Algebraic number

In mathematics, an algebraic number is any number that is a root of a nonzero polynomial in one variable with rational coefficients. I.e., they can be represented using algebraic expressions and equations.

The Algebraic Number Game
This is a game I like to play to aid in understanding algebraic numbers.

Here are the rules:
 * 1) You can only use whole numbers.
 * 2) You may add, subtract, multiply, and/or raise numbers to a whole power.

The objective of the game is to reduce numbers down to 0 without simply just multiplying the numbers by 0.

Prime and Composite Numbers
Let's start with something basic, like my favorite number &mdash; 5. We can subtract 5 from 5 to bring it down to 0.


 * $$5-5=0$$

Note that 5 is a prime number. Let's try a composite number instead, like 6. We can subtract 6 from 6 to bring it down to 0.


 * $$6-6=0$$

Integers
How about integers? Let's try something like –7. Well, all we need to do is add 7 to bring it to 0.


 * $$-7+7=0$$

Now we've brought all of the whole numbers to 0.

Fractions
We've now moved into the realm of fractions. What can we do about those?

Let's say we have the following fraction, assuming a and b are whole numbers and b is not 0.


 * $$\frac{a}{b}\,$$

According to the rules of the game, we can't subtract by the same fraction, since we need to use whole numbers to play the game. So, what can we do to get it to 0?

First, we need to get rid of the denominator. We can do this by multiplying the fraction by b, which cancels out b. This results in simply a.


 * $$b(\frac{a}{b}\,)=a$$

Now, we just subtract a from a, giving us 0.


 * $$a-a=0$$

This will work with every fraction.

Irrational Numbers
What can we do with an irrational number? Surely those can't be algebraic. While it's true that many irrational numbers aren't algebraic (thereby making them transcendental numbers), there are a good handful of algebraic irrational numbers.

Let's try with the square root of 2. Again, we need to use whole numbers to do this.

We can bring the square root of 2 down to 0 by squaring it (i.e. multiplying it by itself) and subtracting 2.


 * $$(\sqrt{2}\,)^{2}=2$$
 * $$2-2=0$$

Now let's try the same method with the Golden Ratio.


 * $$\phi\,=\frac{1+\sqrt{5}\,}{2}\,$$

This won't be easy, but let's see what we can do.

First, we'll get rid of the denominator.
 * $$2(\frac{1+\sqrt{5}\,}{2}\,)=1+\sqrt{5}\,$$

Since we're dealing with a square root, we'll square the whole thing. This requires a bit of FOIL, but it's not too much trouble.
 * $$(1+\sqrt{5}\,)^{2}=(1+\sqrt{5}\,)(1+\sqrt{5}\,)=1+2\sqrt{5}\,+5=6+2\sqrt{5}\,$$

Now, let's subtract 6, since we can do this to get the radical by itself.
 * $$6+2\sqrt{5}\,-6=2\sqrt{5}\,$$

Now, let's square the radical.
 * $$(2\sqrt{5}\,)^{2}=4*5=20$$

Finally, we subtract 20. This brings the Golden Ratio down to 0.
 * $$20-20=0$$

Therefore,


 * $$(2\phi\,^{2}-6)^{2}-20=0$$

or


 * $$4\phi\,^{4}-24\phi\,^{2}+16=0$$

Like I stated before, most irrational numbers are transcendental, so they can't all be brought down exactly to 0. However, there are a good amount of algebraic irrational numbers, so don't lose all hope in finding them in this realm.

Complex Numbers
We've now entered the realm of complex/imaginary numbers, which can be expressed as the following for any complex number c:


 * $$c=a+bi$$

a and b are whole numbers. a is the "real" part, while b is the "imaginary" part, due to i.

Let's bring i down to 0.


 * $$i=\sqrt{-1}\,$$

Well, this is easy...just square it and add 1.


 * $$(\sqrt{-1}\,)^{2}=-1\therefore\,-1+1=0$$

In general, you can bring any complex number down to 0 using the same method we used to bring the Golden Ratio down to 0, except we don't need to get rid of the denominator (there is none, unless you count 1...and that makes no difference). Remember that i2 is equal to –1.


 * $$(a+bi)^{2}=(a+bi)(a+bi)=a^{2}+2abi+b^{2}i^{2}$$


 * $$a^{2}+2abi+b^{2}i^{2}=a^{2}+2abi-b^{2}$$


 * $$a^{2}+2abi-b^{2}-(a^{2}-b^{2})=2abi$$


 * $$(2abi)^{2}=4a^{2}b^{2}i^{2}=-4a^{2}b^{2}\therefore\,-4a^{2}b^{2}-(-4a^{2}b^{2})=0$$

Therefore,


 * $$[(a+bi)^{2}-(a^{2}-b^{2})]^{2}+4a^{2}b^{2}=0$$

There is no way to simplify this. I've tried, and I ended up with 0 = 0. If you want to try to simplify it, go right ahead. I must warn you; it isn't easy, and the resulting equation is very long.

What's the big deal?
I realize that these numbers seem entirely unrelated. What do irrational numbers, complex numbers, integers, and fractions have in common? We know they're algebraic, but why are they algebraic?

I can answer this with just four words: Replace them with x.


 * $$x-5=0$$


 * $$x-6=0$$


 * $$x+7=0$$


 * $$bx-a=0$$


 * $$x^{2}-2=0$$


 * $$(2x^{2}-6)^{2}-20=0$$


 * $$4x^{4}-24x^{2}+16=0$$


 * $$x^{2}+1=0$$


 * $$[x^{2}-(a^{2}-b^{2})]^{2}+4a^{2}b^{2}=0$$

Now, you can use algebra to solve for each x value. This means that all the numbers, as well as many others, can be found using algebra. Hence, they're algebraic!

Like I said, not all numbers are algebraic. In fact, much of the real numbers are not algebraic, which means they're transcendental. All complex numbers are algebraic, though most real numbers are not. Among these are e and π.