<PRE><B>Question 362693</B>
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I don&#39;t know how to &quot;calssify&quot; anything, including numbers.  Now if you actually meant to ask about how to &lt;i&gt;&lt;b&gt;classify&lt;/b&gt;&lt;/i&gt; numbers, then perhaps I can help you.


&lt;i&gt;&lt;b&gt;Natural Numbers&lt;/b&gt;&lt;/i&gt;.  Depending on who you talk to it is either the set of counting numbers, that is 1, 2, 3,... and so on, OR it is the set of counting numbers including zero, i.e., 0, 1, 2, 3, ...


&lt;i&gt;&lt;b&gt;Whole Numbers&lt;/b&gt;&lt;/i&gt;.  The set of counting numbers and zero.  Just another way of saying the Natural Numbers if you are one of those people who think that zero is a natural number.


&lt;i&gt;&lt;b&gt;Integers&lt;/b&gt;&lt;/i&gt;.  The set of whole numbers and their additive inverses, so:  ..., -3, -2, -1, 0, 1, 2, 3, ...


&lt;i&gt;&lt;b&gt;Rational Numbers&lt;/b&gt;&lt;/i&gt;.  The set of numbers of the form *[tex \Large \frac{p}{q}] where *[tex \Large p] and *[tex \Large q] are integers.  They are called &lt;i&gt;Ratio&lt;/i&gt;nal Numbers because they are formed by a &lt;i&gt;ratio&lt;/i&gt; of integers.


&lt;i&gt;&lt;b&gt;Irrational Numbers&lt;/b&gt;&lt;/i&gt;.  All real numbers that are not rational, that is that cannot be expressed as the quotient of two integers.  There are two types of Irrational Numbers:  Algebraic numbers are a root of a polynomial equation of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \alpha_0x^n\ +\ \alpha_1x^{n-1}\ +\alpha_2x^{n-2}\ +\ \cdots\ +\ \alpha_{n-2}x^2\ +\ \alpha_{n-1}x\ +\ \alpha_n\ =\ 0]


Whereas Transcendental numbers are not roots of such an equation.  *[tex \Large \sqrt{2}] is an example of an algebraic number.  *[tex \Large \pi] is an example of a transcendental number.


&lt;i&gt;&lt;b&gt;Real Numbers&lt;/b&gt;&lt;/i&gt; Everything discussed above all wrapped up in one great big set.  An interesting fact:  Pick any two irrational numbers and you can find an infinite number of rational numbers between them.  Pick any two rational numbers and you can find an infinite number of irrational numbers between them -- and furthermore, an infinity of them will be algebraic and an infinity of them will be transcendental.


&lt;i&gt;&lt;b&gt;Imaginary Numbers&lt;/b&gt;&lt;/i&gt;  Actually, that should be Imaginary Number (singular) since there is only one of them.  The imaginary number is *[tex \Large i] and it is defined as *[tex \Large i^2\ =\ -1]


&lt;i&gt;&lt;b&gt;Complex Numbers&lt;/b&gt;&lt;/i&gt; Numbers of the form *[tex \Large a\ +\ bi] where *[tex \Large a] and *[tex \Large b] are real numbers and *[tex \Large i] is the imaginary number defined above.
 



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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