Basic ideas and concepts:


Additional Remarks:

1) As no one can write down pi or any other irrational number,  we do not use irrational numbers but their rational approximations by truncating a few digits, say 3 or ten, but always a few of them. These approximations turn out to be good enough for all practical purposes.

2) There are infinitely many irrational numbers. This is a necessary clarification to the ambiguity of text books and teachers who demonstrate that all integers can be considered rationals, as well as all other fractions of the sort p/q where q is different from zero; then a few irrationals are meant to be mentioned: √2, , Euler number. Let’s add here: the square root of any non perfect square is an irrational number. The same is valid for the cubic roots of any non perfect cube. Summary: the nth root of a number ( x1/n) is irrational unless x is the nth power  of an integer. Most logarithms and values of trigonometric functions are also irrational numbers. There are also irrational numbers that cannot be found as solutions (roots) of integer polynomials. This kind of numbers are meant to transcend algebra, and consequently were called “transcendental” numbers. √2 is an algebraic number, the solution or root of the equation: x2-2 = 0; rational numbers are algebraic numbers of degree one: 2x-1 = 0 the solution of the equation is ½. George Cantor demonstrated in 1874 that the set of irrational numbers is larger (uncountable) than the set of rational numbers (which is a countable set). Not all infinities are the same: there are more irrational numbers than rational numbers.

3) How can it be that irrational numbers and even rational numbers never end? Because space is continuous and extensively infinite in depth.

4) Irrationality of √2. Can this number be represented as the ratio of two integers?

Let's assume √2 is a rational number.  Then we can write it √2  = p/q where p, q are integers, q different form zero, and also p and q have no common divisors (p/q is reduced to its lowest terms).

By squaring both sides of the equation follows that,

 2 = p2/q2,  

 p2 = 2q2

So p2 is a multiple of two and therefore p itself is a multiple of 2.

Then since p is a multiple of two, or even number, we can represent it as 2k, this is, p = 2k

Substitute p = 2k into:

 p2 = 2q2

And we have

(2k)2 = 2q2 which leads to 2k2= q2

Therefore q2 is even, q is even: contradiction! Both numbers p and q cannot be even or multiple of two since we established as a condition that this ratio p/q had been reduced to its lowest terms.

Proving that other square roots are irrational takes more or less the same steps than the √2.

5) Irrationality and logaritms: Is log2 rational?

log 2 (base 10) is equal to 10log2 = 2

Assuming log 2 is rational the log can be equated to the ratio of two integers p/q:

log 2 = p/q and since the logarithm can be expressed in the exponential form:

10p/q = 2  then by raising both sides to the power of q we obtain:

10p = 2q

Which is impossible: no power of ten raised to an integer is equal to a power of two raised to another integer. Contradiction. Therefore, log 2 and similarly log 3 etc are irrational numbers. However, log2 8 = 3 a rational number, simply because 8 is a power of 2. Only when the number whose log is sought is a power of the base of the log, logaritms are rational numbers.