the student can demonstrate a measurable level of command of all of the objectives. Students who earn a D are not able to demonstrate proficiency.

A student who struggles with the material does not deserve an A, even if they worked harder than those who earned an A.  This might seem unfair, but unless the objective of the class is to teach the value of hard work, to reward the hardworking, but barely proficient, student with a label of mastery is to cheat the student and cheapen the merit of your class.

Do these definitions mean that a lazy kid that get 95% on the final exam deserves an A, but that a hard working kid that gets a 52% on the same final deserves an F?  I say, with a few qualifications, yes.

Is this really fair to the student who works hard but has not yet realized an appropriate level of mastery to be awarded a passing grade? (I used the phrase, “has not yet,” instead of, “cannot,” to acknowledge the belief that students can learn, and if they are motivated and working, the only question will be the time scale of when they learn the material.)

I would say, for a math class, that the best thing that can happen is they are awarded the appropriate grade, an F.  Consider if this student is given a passing grade and the class is a prerequisite course?  They’re truly set up for failure in the subsequent class.

There is perhaps no worse example of bad teaching that remains within legals bounds than to inappropriately assign grades to students.  If a student deserves a C based on ability, but is given an A based on effort, they will believe they are doing everything right and do not need to improve in order to achieve similar success in subsequent courses.

But to give a student who possesses mastery a failing grade in a class because of lack of work ethic is to teach the student that passing classes is a matter of compliance.  Behave and you’ll be rewarded.  Those kids are taught that grades are not a reflection of knowledge or ability, and that means that education is not about learning.  To me, this is an injustice.

I do not believe in the efficacy of these objective lessons.  That would be, failing a student based on the notion that they do not deserve to pass because they are lazy. I believe that given meaningful and challenging opportunities, most of these highly intelligent, but seemingly lazy, students will show themselves to be hard working with amazing focus and direction and incredible capacity for quality work.

What about percentages.  Is it appropriate that an 80% is a B, if a B means fluency?

When I first began teaching I would have said, absolutely, a student does not deserve an A if they scored an 87% on their test.  Since then I’ve changed my mind.  Some topics require higher than 90% accuracy to be awarded an A, while with other topics, mastery might be far below 90%.

The level of complexity, variability of solutions and length of assessment all must be considered.  This is why sometimes a grading rubric is far superior to assigning grades based on a percentage of correctness or completion.

I teach a curriculum that is designed and tested by Cambridge University, the IGCSE test is what students take.  They have a very different way of assigning and defining grades than we use here in the United States.  Without going into details about how they do the specifics, they assign large portions of credit based on evidence of appropriate thinking.  In other words, if a student demonstrates understanding they will receive passing credit.  But, to achieve a high grade, mastery is truly measured.  And yet, in math at least, the percentages of correctness for mastery are usually in the mid-70’s.  This is because the nature of the questions asked are often non-procedural and the method of solution is not clear, students cannot be trained on how to answer the questions they face on IGCSE exams.

How a student can earn a grade varies, or should, depending on subject and age, and perhaps even minor topic within the subject.  I believe that separating student work into weighted categories is an appropriate method of helping make transparent to the student how their grade will be assigned.  It also by-passes the tricky question of, “What is a point?”  For me, a homework assignment is worth 5 points, they’re assigned daily, except Fridays, for a total of 20 points for the week.  Yet, a quiz might only be worth 12 points, but will be a far more accurate representation of student’s ability on the topic.

By assigning weights to the categories, this can be easily balanced.  This begs the question, how do you weight the categories?

But what about the student who works, performs all assigned tasks, but can only demonstrate a level of understanding best described as “Lacking Proficiency?”  Shouldn’t hard work be rewarded?

And whatever your beliefs on these questions, would your opinion change depending on the age of the student, or perhaps the subject?  Should a Chemistry student be rewarded for effort in the same way they’d be rewarded for effort in a Dance class?

At some point, nobody cares about potential or effort.  If a child’s mother wants his room clean, she knows he has the potential to clean it, but if he fails to do so, the potential matters not.  And if he’s really trying to get it done, but cannot master the discipline to carry through the task, does the effort really matter?

Here is how I set up my grades for high school.  It is nuanced and complicated, but I’ll give the outline.  Note that for college classes I use a different system.

In high school I weigh categories of grades and have changed the percentages and categories over time until I settled on what seems to work best.  These work for my students because it seems to motivate the lazy-smart students and also rewards the hardworking – low aptitude student, because if they remain persistent, they will learn.

Tests – 40%
Quizzes – 25%
Homework – 25%
Other – 10%

I believe extra credit should be awarded for students that perhaps help others, or for extraordinary performance.  However, a student should NOT be allowed to raise their grade through extra credit.  That is, at the end of the term a student is given a pile of work, that if performed, will raise their grade.  This is bad teaching!

The difference between a quiz and a test is similar to the difference between a doctor’s check-up versus an autopsy.  The quiz is a chance to see how things are going and adjust accordingly.  The test is final.  In high school I award credit for homework based on completion, but do not accept late homework.

Rewarding Effort?

While I wish that effort equaled success, it doesn’t always work that way…depending on how you define success.  For example, I can try as hard as possible to paint a world-famous landscape, but will likely fail if my measure of success is producing a world-famous piece of art. That said, I believe there is a reward beyond measure only discovered with true effort.  Our potential, our best, is not static, it changes.  It changes in respect to our current level of effort.  We can never fulfill our potential, you see.  It is always slightly above how hard we are trying.  So, if you’re not really trying, your potential decreases, but if you’re pushing your limits, the limits themselves stretch.  That is the real downfall of those with an inherent talent that never learn to push themselves.  Their potential decreases, dropping down to just higher than their level of effort.

I greatly reward effort, encourage it and makes positive examples of how effort promotes success.  However, I do not assign grades to effort.  How hard someone needs to try in a given subject to be successful varies entirely upon the student’s aptitude.  And suppose you have a truly gifted student, they could be great, if they learn to work hard, right?

Well, perhaps, but there’s more than work ethic involved in greatness.  What role does passion play?  Take a great young musician and over-structure their training and practice, they’ll burn out.  You’ll snuff their passion.

I asked the boy whose situation started this whole conversation if he felt he deserved to be in summer school.  Before he answered I explained that I didn’t have an expected answer, I didn’t really know if he belonged in summer school or not.  Without hesitation, he said he did deserve summer school, because, he said, he was lazy.

So maybe the kid will learn that if he’s lazy he gets punished.  But he also learns that grades are arbitrary, with respect to ability.

I do not like objective lessons, do not believe them to be effective.  I prefer a punishment that fits the crime, but also one that redirects the offender, allows them to correct their action.

I cannot say in this child’s case specifically, I was not there and I am not judging his teacher, but perhaps a quicker punishment that redirected him could have also taught him that being lazy was unacceptable and at the same time also allowed him to see grades as a reflection of his abilities.

All that said, this is highly contentious and varies incredibly depending on particular situations of students.

Let me know what you think, agree or disagree.  Leave me a comment.

## Exponents Part 2

two

Exponents Part 2

Division

In the previous section we learned that exponents are repeated multiplication, which on its own is not tricky. What makes exponents tricky is determining what is a base and what is not for a given exponent. It is imperative that you really understand the material from the previous section before tackling what’s next. If you did not attempt the practice problems, you need to. Also watch the video that review them.

In this section we are going to see why anything to the power of zero is one and how to handle negative exponents, and why they mean division.

What Happens with Division and Exponents?

Consider the following expression, keeping in mind that the base is arbitrary, could be any number (except zero, which will be explained soon).

${3}^{5}$

This equals three times itself five total times:

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

Now let’s divide this by 3. Note that 3 is just 31.

$\frac{{3}^{5}}{{3}^{1}}$

If we write this out to seek a pattern that we can use for a short-cut, we see the following:

$\frac{{3}^{5}}{{3}^{1}}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{3}$

If you recall how we explored reducing Algebraic Fractions, the order of division and multiplication can be rearranged, provided the division is written as multiplication of the reciprocal. That is how division is written here.

$\frac{{3}^{5}}{{3}^{1}}=\frac{3}{3}\cdot \frac{3\cdot 3\cdot 3\cdot 3}{1}$

And of course 3/3 is 1, so this reduces to:

$\frac{{3}^{5}}{{3}^{1}}=3\cdot 3\cdot 3\cdot 3={3}^{4}$

The short-cut is:

$\frac{{3}^{5}}{{3}^{1}}={3}^{5-1}={3}^{4}$

That is, if the bases are the same you can reduce. Reducing eliminates one of the bases that is being multiplied by itself from both the numerator and the denominator. A general form of the third short-cut is here:

Short-Cut 3: $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$

This might seem like a worthless observation, but this will help articulate the very issue that is going to cause trouble with exponents and division.

$\frac{{3}^{5}}{{3}^{1}}={3}^{5}÷{3}^{1}$ .

But that is different than

${3}^{1}÷{3}^{5}$

The expression above is the same as

$\frac{{3}^{1}}{{3}^{5}}$

This comes into play because

$\frac{{3}^{1}}{{3}^{5}}={3}^{1-5}$,

and 1 $–$ 5 = -4.

Negative Exponents?

In one sense, negative means opposite. Exponents mean multiplication, so a negative exponent is repeated division. This is absolutely true, but sometimes difficult to write out. Division is not as easy to write as multiplication.

Consider that 3-4 is 1 divided by 3, four times. 1 ÷ 3 ÷ 3 ÷ 3 ÷ 3. But if we rewrite each of those ÷ 3 as multiplication by the reciprocal (1/3), it’s must cleaner and what happens with a negative exponent is easier to see.

$1÷3÷3÷3÷3\to 1\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}$

This is classically repeated multiplication. While one times itself any number of times is still one, let’s go ahead and write it out this time.

$1\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\to 1\cdot {\left(\frac{1}{3}\right)}^{4}$

This could also be written:

$1\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\to 1\cdot \frac{{1}^{4}}{{3}^{4}}$

The second expression is easier, but both are shown here to make sure you see they are the same.

Since 1 times 14 is just one, we can simplify this further to:

$1\cdot \frac{{1}^{4}}{{3}^{4}}=\frac{1}{{3}^{4}}.$

Negative exponents are repeated division. Since division is hard to write and manipulate, we will write negative exponents as multiplication of the reciprocal. In fact, if instructions say to simplify, you cannot have a negative exponent in your final answer. You must rewrite it as multiplication of the reciprocal. Sometimes that can get ugly. Consider the following:

$\frac{b}{{a}^{-5}}$

To keep this clean, let us consider separating this single fraction as the product of two rational expressions.

$\frac{b}{{a}^{-5}}=\frac{b}{1}\cdot \frac{1}{{a}^{-5}}$

The b is not a problem here, but the other rational expression is problematic. We need to multiply by the reciprocal of $\frac{1}{{a}^{-5}}$, which is just a5.

$\frac{b}{{a}^{-5}}=\frac{b}{1}\cdot \frac{{a}^{5}}{1}={a}^{5}b$.

This can also be considered a complex fraction, the likes of which we will see very soon. Let’s see how that works.

$\frac{b}{{a}^{-5}}$

Note: ${a}^{-5}=\frac{1}{{a}^{5}}$

Substituting this we get:

$\frac{b}{\frac{1}{{a}^{5}}}$

This is b divided by 1/a5.

$b÷\frac{1}{{a}^{5}}$

Let’s multiply by the reciprocal:

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

Now we will rewrite it in alphabetical order (a good habit, for sure).

${a}^{5}b$

Let us consider one more example before we make our fourth short-cut. With this example we could actually apply our second short-cut, but it will not offer much insight into how these exponents work with division.

This is the trickiest of all of the ways in which exponents are manipulated, so it is worth the extra exploration.

$\frac{2{x}^{-2}{y}^{-5}z}{{2}^{-2}x{y}^{3}{z}^{-5}}$

As you see we have four separate bases. In order to simplify this expression we need one of each base (2, x, y, z), and all positive exponents. So let’s separate this into the product of four rational expressions, then simplify each.

$\frac{2{x}^{-2}{y}^{-5}z}{{2}^{-2}x{y}^{3}{z}^{-5}}\to \frac{2}{{2}^{-2}}\cdot \frac{{x}^{-2}}{x}\cdot \frac{{y}^{-5}}{{y}^{3}}\cdot \frac{z}{{z}^{-5}}$

The base of two first:

$\frac{2}{{2}^{-2}}\to 2÷{2}^{-2}$

We wrote it as division. What we will see is dividing is multiplication by the reciprocal, and then the negative exponent is also dividing, which is multiplication by the reciprocal. The reciprocal of the reciprocal is just the original. But watch what happens with the sign of the exponent.

First we will rewrite the negative exponent as repeated division.

$2÷\frac{1}{{2}^{2}}$

Now we will rewrite division as multiplication by the reciprocal.

$2\cdot {2}^{2}={2}^{3}$

Keep in mind, this is the same as 23/1.

We will offer similar treatment to the other bases.

Consider first $\frac{{x}^{-2}}{x}=\frac{{x}^{-2}}{1}\cdot \frac{1}{x}$

Negative exponents are division, so:

$\frac{{x}^{-2}}{x}=\frac{{x}^{-2}}{1}\cdot \frac{1}{x}$

Notice the x that is already dividing (in the denominator) does not change. It has a positive exponent, which means it is already written as division.

$\frac{{x}^{-2}}{1}\cdot \frac{1}{x}\to \frac{1}{{x}^{2}}\cdot \frac{1}{x}=\frac{1}{{x}^{3}}$

This is exactly how simplifying the y and z will operation.

$\frac{{2}^{3}}{1}\cdot \frac{1}{{x}^{2}\cdot x}\cdot \frac{1}{{y}^{5}\cdot {y}^{3}}\cdot \frac{z\cdot {z}^{5}}{1}$

Putting it all together:

$\frac{2{x}^{-2}{y}^{-5}z}{{2}^{-2}x{y}^{3}{z}^{-5}}=\frac{{2}^{3}{z}^{6}}{{x}^{3}{y}^{8}}$.

Short-Cut 4: Negative exponents are division, so they need to be rewritten as multiplication by writing the reciprocal and changing the sign of the exponent. The last common question is what happens to the negative sign for the reciprocal? What happens to the division sign here: $3÷5=3×\frac{1}{5}$. When you rewrite division you are writing it as multiplication. Positive exponents are repeated multiplication.

${a}^{-m}=\frac{1}{{a}^{m}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{{a}^{-m}}={a}^{m}$

This is the second to last thing we need to learn about exponents. However, a lot of practice is required to master them fully.

To see why anything to the power of zero is one, let’s consider:

${3}^{5}$

This equals three times itself five total times:

${3}^{5}=3"#x22C5;3\cdot 3\cdot 3\cdot 3$

Now let’s divide this by 35.

$\frac{{3}^{5}}{{3}^{5}}$

Without using short-cut 3, we have this:

$\frac{{3}^{5}}{{3}^{5}}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3}=1$

Using short-cut 3, we have this:

$\frac{{3}^{5}}{{3}^{5}}={3}^{5-5}$

Five minutes five is zero:

${3}^{5-5}={3}^{0}$

Then 30 = 1.

Τhe 3 was an arbitrary base. This would work with any number except zero. You cannot divide by zero, it does not give us a number.

The beautiful thing about this is that no matter how ugly the base is, if the exponent is zero, the answer is just one. No need to simplify or perform calculation.

${\left(\frac{{3}^{2x-1}\cdot {e}^{\pi i}}{\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}}\right)}^{0}=1$

Let’s take a quick look at all of our rules so far.

 Short-Cut Example ${a}^{m}\cdot {a}^{n}={a}^{m+n}$ ${5}^{8}\cdot 5={5}^{8+1}={5}^{9}$ ${\left({a}^{m}\right)}^{n}={a}^{mn}$ ${\left({7}^{2}\right)}^{5}={7}^{10}$ $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ $\frac{{5}^{7}}{{5}^{2}}={5}^{7-2}={5}^{5}$ ${4}^{-3}=\frac{1}{{4}^{3}}$ ${a}^{0}=1$ 50 = 1

Let’s try some practice problems.

Instructions: Simplify the following.

1. ${\left({2}^{8}\right)}^{1/3}$ 2. $3{x}^{2}\cdot {\left(3{x}^{2}\right)}^{3}$

3. $\frac{5}{{5}^{m}}$ 4. $\frac{{5}^{2}{x}^{-3}{y}^{5}}{{5}^{-3}{x}^{-4}{y}^{-5}}$

5. $7÷7÷7÷7÷7÷7÷7÷7$ 6. $9{x}^{2}y÷9{x}^{2}y$

7. $9{x}^{2}y÷\left(9{x}^{2}y\right)$ 8. ${\left({x}^{2}\cdot 2{x}^{6}\right)}^{2}\cdot {\left({x}^{2}\cdot 2{x}^{6}\right)}^{-2}$

9. ${\left({a}^{m}\right)}^{n}\cdot {a}^{m}$ 10. $\frac{{\left(3{x}^{2}+4\right)}^{2}}{{\left(3{x}^{2}+4\right)}^{3}}$

## 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 Smallest Things Can Cause Huge Problems for Students

preemptive

Pre-Emptive Explanation

It is often the case, for the mathematically-insecure, that the slightest point of confusion can completely undermine their determination. Consider a beginning Algebra student that is learning how to evaluate functions like:

$\begin{array}{l}f\left(x\right)=3x-{x}^{2}+1\\ f\left(2\right)\end{array}$

A confident student is likely to make the same error as the insecure student, but their reactions will be totally different. Below would be a typical incorrect answer that students will make:

$\begin{array}{l}f\left(2\right)=3\left(2\right)-{2}^{2}+1\\ f\left(2\right)=6+4+1\\ f\left(2\right)=11\end{array}$

The correct answer is 3, and the mistake is that -22 = -4, because it is really subtract two-squared. And when students make this mistake it provides a great chance to help them learn to read math, especially how exponents are written and what they mean.

Here’s what the students actually read:

$\begin{array}{l}f\left(x\right)=3x-{x}^{2}+1\\ f\left(2\right)=3\left(2\right)+{\left(-2\right)}^{2}+1\end{array}$

A confident student will be receptive to this without much encouragement from you. However, the insecure student will completely shut down, having found validation of their worst fears about their future in mathematics.

There are times when leaving traps for students is a great way to expose a misconception, and in those cases, preemptively trying to prevent them from making the mistake would actually, in the long run, be counter-productive. Students would likely be mimicking what’s being taught, but would never uncover their misconception through correct answer getting. Mistakes are a huge part of learning and good math teaching is not about getting kids to avoid wrong answers, but instead to learn from them.

But there are times when explaining a common mistake, rooted in some prerequisite knowledge, is worth uncovering ahead of time. This -22 squared is one of those things, in my opinion, that is appropriately explained before the mistakes are made.

## Math Can Not Be Taught, Only Learned

Math is something that cannot be taught, but can be learned.  Yet, math is taught in a top-down style, as if access to information will make a student successful, and remediation is rehearsal of that same information.  Earnest students copy down everything, exactly like the teacher has written on the board, but often still struggle and fail to comprehend what is happening.  I argue that if copying things down was a worthy exercise, why not just copy the textbook, cover to cover.  Of course such an activity would yield little benefit at all because math is a thing you do more than it is a thing you know.  Math is only partly knowledge based and the facts are rarely the issue that causes trouble for students.  I’d like to propose that you, either parent, student, administrator or teacher, considers math in a different light and perhaps with some adjustment the subject that caused such frustration will be a source of celebration.

There are many things that cannot be taught but can be learned.  A few examples are riding a bike, playing an instrument, creative writing and teaching.  Without question knowledge is a key component to all of these things, but it is rarely the limiting factor to success or performance.  Instead, the skill involved is usually the greatest limiting factor.  I argue that to learn these things a series of mistakes, incrementally increasing in complexity, must be made in order to learn.  Let’s see if this will make more sense with a pair of analogies.

First, watching someone perform something that is largely skill-based is of little use.  Consider driving a car.  A fifteen year old child has spent their entire life observing other people drive.  And yet, when they get behind the wheel for the first time, they’re hopelessly dangerous to themselves and all others on, or just near, the roads!

Second considering learning to ride a bike.  Sure, the parts of the bike are explained to the child, but they have to get on and try on their own.  The actually learning doesn’t really occur until the parent lets go (letting go is huge!) and the child rolls along for a few feet until they fall over.  Eventually they get the hang of the balance but then crash because they don’t know how to stop.  After they master braking they crash because they don’t know how to turn.  And then speed, terrain, and other obstacles get thrown in the mix.  Each skill must be mastered in order.  Preemptively explaining the skills, or practicing them out of context does not help the child learn to ride a bike.  They must make the mistakes, reflect, adjust and try again.

What a math teacher can provide is the information required, but more importantly feedback, direction and encouragement.  If a student understands that making mistakes isn’t just part of learning, but that a mistake is the opportunity to learn (and without it only imitation has occurred), and a teacher helps provide guidance, encouragement and feedback, then both parties will experience far greater success.  When a math teacher completes a problem for a student it is similar to an adult taking the bicycle away from the child and riding it for them.  When a student gives up on a problem, it’s as if they stopped the car and got out, allowing the adult to drive them home.

The job of math teacher is perhaps a bad arrangement of words.  Coach, mentor or sponsor is perhaps more appropriate.  There is no magic series of words, chanted under any circumstance, that will enlighten a struggling student.  The frustration making mistakes should be cast in a different light, a positive light.  The responsibility of learning is entirely on the student.  They cannot look to teachers, friends or tutors for much beyond explanation of facts.

In a future post I will explain how too much direction and top-down teaching of math promotes failure of retention and inability to apply skills in new applications.  But for now, please consider that math cannot be taught.  A teacher cannot teach it, but can help a student to learn.