# Abel's Proof: A Gentle Introduction to Mathematics

Mathematics Can Be Interesting... Really

Mathematics is not something that we normally associate with sublime beauty. The way mathematics is taught, it is a subject that many if not most people would prefer to avoid. Just open any Dover math book that you can find your local book store and you may be hard pressed to find any passage that makes any sense. With its barrage of definitions, lemmas, theorems, and QED's, it is a wonder that it claims to be written in English.

And yet, if you learned the details behind all those proofs. If you start being able to distinguish between textbook lemmas and the good stuff, there is a surprising excitement that emerges. It is like looking at a Picasso, getting annoyed by the abstraction, and then suddenly realizing that the artwork represents a maximum of effect with a minimum of detail. In other words, it is elegance of the highest level.

In today's entry, I will walk you through one of the highest achievements of mathematics: Abel's Impossibility Proof. I will work hard to make the journey pleasant and gentle. If you have any questions, feel free to post them to this hub. To provide the context for the proof, it is necessary to explore that some important ideas from the past.

A Very Short History of Math

All human societies have words for simple numbers. It takes deep mathematical insight to realize that 0 is also a number. With this realization, it becomes possible to represent (or at least to approximate) all numerical values with a finite set of digits. We call 0 to 9, the Arabic numerals but they were pretty much invented in India (in fact, they should be called Hindu-Arabic numerals). They were later popularized by the great Arab mathematicians (the word algebra after all is Arabic in origin). The Greeks and Romans had a plenitude of numbers and this made even basic multiplication almost impossible. The Arabic numerals greatly changed the possibilities for doing advanced mathematics.

Another important innovation was the invention of algebraic notation where we represent an unknown value by a letter such as x. The ability to state mathematical relationships as formulas greatly improves the ability to introduce abstractions. Consider the state of algebra before the use of 0 and the use of algebraic notation. The ancient Greek mathematician Diophantus (around 200AD) is often considered the father of algebra. He wrote a classic book (the translated book is available online here) where he asked questions about unknowns values. Since he didn't have the arabic numerals and the Greeks were partial to geometry, he asked these questions using geometric language. For example, here's a problem that he posed: find three squares where one is equal to the sum of the other two.

Diophantus's question becomes easier to solve if we use algebraic notation. Just as 0 enabled us to simplify the amount of numbers we need, algebraic notation enables us to restate mathematical questions in a simpler way.

First, we can treat unknown values as numbers. We can call them x,y,z. The first person to do this was Muhammad ibn Musa al-Khwarizmi (780 - 850 AD) in his famous book Al-Jabr (Although the full name in English is The Compendious Book on Calculations by Completion and Balancing). Al-Khwarizmi is also considered by many to be the father of Algebra. In addition to x,y,z, he used notation for exponents. An unknown square becomes x2. If you would like a gentle introduction to powers and exponents, review here before continuing.

So, now, armed with letters for unknowns and notation for exponents, we can solve Diophantus's problem. "Find three suqare where one is equal to the sum of the other two" can be restated as find x,y,z such that x2 + y2 = z2. (By the way, the answer is 3,4,5 since 9 + 16 = 25.

Any expression like x2 + 3x + 2 is called a polynomial. Each grouping of x and a number is called a monomial. When you match a polynomial with an equal sign and another value, you get a mathematical equation. Al-Khwarizmi on a Soviet Union stamp issued on September 6, 1983

Using Algebraic Notation

With this notation, it now became possible for mathematics to take off. Al-Khwarizmi showed the world how to solve what are known as the quadratic equations. These are equation of one unknown value where no power is greater than 2. A typical example is: 5x2 + 3x + 2 = 10. Find x (the answer in this case is x=1). Al-Khwarizmi showed that there is a formula that solves all possible quadratic equations.

If you raise the power to 3 or 4, the problem is significantly harder. Among Western mathematicians, it was not until the very famous Italian mathematician Girolamo Cardano (1501 - 1576 AD) that a solution is found for equations of one unknown where no power is greater than 3. This equation, known as the cubic equation, turns out to be more difficult than the quadratic. But its solution led very quickly to the solution of quartic equations by Lodovico Ferrari (1522 - 1565 AD). Quartic equations are equations of one unknown value where no power is greater than 4. For mathematicians, these were very exciting times.

But if there's a general solution for equations with a power of 2, power of 3, and power 4, shouldn't there also be a general solution for equations with a power of 5? Despite the efforts of great geniuses of mathematics (Francois Viete, Rene Descartes, Isaac Newton, Leonhard Euler, Joseph Lagrange, and Carl Friedrich Gauss), no general solution to the 5th power (known as the quintic) was found.

The Reason It's So Hard: It Can't Be Done

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In 1799, an Italian medical doctor trained in mathematics proposed that quintic equations and any equations with powers greater than 5 had no general solution. Paolo Ruffini (1765 - 1822 AD) presented his argument in a book which he sent to all the famous mathematicians of his day. Not surprisingly, mathematicians were not ready to believe that the problem was impossible. They ignored his argument, his methods, and continued to look for a general solution.

In retrospect, poor Ruffini did have a small gap in his reasoning but other than this small gap which would later be filled by Niels Abel, his argument was sound. He was an outsider but he was right and no one listened. In fairness to the mathematicians, his book was long and used nonstandard terms.

The foolproof that there is no general solution to the quintic equation would have to await the entry of a young genius from Norway named Niels Abel (1802 - 1829 AD). Independently of Ruffini, he came up with his own proof and this one was eventually accepted by the mathematical community. Unfortunately for Abel, he died before he received the full credit that he deserved. Today, one of the highest prizes a person achieve in mathematics is called the Abel Prize.

Using algebraic notation, Abel's proof says that there is no general solution to an equation such as: 5x5 + 4x4 + 3x3 + 2x2 + x = 15. Yes, we can figure out that the answer is x=1 but we can't do it in the same way that we can for the quadratic equation. If our equation is x2 + x + 1 = 3, then we can solve it using Al-Kwarizmi's solution to the quadratic equation.

To understand the solution to the quadratic equation, we also need to know about square roots. A square root is just the inverse of a square. If 32 = 9, then sqrt(9) = 3. We can speak about square roots, cube roots, quadriatic roots, etc. A root is just the inverse of an exponent. Mathematicians call all these roots "radicals" or "nth roots". For purposes of my discussion today, I will stick to radicals.

Now, with the idea of square roots, we can state the solution to all quadratic equations. For any equation ax2 + bx + c = 0, the solution is:

x = [-b ± sqrt(b2 - 4ac)]/(2a)

To understand the equation, let's use it.

The General Solution to the Quadratic Equations

So, let's take a minute to explore the above formula in more detail. Let's use it to find the solutions to the equation:

x2 + x + 1 = 3.

First, we need to change it into ax2 + bx + c = 0 form.

To do this, we subtract 3 from each side to get:

x2 + x - 2 = 0.

Now, we can map it to a,b,c so that we have:

(a = 1, b=1, c=-2).

Now, we plug it into our formula:

x = [-1 ± sqrt(12 - 4(1)(-2)]/(2(1))

We can simplify the above equation to get:

x = [-1 ± sqrt(1 + 8)]/

Now, sqrt(9) = 3 since 3*3=9 so we get two answers.

[-1 + 3]/2 and [-1 - 3]/2

This means that the two answers are 1 and -2.

The above formula can be used to solve all quadratic equations.

Since it uses square roots, we say that the quadratic equation is solvable by radicals. The solutions for the cubic and quartic equations are more significantly more complicated. Still, since they have solutions that use cubic roots and square roots, we say that they are also solvable by radicals.

Well, shouldn't quintic equations also be solvable by radicals? It turns out that they are not. Something special happens at powers of 5. Let's explore Abel's thinking one step at a time.

Insight #1 Behind Abel's Proof: Sets

The first step in understanding Abel's argument is to view numbers in terms of sets and operations on sets.

Let's start by talking about sets. A set is any collection of numbers. The set of positive whole numbers is 1,2,3.... The set of negative whole numbers is -1,-2,-3,... The nice thing about sets is that they are often infinite so in using them, we have the ability to make statements about all numbers.

For mathematicians, there are three types of sets that are particularly important. They are the rational numbers, the real numbers, and the complex numbers. The rational numbers were the favorite of the Greeks. For the Greeks, they were always positive but for us, they are any positive or negative number that is either an integer or can be represented as a fraction. 1/2 is rational. 1/3 is rational. 5 is rational. In fact, all terminating decimals are rational.

Are all numbers rational? The famous mathematician Pythagorus thought so. He organized a secret society of mathematicians devoted to the study of the rationality of numbers. Unfortunately, Pythagorus soon found out that some numbers cannot be represented as fractions. It turns out for example that many radicals are irrational. For example, the sqrt(2), no matter how hard you try, cannot be represented as a fraction (more about this later).

The set of real numbers include rational numbers and all the irrational numbers. So, pi is a real number even if it is not a rational number and all the radicals are real numbers. But it turns out there are more numbers.

Let's go back to Cardano's solution to the cubic equation. Rafael Bombelli (1526 - 1572 AD) discovered that if he assumed that -1 had a square root, then a whole universe of numbers was uncovered. Here's why he did it. If you multiply two negative numbers together, you get a positive number. If you multiply two positive numbers together, you also get a positive number. How can the product of the same number be a negative number? Enter Imaginary Numbers or Complex Numbers. These numbers evolve around a number i. i is very similar to the real numbers except when you multiply it with itself, the result is -1. Below is the mathematics of i.

These are the same as the real numbers:

i + i = 2i

i - i = 0

i/i = 1

Now here's the strange one:

i*i = -1

So, what's the square root of -4? It can't be a real number but if we use i, we can find a solution. sqrt(-4) = 2i since:

(2i)*(2i) = (2*2)*(i*i) = 4*(-1) = -4 Euler's Identity: One of the most famous equations of all time.

Insight #2: Operations

How did we define i? Well, we made up a new number called i. But why was that good enough? Why couldn't we make up another number called j? The reason is that we clearly defined the operations with i.

In fact, for each of the sets there is a large number of assumptions about what we can do with these numbers . By clearly defining the operations for type of number, we are free to use the sets for very advanced algebraic notation. Indeed, if you are really good and bring trigonometric equations and calculus into the picture, you can end up with Euler's very nifty equation: e + 1 = 0 (see here for more details if you are interested).

Here are some of the basic assumptions:

We assumed that if you add or multiply any two numbers together, you get one and only one answer and it just so happens to be in the existing set.

(2) Order of Numbers in Operation

We assumed that order of addition or multiplication doesn't matter. In other words, we assumed that a + b = b + a and ab = ba.

That's how we got from (2i)*(2i) = (2*2)*(i*i)

Indeed, having clear and precise definitions of operations is why mathematicians are so confident that imaginary numbers such as i really work.

Now, if we are very specific about what types of operations we allow, then we get to a third concept: fields.

Insight #3: Fields

A field is simply a set that has certain well-defined operations.

A mathematical field is a set that supports the following operations

if a=b, then for any value c:

1a. a+c=b+c

1b. ac=bc

2. a+b = b+a

3. (a + b) + c = a + (b + c)

4. a + 0 = a

5. a + -a = 0

6. a(bc)=(ab)c

7. a(b+c) = ab + ac

8. ab=ba

9. a*1=a

10. if a ≠0, then a*(1/a) = 1.

Now, the concept of fields may not seem that exciting. But it turns out that this is really a key insight.

I should point out that the Complex Numbers, Real Numbers, and Rational Numbers are all fields by the above definition. On the other hand, the whole numbers and the integers are not. It's Assumption #9 that gets them in trouble. For integers, there's not always a 1/a that is an integer. Sure 1/1 is an integer but 1/2 is not.

Closely related to the concept of a field is the concept of a subfield. If A,B are both fields with the same operation and A is a subset of B, then A is also a subfield of B. From this viewpoint, the Rational Numbers are a subfield of the Real Numbers which are a subfield of the Complex Numbers.

By reasoning with fields, we can make definitive statements about not only rational numbers, real numbers, and complex numbers, but also any other number fields that we can generate. This will become very important in proving that there is no general solution for the quintic equation.

There is one more very important idea related to fields.

Insight #4: Field Extension

A field extension is a generalization from what Bombelli did. How do we go from the real numbers to the complex numbers, you add a single value i to the real numbers.

From this perspective, we can think of a the field C as an extension of the field R. In mathematical notation, we write this as follows C = R(i).

We did the same thing when it came to the solution to the quadratic equation. We used sqrt(b2 - 4ac).

So what do Field Extensions have to do with Abel's Proof

Field extensions turn out to be very important in solving mathematical problems. It comes down to one more idea: reducibility.

In a nutshell, if you've done your high school algebra, reducibility means that an equations breaks apart into smaller pieces that are easy to use. Here's a simple example:

(x2 - 4) is reducible into (x - 2)(x + 2) since (x - 2)*(x+2) = x*x -2x + 2x + (-2)*(2) = x2 - 4.

On the other hand, certain equations are not reducible at least as far as the real numbers go. For example (x2 + 4) is not reducible in real numbers. But, this is where field extensions come in. If we open up the possibility to i, then the equation becomes reducible.

(x2 + 4) = (x -2i)(x + 2i) since (x-2i)*(x+2i) = x*x -(2i)x + (2i)x + (-2i)(2i) = x2 + 4.

Do you see it? By allowing field extensions, we now have two ways to tackle an algebra problem. Either we use the standard algebraic operations allowed by a field (addition, multipication, etc.) or we use field extensions (square root, cube root, etc.). That's it. There's no other way. This is one of the major ideas that Abel was able to show.

Abel uses a proof by contradiction. This is one of the favorite tactics of mathematicians. The idea is that you make a bad assumption, show that it leads to a contradiction, and then back out of it.

Proof By Contradiction: A Warm Up

I will use a Proof by Contradiction to show that sqrt(2) is not a rational number -- that is, it cannot be represented as a fraction.

Theorem: sqrt(2) cannot be represented as a fraction

Proof:

(1) Assume that sqrt(2) is representale as a fraction.

(2) Let's call this fraction a/b where a,b are integers.

(3) Since it's a fraction, we can assume that a/b is the smallest representation of this fraction.

We can do this since fractions have multiple forms 1/2 = 2/4 = 3/6 etc. But there is only 1 form that is the smallest and that's 1/2.

(4) Now since a/b = sqrt(2), let's square both sides to get:

a2/b2 = 2

(5) If we multiple b2 to both sides, we get:

a2 = 2b2

(6) Now, it is clear that a2 is an even number. So a is an even number.

(7) Since a is even, there must exist another number a' such that a = 2a'

(8) So, now, let's substitute (2a') for a to get:

(2a')2 = 2b2

(9) Now (2a')2 = 4a'2 so that:

4a'2 = 2b2

(10) Dividing both sides by 2 gives us:

2a'2 = b2

(11) And lo and behold, b2 is also even so that there exists a smaller number b' such that b=2b'

(12) But wait. We have a contradiction. We assumed that a/b is the smallest fraction. But if a=2a' and b=2b', then we have a/b=(2a')/(2b') = a'/b' where a' is smaller than a and b' is smaller than b. This contradictions step#3.

(13) Therefore, we reject our assumption in step #1 and we get to say that we disproved that sqrt(2) can be represented as a fraction.

We have completed a proof by contradiction!

Insight #5: Abel's Criteria

To use all these ideas, Abel establishes a criteria for whether a general solution exists. This is a test that verifies that a certain algebraic equation is solvable or not solvable by radicals.

Abel's Criteria: A polynomial is solvable by radicals if and only if it is reducible by field extension and by the standard operations of a field.

Remember, reducible just means that we can break down the equation into simpler terms. If you think about, a general solution to an algebraic equation just breaks the equation down into simpler terms and moves x to one side of the equation and all the other values to the other side.

For example, in each the solution to the quadratic equation, cubic equation, and quartic equation, we really did two things. We uses a field extension to split apart the terms and used field operations to move the x's on one side of the equation and the coefficients on the other side.

To prove my point, let's look at the derivation of the quadratic equation from before. To generalize our result, we'll use letters to represent the coefficients.

(1) ax2 + bx + c = 0

(2) Now, first, we multiply 4a to both sides (Field Operation #1a, see above) so that:

4a2x2 + 4abx + 4ac = 0

(3) Next, we subtract 4ac from both sides (Field Operation #1b, see above)

4a^2x^2 + 4abx = -4ac

(4) Now, we add b2 to both sides (Field Operation #1b, see above)

4a2x2 + 4abx + b2 = b2 - 4ac

(5) Being mathematicians we know that (2ax + b)2 = 4a2x2 + 4abx + b2.

(6) This means that we can refactor to the following:

(2ax + b)2 = b2 - 4ac

(7) Now, we take the square root of each side.

(2ax + b) = sqrt(b2 - 4ac)

(8) Now, normally, the sqrt(b2 - 4ac) might give us a problem but we do a field extension. We add sqrt(b2-4ac) to our field. So, for this point on, we are allowed to do all of our field operations on the additional value of sqrt(b2-4ac).

(9) Now, we subtract b from both sides. (Field Operation #1b, see above) so that we have:

2ax = sqrt(b2 - 4ac) - b

(10) Finally, we divide both sides by 2a to get our solution:

x = [sqrt(b2 - 4ac) - b]/2a

Since this is a real math proof. We can end our proof with QED if we like. That's just mathematical shorthand for "I'm done." It literally stands for"quod erat demonstrandum" which means "that which was to have been demonstrated".

Insight #6: Certain Polynomials are not reducible in any field extension

If you know your mathematics, then you know that all algebraic equations have solutions. The Fundamental Theorem of Algebra tells us that a polynomial of one unknown that has a power of 5 always have 5 solutions.

This means that a quadratic equation always has two solutions. A cubic equation always has three and so on. And these solutions will either be real numbers or complex numbers. This major insight was first proven by the German mathematician Carl Friedrich Gauss (1777-1855) in a book that he wrote when he was 21.

Now, general solutions don't support all the combinations. Leopold Kronecker (1823 - 1891) using Abel's work discovered that a polynomial with an odd, prime degree is only reducible to a field extension if has just one real root (and the others are complex roots) or if all of its roots are real.

The details on this point are subtle but the result leads us to the proof. This is especially useful since 5 is an odd prime so Kronecker's Principle governs quintic equations.

Abel's Proof for Quintic Equations

Now, that we have Abel's Criteria, we can make the proof. Here's the high level flow of the proof by contradiction.

(1) Assume that all quintic equations are solvable by radicals.

(2) Then, they must all be reducible using field extensions. [See Abel's Criteria above]

(3) But certain polynomials are not reducible using field extensions.

Consider the polynomial:

x5 - ax - b = 0

Using analysis (such as Sturm's Theorem), we find that it possesses three real roots and two complex roots. Using Kronckecker's Principle above, we can see that it is not reducible to any field extension.

(4) Using Abel's Criteria, it follows that at least one quintic polynomial is not solvable by radicals.

(5) But, this contradicts our assumption in step #1 that all quintic equations are solvable by radicals.

(6) Therefore, we have a contradiction and we can reject our assumption in step #1.

QED

If you've made it this far, pat yourself on back. Great job!

If you want to see Abel's proof in all its details, I highly recommend two books. For the general reader, check out Abel's Proof by Peter Pesic.

For the mathematician, check out 100 Great Problems of Elementary Mathematics by Heinrich Dorrie.

KK on December 26, 2013:

THANK YOU SO MUCH! YOU DON'T KNOW HOW MUCH THIS ARTICLE MEANS TO ME! BE A TEACHER!

mike on April 02, 2012:

only the old experienced hubbers will include a link in their hubs ( if the link reserves to be included in the hub).

here is good link to an awesome article

http://science-and-mathematics.blogspot.com/2012/0...

thanks

Sunil Laudari on January 27, 2012:

The book 'ABEL'S PROOF' is so exciting that it makes the sense of quantum approach to the mathematical problems

goznevi on December 15, 2011:

I love every line

here some typos:

... Using Kronckecker's Principle above - Kronecker

... to a field extension if (it) has just one real root

... We use(d) a field extension to split apart the terms and used...

This is not clear:

..It's Assumption #9 that gets them in trouble. For integers, there's not always a 1/a that is an integer...

Assumption #9 doesn't talk about the inverse operation but rather about the identity operation

Thanks for helping me understand once more how beautiful math is

Yours

alfredo.duran

ushnafarooqi@yahoo.com on December 07, 2011:

i like maths and those who invented maths i can say they were very sharp and obviously intelligent

Anon on June 20, 2011:

"Using Kronckecker's Principle above, we can see that it is not reducible to any field extension."

Should be:

Using Kronckecker's Principle above, we can see that it is not reducible to any real field. It is however reducible over a complex field (all polynomials are).

sujit on March 27, 2011:

Hi Larry,

Thanks a lot...I have searched the whole internet to find a clear explanation for Extension Field.Finally I found here.Also,how ignorant i feel when I see the same can be explained just in plain high school algebra,which i have ignored long back. :(

larryfreeman (author) from Fremont, CA on November 15, 2010:

Hi Atul Bhatia:

Here's a complete proof of theorem from my blog:

http://fermatslasttheorem.blogspot.com/2008/10/abe...

Atul bhatia on November 14, 2010:

I want Abel proof of impossibility theorem in details from a to z can you provide me.

zaheer on October 19, 2010:

i like mathematics and also i get M.Sc degree in this field but still i have some problem, becouse in our school and colleges dont explaine every things clearly, please any mathematician contact me to solve, plzzzzzzz. (zaheerk269@gmail.com)

Will Apse on March 17, 2010:

I studied mathematics until I was 18. I was genuinely interested but also incredibly frustrated that nothing was ever explained properly and we were simply expected to accept piles of formulas and remember them. I can't say I followed everything here to the end (its late) but it renewed my faith that mathematics does make sense.

There is a huge need for this kind of thing so thanks for taking the trouble.

hirak on January 29, 2010:

Good job,but I think you should mention SRIDHAR ACHARYA for inventing roots of quadratic equation.....

Manna in the wild from Australia on November 01, 2009:

Nice work. Here are some typos:

"we can treat unknown values as numbers."

do you mean as letters?

""Find three suqare"

=> square.

"the answer is 3,4,5 since 9 + 16 = 25." is *one* answer, not t *the* answer.

"A typical example is: 5x2 + 3x + 2 = 10. Find x (the answer in this case is x=1)" This is one of two answers, The other root is -1.6.

I think it was one of Pythagoras's students who found or at least insisted upon publication of the discovery of irrational numbers and there was bloodshed as a result.

""we can treat unknown values as numbers."

do you mean as letters?

""Find three suqare"

=> square.

"the answer is 3,4,5 since 9 + 16 = 25." is *one* answer, not t *the* answer.

"A typical example is: 5x2 + 3x + 2 = 10. Find x (the answer in this case is x=1)" This is one of two answers, The other root is -1.6.

I think it was one of Pythagoras's students who found or at least insisted upon publication of the discovery of irrational numbers and there was bloodshed as a result.

"From this perspective, we can think of a the field C"

a??

"only 1 form " should read "only one form"

"representale" == representable.

"(5) If we multiple " ==> multiply

larryfreeman (author) from Fremont, CA on September 17, 2009:

Hi rv,

Thanks very much for your comment on the definition of subfield!

I have revised the hub to address this point.

Cheers,

-Larry

rv on September 17, 2009:

Very informative presentation - one that stimulates curiosity for "further reading". I hope more high school students get to read and appreciate it.

I would just point out that the statement "If A,B are both fields and A is a subset of B, then A is also a subfield of B" is not entirely correct, or rather it leaves out a very important detail: the operations under which B is a field are those inherited from A.

Otherwise, e.g. one can define "strange" operations on the numbers of the type (rational)*sqrt(2) as such: a*sqrt(2)+b*sqrt(2)=(a+b)*sqrt(2) (nothing new so far) and [a*sqrt(2)]~[b*sqrt(2)]=(a*b)*sqrt(2) (here ~ denotes a special multiplication, while * is just the usual multiplication of real numbers). All the axioms are verified (the "1" becomes 1*sqrt(2) and the inverse of 3*sqrt(2) is (1/3)*sqrt(2) etc.) but the smaller field is not a subfield of the real numbers, since the multiplication is different.

i from Earth on June 06, 2009:

Thank You Larry for recommendations and usefull links to amazon.com

've bought both of them and waiting for more!

larryfreeman (author) from Fremont, CA on October 06, 2008:

Hi Cheryl,

Thanks for the feedback.  To prove that  Z/nZ is a commutative ring, you need to show that it satisfies the following rules:

Multiplication rules (Commutative, Associative, Distributive)

I have more details on my blog:

http://mathrefresher.blogspot.com/2006/05/fields-a...

-Larry

Cheryl on October 03, 2008:

Hi, Larry. I loved this hub. Very informative. I have a question from the &quot;refresher&quot;, though. To prove that Z/nZ is a commuative ring doesn't require anything more to PROVE that the commutative, distributive, and associative properties hold?

larryfreeman (author) from Fremont, CA on September 02, 2008:

Luke,

Thanks for pointing out. I just fixed it. :-)

-Larry

Luke Worth on September 02, 2008:

He's probably talking about field operation number 7.

larryfreeman (author) from Fremont, CA on July 21, 2008:

Hi EpicTruth,

Thanks for noticing! :-) Where do you see the mistake? I am very glad to correct any typos or errors on my part.

Cheers,

-Larry

epictruth from Frisco on July 21, 2008:

I found only one mistake in all your equations but I'll forgive you.

larryfreeman (author) from Fremont, CA on July 20, 2008:

Great title. I probably would have used that one if I had thought of it. :-)

Jason Menayan from San Francisco on July 20, 2008:

Wow! Your hub should be retitled &quot;History of Algebra in 20 minutes&quot; :-)