Beal's Conjecture And The Problem of Multiplying Irrational Numbers

James Constant

math@coolissues.com

Earlier, I announced that Beal's conjecture is disproved for the same reasons Fermat's last theorem is proved. See http://www.coolissues.com/mathematics/Beal/beal.htm Since then, solutions z,x,y are claimed to exist for Beal's conjecture. The present work reconsiders Beal's conjecture and looks at the related problem of finding the product of two irrational numbers. Assume

z^m=x^a+y^b . . . . . . . . . . . . . . . . . . . . to be proved . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)

where z,x,y,m,a,b are any numbers and m,a,b>2.

What constitutes proof of (1)? Unlike the Pythagorean theorem, (1) has no geometric proof and it has no algebraic proof because it is a single equation with three unknowns. There are sets of z,x,y and sets of m,a,b from which we are to find some sets z,x,y,m,a,b as "solutions", if any. In (1) all we have is a single equality and some stated conditions on sets of z,x,y and m,a,b. Therefore, proof here are answers to the questions, first, does equality (1) necessarily exist under the stated conditions and, second, if it necessarily exists does it sufficiently answer the question whether any set(s) of z,x,y,m,a,b exist under the stated conditions.

Fermat's last theorem (FLT) is the negation of (1) when z,x,y,m.b,a are integers and m=a=b>2. Then, (1) exists only if (x^m+y^m)^1/m is an integer. FLT is proved because when (x^m+y^m)^1/m is written as

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)

and the parenthesis term expanded in a binomial series, (2) is shown to be an irrational number, the product of a rational number x and irrational number (1+(y/x)^m)^1/m. See Proof of Fermat's Last Theorem at http://www.coolissues.com/mathematics/Fermat.fermat.htm When this same method is applied to Beal's problem the result is a product of two irrational numbers which may, until proven, be a rational or irrational number.

Beal's conjecture (BC) states if (1) exists with positive integers z,x,y,m,a,b and m,a,b>2 then z,x,y have a common factor f>1. Thus, Beal's equality (BE) exists only if (x^a+y^b)^1/m is a positive integer. In equations (3)-(7) I assume no common factors in z,x,y. To distinguish from spurious proofs unrelated to BC, I also assume no common terms. See ADDENDUM 1 below. When (x^a+y^b)^1/m is written as

. . . . . . . . . . . . . . . . . . . . . . . . k₁= integer .. . . . . . . (3)

and the parenthesis term expanded in a binomial series, (3) is a product of an integer x^a/m (when x=k₂^cm k₂,c integers) and an irrational number (1+y^b/x^a)^1/m or is the product of two irrational numbers x^a/m (when k₂,c integers) and (1+y^b/x^a)^1/m. When y^b/x^a>1 the parenthesis term diverges and has no value. Accordingly, as a product of a rational and irrational numbers, k is an irrational number and the negation of (1) is proved. However, as a product of two irrational numbers, it remains open to prove that k may be a positive integer, rational or irrational number..

To find integer solutions of (3) set

. . . . . . . . . . . . . . .

. . . . . . . . . . . x,y,m,a,b,c.k₁,k₂,r₁,r₂,m= positive integers . . . . . . . . m,a,b>2 . .. . . . . . . (4)

The product of irrational numbers in (4) is a positive integer when

.

. . . . . . . . . . .p₁,p₂,q₁,q₂ integers . . . . . . =0, integer . . . . . . (5)

The existence of equalities (1) and (3), therefore, requires necessary but not sufficient conditions (4),(5). Equations (4) and (5) simply state that there are an infinite number of points (products) on the circumference of a circle only two of which are necessary but not sufficient for a rational solution (1),(3) to exist. Sufficiency, to be proved, requires finding whether product k exists under the stated conditions. Assume

. . . . . . . . . . . . . . . to be proven

. . . . . . . . . . . . . . . . c,k₂ integers . . . . . . . (6)

Equation (6) is the fundamental problem of multiplying irrational numbers. Here

. . . . . . . . . . . . . . .c,k₂ integers. . . . .. . . (7)

from which I conclude by negation that, since logs are rational numbers, k is an irrational number and thus Thus, when no common factors in z,x,y exist and no common terms exist, BC is disproved because, first, for most sets z,x,y,m,a,b (1),(3) do not necessarily exist and, second, when (1),(3) do necessarily exist they are sufficient because, in (6),(7) they find that product k is an irrational number.

The foregoing procedure does not change if a common factor f >1 exists. In such case use (1) and (3)-(7) by replacing z,x,y by fz,fx,fy respectively. Assume

(fz)^m=(fx)^a+(fy)^b . . . . . . . . . . . .f>1 . . . . . . . . . . . to be proved . . . . . . . . . . . . . . . . . . . (8)

from which it is obvious that f is a positive integer. In equations (8)-(14) I assume no common factors in z,x,y. To distinguish from spurious proofs unrelated to BC, I also assume no common terms See ADDENDUM 1 below. Rewrite (fx)^a+(fy)^b as

. . . . . . . . . . . . . . . . . . k₁= integer . . . . . . . (9)

and assume

. . . . . . . . . . . . . . . . . . .to be proven . . . . . . (10)

Equation (10) is the fundamental problem of multiplying irrational numbers. Here

. . . . . . . fx positive integer . . . . . . .. . . . . . (11)

in which log_e(1+(fy)^b/(fx)^a)^1/m is a rational number. See ADDENDUM 2 below. Accordingly, I distinguish two cases

Case 1: log_e(fx)^a/m is a rational number; (fx) not a power of m

. . . . . . . . . . . . . . . .k₂,k₃ integers . . . . . . . . . . . . . . . . . . . . . . . (12)

and, therefore, k is an irrational number. Case 1 is the same case discussed in equation (7) and thus when common factors f>1 exist. Thus, when common factors f>1 exist, when no common factors in z,x,y exist and no common terms exist, BC is disproved for the same reasons I give in discussing (7).

Case 2: log_e(fx)^a/m is an irrational number; (fx) is a power of m

. . . . . . . . . . . . . . . . . . .k₂,k₃ integers . . . . . . . . . . . . . . . . . . . . . (13)

and, therefore, k is an rational number, not necessarily a positive integer. In Case 2, I conclude by affirmation that f,k are assumed transcendental and thus irrational numbers and thus k=fz only if k is a positive integer.Thus, when common factors f>1 exist, when no common factors in z,x,y exist and no common terms exist, BC is disproved because, first, for most sets z,x,y,m,a,b (8),(9) do not necessarily exist and, second, when (8),(9) do necessarily exist they are sufficient because, in (10),(11),(13) they find that factor f and product k are irrational numbers.

Case 2 arises when prime numbers are involved. Since fz=k only if k is a positive integer

. . . . . . fz,fx positive integers . . .. . . . . . . . (14)

in which, since log_z(fx)^a/m is an irrational number and log_z(1+(fy)^b/(fx)^a)^1/m is a rational number, log_zfz must be an irrational number to maintain the equality. This can happen only if z is but f,x is not a prime number or vice versa. In either case, z,f,x are assumed being transcedental and thus irrational numbers, an impossibility.

While some sets z,x,y, therefore, may satisfy BE if k is a positive integer they cannot satisfy BC because they assume z,x, common factors f and products k are irrational numbers. Thus, some sets z,x,y with prime numbers, while providing "solutions" for BE, fall outside Beal's conditions for fz,fx,fy being integers. Equations (10),(11),(13),(14) do not preclude equations (8),(9) from having solutions. They only preclude positive integers fz,fx,fy from being assumed also as irrational numbers, an impossibility.

________________________________

ADDENDUM 1. Some reviewers claim that "solutions" to BE are not very scarce. See athttp://planetmath.org/encyclopedia/BealsConjecture.html However, they assume what is to be proven. Each equation in the cited reference assumes an equality without proof of origin in BC. If common factors are canceled, each equation originates from an equality other then BE. Some examples, of sets which do not originate from BC, are 3⁵ = 6³ + 3³ and 9⁷ = 27⁴ + 162³. If common terms are canceled, both equations reduce to

3²=1 + 2³ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(15)

which falls outside BC because no common factors exist and because m=2 is not the required m>2. Additionally, while numerically correct, (15) falls outside BC because it assumes prime number 3 is an irrational number, an impossibility.

ADDENDUM 2. In (6),(7), I set is an irrational number for rational , and in (10),(11), I set as a rational number for irrational . Rational and irrational numbers are like hip joined twins. The exponential function is a continuous function from minus to plus infinity along which has both rational and irrational values. It is proven that when =r (rational) =i (irrational). Logically we can say that when =i =r because it cannot be rational.

Mathematically, the continuity of is an immediate consequence of the definition of the inverse function =logy (y>0). Function y is continuous for all values of , and which for rational values of coincides with . For irrational values of , y coincides with in the limit. Thus,

. . . . . . . . . . . . . . . . i=irrational, r=rational numbers . . . . . . . . (16)

so that when =r, e^r=i and when =i, eⁱ=r.

Copyright © 2008 James Constant

By same author: http://www.coolissues.com/mathematics/sameauthor.htm