When written as decimals, irrational numbers do not terminate or repeat ??!

what does this mean ?!

Think of Pi (3.1415926...), it goes on forever without repeating or terminating - it would be irrational. But if you were to write 4/3 in decimal form: 1.3333333... it would be rational, because it is repeating the 3's. Or take 9/13 in decimal form: 0.692307692307... it is repeating even though it's not a single number that repeats, it is the 692307 that repeats itself. This too is rational.

Another way of expression this is if a number x can be written in the form p/q where p and q are integers reduced to lowest terms, then its decimal eqivalent will either terminate in a 5 if the denominator is of the form 2^a*5^b. For example, if n = 3/8, the denominator 8 =2^3
(Here a=3 and b =0), then
3/8 = 0.375

If the denominator has neither 2 nor 5 as a divisor, then the decimal representation has repeating sequences of digits. For example,
4/13 = 0.307692 307692 307692 ..............
where the sequence 307692 goes on forever.
Now a number x is irrational if it cannot be expressed as a fraction p/q (remember, p and q need to be integers).

An example of an irrational number is the square root of 2. It is impossible to express sqrt(2) as a fraction where the numerator and denominator are integers. A proof of this is as follows:

(You can skip this bit if you want)
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Suppose we assume that we can write sqrt(2) = p/q (where p/q have no common factor - ie reduced to lowest terms). Then squaring both sides, we get
2 = p^2 / q^2
or
2q^2 = p^2
Since p and q have no common factor, 2 must divide p. In other words p = 2k for some integer k. Therefore
2q^2 = (2k)^2
ie
2q^2 = 4k^2
q^2 = 2k^2
Since p and q have no common factor and k divides p, k cannot divide q. Which means that 2 must divide q.
But this is impossible because p and q have no common factor!
The conclusion is that sqrt(2) cannot be written in the form p/q.