Irrational Numbers Proof ```Name: Huen Yeong K. Status: student Age: N/A Location: N/A Country: N/A Date: N/A ``` 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? Replies: 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 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: 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 ``` Click here to return to the Mathematics Archives

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