Ask A Scientist© Mathematics Archive

name        Huen Yeong K.
status      educator

Question -  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?
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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
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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
contradiction:
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
integer.
     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: R<1/(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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