Wednesday, December 1, 2010

More talk about numbers

At the end of this post we had arrived at the rational numbers and were able to derive four mathematical operations. At first sight this might look like an impressive achieving, taking into account the tools we set ourselves to use, but due to some shortcomings that plague our construction we think that things can get better.

— 1. Problems —

  1. Our operations are binary

    By that we mean that our mathematical operations only make sense when applied to two numbers. So for instance $ {2+3+4} $ has no answer in our current state of affairs. But worry not dear reader because things can be formalized in a fast and straightforward way. And there's nothing too fancy, or technical in this formalization: it is nothing more than common sense put to action.

    How do we add $ {2+3+4} $? We just do $ {2+3=5} $ and $ {5+4=9} $. So the correct formalization of $ {a_1 + a_2 + a_3 + \ldots + a_n} $ just is:

    $ \displaystyle ((((a_1 + a_2) + a_3) + \ldots )+ a_n) $

    That is to say that we define addition of $ {n} $ numbers to be done two by two (this means that addition is still binary but can be done repeatedly).

    Of course that all of this is valid for the three remaining operations even though we haven't explicitly said so. Thus this is one possible problem that was taken care of.

  2. Another problem is that we have two inverse operations and due to the existence of them we could expand the set of the available numbers. But we never said nothing about what happens to the properties of the operations while we continuously do that. And for that matter we never said anything at all about the properties of the operations in the first place!

  3. Another problem that we have is that our number system is rather incomplete.

— 2. Properties of the mathematical operations —

To solve our second problem it's time for us to talk about the properties of addition, multiplication, subtraction, and division.

  1. Associativity

    $ \displaystyle (m+n)+p = m+(n+p) $

    $ \displaystyle (m \times n) \times p = m\times (n \times p) $


  2. Commutativity

    $ \displaystyle m+n=n+m $

    $ \displaystyle m \times n = n \times m $


  3. Distributivity

    $ \displaystyle m \times (n+p)= m \times n + m \times p $

  4. Neutral Element

    For every natural number it is:

    $ \displaystyle m+0 = m $

    and

    $ \displaystyle m \times 1 = m $

If we think about addition and multiplication in the simple terms of the first post in this series it is easy to see why these properties hold. But remember that those simple terms apply to the natural numbers. As we went along in enlarging the set of numbers that we were working with we never cared about what happened to those properties.

Do they still hold in $ { \mathbb{Z} } $ and $ { \mathbb{Q} } $?

We can take the lazy out and just define those sets as being sets in which the properties hold or we can check how things really are. Even though I'll take the lazy way out I want readers of this blog to know that things can be done in an intellectually satisfying way.

— 3. Powers —

To solve our third problem we'll just continue to use our ambition and see where it leads us.

Just like we introduced multiplication as a device of allowing us to write more succinctly a sum of $ {n} $ equal numbers we will now introduce the power operation as a more succinct way of writing a product of equal factors.

So for instance we'll write $ {2^5} $ for $ {2 \times 2 \times 2 \times 2 \times 2} $. In general if we have a number $ {a} $, this number is called the base, multiplying with itself $ {n} $ times, this number is called the exponent, we'll write $ {a^n} $ for the end result.

Now that we have introduced powers as a shorthand notation for multiplication of equal factors we have to know how this new entity behaves with the previous four operations.

When we are summing (subtracting) powers we have to first calculate the result of the powers and after this we sum (subtract) those same results. For example:

$ \displaystyle 2^3+ 3^2 = 8+9=17 $

$ \displaystyle 4^2 + 3^2 = 16+9=25 $

$ \displaystyle 5^3-10^2=125-100=25 $

But when we are multiplying or dividing there are some rules that can be used in order to get the right result in a faster way.

— 3.1. Multiplying powers —

Unlike addition (subtraction) of powers when one is multiplying powers we can do it in a more cleaner way in some number of cases:

  1. Same base

    For instance:

    $ \displaystyle 2^3 \times 2^4=2\times 2 \times 2 \times 2 \times 2 \times 2 \times 2=2^7 $

    Thus the final power has the same base as the initial ones and its exponent is just the sum of the exponents of the initial power.

    Of course that the numbers $ {2} $, $ {3} $ and $ {4} $ that we choose initially aren't in any way special and as a general rule it always is:

    $ \displaystyle a^n\times a^m = a^{n+m} $


  2. Same exponent

    For instance:

    $ \displaystyle 3^3 \times 4^3= 3 \times 3 \times 3 \times 4 \times 4 \times 4= (3\times 4) \times (3\times 4) \times (3\times 4)= 12^3 $

    Once again the numbers we initially choose have nothing special about them so we can state in full generality that

    $ \displaystyle a^n\times b^n = (a\times b)^n $

Yes we have stated these two properties for binary multiplication but they can be extended to any number of factors without any problems whatsoever.

— 3.2. Dividing powers —

In the case of division we'll just state the theorems and won't bother to present any example since this can be done very easily by an interested reader once he grasps the idea behind the examples used for multiplication.

  1. Same base

    $ \displaystyle a^n/a^m=a^{n-m} $


  2. Same exponent

    $ \displaystyle a^n/b^n=(a/b)^n $

— 4. Here comes trouble! —

Since we haven't abandoned our principle of ambition thus far the questions that are being begged to be answered are if we can introduce the inverse operation of a power? and where such course of action might take us?

The answer to the first question is an emphatic yes. The answer to second question is that the inverse operation of a power takes us to wonderful places and open the floodgates of some wonderful mathematics.

The answer of these two questions will be the subject of our next, and possibly final, installment and in these series of posts.

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