An irrational number cannot be represented as a fraction composed of integer parts. EG [PI], [e]. There are an infinite number of irrational numbers between each [rational] number, and there are an infinite number of rational numbers. Which you might think means there are more irrational than rationals [[ [CL] feels compelled to point out that this is '''not''' a correct proof; notice, for example, that between each two distinct irrationals there are infinitely-many rationals]]. See [Cantor] for the transfinite numbers. Any number whose decimal representation recurs at any point is rational. This is proved since: Make PRE = the decimal representation up to where the number starts repeating, and replace the repeating part with 0000... then the number PRE can be represented as PRE*1000..../1000.... (ie a fraction, using as many digits in the multiplier as needed to make PRE an integer). Eg 3.4352141414... = 34353/10000 + 0.141414/10000.... The repeating part can be made into a fraction by taking the repeat and dividing by 99999.... (as many 9s as are in the repeated part). For example: .77777... = 7/9. 0.7171.. = 71/99; 0.142857142857... = 142857/999999, .00000717171... = 71/99/100000 and so on. So any number which has a recurring representation can be written as a sum of 2 fractions: Val = PRE + rept/999.... and is therefore a rational number. Well that is a simple version of the proof, but largely correct. In finite digit arithmetic (eg in a coding language such as [C++] or [Fortran] or [TCL]) there are only a finite number of digites representing a value. Most computer languages therefore use rational numbers ONLY, and operate on rational numbers too. Thus computers really work in integers (though some algebra packages such as Mathematica, Maple etc) do have some understanding of irrationals. [GWM] ---- also qv [Perl Syntax] ---- [Category Mathematics]