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A student asked me whether ln(k) where k is integer which ranges from 2 to infinity are all irrational numbers. Are they? Is there a proof?

A logarithm is the inverse function of raising a number to a power. It asks the following: I have two numbers, "N" and "b". I ask the question: What power (exponent), for the moment call it "a", must I raise "b" so that the result is equal to N. That is: N=b^a. The number "a" is given a different symbols. We rewrite "a" as a=ln[b](N). This reads: The number "a" is the logarithm of "N" to the base "b". In principle "b" can be any number, except 0 and 1 of course.

The number "e" =2.718281828... pops up many places in math. It is not only irrational, that is, it cannot be expressed as the ratio of two integers, say p/q. It is transcendental, which means that it is not the root (solution) to any polynomial of any degree with integer coefficients. This condition is more restrictive than saying a number is irrational. For example, the polynomial of degree '2', X^2 - 3=0, has the solution X = sqrt(3), which is irrational.

We write ln(N) = a, which means what power "a" must I raise the number "e" in order to obtain the number "N". The answer is always irrational for integers "N" except for N=1, because ln(1)=0. The function ln(N) may even be transcendental.

Rational numbers have the property that when written in decimal form, it will be a repeating decimal, for example 3/11 = 0.27272727... always repeating the sequence "27". The opposite is also true: any repeating decimal is a rational number, that is, it can be written as some ratio p/q. One might look at the number "e" and think it has a repeating sequence "1828" but when you evaluate it to more significant figures that repeat sequence does not keep going.

Vince Calder

It is not correct to assign properties to the sum of an INFINITE series
based on the behavior of a partial FINITE series of the same form. An
example is the INFINITE SUM (1/x). Any FINITE SUM (1/x) is a rational number
[1/2 + 1/3 + 1/4 = 13/12]. But, the INFINITE SUM (1/x) is divergent, that
is, unbounded.

     It is possible to prove that the number 'e' is irrational, by
Consider now INFINITE SUM expression of the number 'e':
e = e^1 = 1 +1/1! + 1/2! + ... + 1/n! + ...
Assert: 'e' is rational, that is it can be written as a ratio, denoted: N/D
where the numerator 'N' and the denominator 'D' are positive integers
greater than unity.
     The product of D factorial, D!, and 'e' is:
D!*e = D! + D!/1! + D!/2! + D!/3! + ... + D!/D! + other terms denoted by R,
the remainder.
     If 'e' is rational, then D!*e is also rational, since D! is a positive
     Examine the other terms of the product D!*R:
D!*R = [D!/(D+1)! + D!/(D+2)! + D!/(D+3)! +...]
     So R = [1/(D+1) + 1/(D+1)(D+2) +...]
since in each ratio of factorials the first D! in the numerator and
denomenator cancel.
     So, R = [1/(D+1) + 1/(D+1)(D+2) +...] < [ 1/(D+1) + 1/(D+1)(D+1) +...]
which is obtained by substituting (D+1) for each term in the sum of
reciprocals in the expression for R. This is true because (D+j) < (D+1) for
any term for j>1.
     Factoring out the first common term: 1/(D+1) gives:
D!*R = 1/(D+1) * [ 1 + 1/(D+1) + 1/(D+1)^2 + 1/(D+1)^3 + ... + 1/(D+1)^j +
     Now the term in the brackets [....] is just the geometric series
expansion of the form: [1+ x + x^2 + x^3 + ...] where x = 1/(D+1). This
INFINITE SERIES sums to: 1/(1 - x ).
     So: R1/(D+1) * [ 1/{1 - (1/(D+1)}] = 1/(D+1) * [ (D+1)/D ] = 1/D.
     Summarizing: 0 < R < 1/ D. From the original definition of R,  both R
and D are positive integers. This is a contradiction. Hence the original
assertion that
'e' is rational is false. Thus 'e' is irrational.

Vince Calder

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