## a^b

Posted by Zorglub on May 23, 2012 – 5:39 pm

Let a and b be two irrational numbers. Is it possible that a^b is rational ?

Posted by Zorglub on May 23, 2012 – 5:39 pm

Let a and b be two irrational numbers. Is it possible that a^b is rational ?

May 23rd, 2012 at 6:14 pm

Yes

I know at least one “general” solution.

May 23rd, 2012 at 7:38 pm

sqrt(2)^x=3, x and sqrt(2) is irrational, 3 is rational

May 23rd, 2012 at 8:16 pm

Hi Kel9902. You need to prove that x (and sqrt2) is irrational.

I know a proof that x is irrational, but I’ll keep that secret for a while.

I assume that everyone knows how to prove that sqrt2 is irrational. That proof is very different to that for x.

May 24th, 2012 at 4:57 am

As it’s an aside, I’ll prove that √M is irrational unless M is a square.

I’ve never seen this proof before, so I’m quite proud of it. I only thought of it after my last post.

Let √M = P/Q where gcd(P,Q) = 1. Multiply both sides by Q and square

=> P

^{2}= M Q^{2}.P, Q and M may be represented by their prime factors. Let P = p

_{1}^a_{1}p_{2}^a_{2}…,Q = q

_{1}^b_{1}q_{2}^b_{2}… and M = m_{1}^c_{1}m_{2}^c_{2}… Substituting =>p

_{1}^2a_{1}p_{2}^2a_{2}… = m_{1}^c_{1}m_{2}^c_{2}… q_{1}^2b_{1}q_{2}^2b_{2}…By the fundamental theorem of arithmetic, this can only be satisfied if

p

_{1}^2a_{1}p_{2}^2a_{2}… = m_{1}^c_{1}m_{2}^c_{2}… and q_{1}^2b_{1}q_{2}^2b_{2}… = 1 and soM = P

^{2}and Q = 1. Otherwise √M is irrational.If the prime representation is in standard/canonical form, then m

_{i}= p_{i}^{2}Similar considerations apply for the nth root of M.

May 24th, 2012 at 2:17 pm

From Kel’s problem … √2 ^ x = 3

√2 is irrational because it cannot be represented as a ratio of two integers.

3 is rational because it can be. (3/1)

The question is whether or not x is irrational.

2 ^ (x/2) = 3

log(base2)(x/2) = log(base2)(3)

x/2 = log(base2)(3)

x = 2 log(base2)(3)

x = 2 log(base10)(3) / log(base10)(2)

x = 3.1699250014423123629074778878956 (and maybe more)

Even if I’ve solved for x correctly I do not know how to prove that this is irrational, even if I have the thought that it is.

if 2 log(base2)(3) is irrational, then this is proof that a and b can be irrational and a^b could be rational.

May 24th, 2012 at 2:27 pm

proof that log(base2)(3) is irrational.

[found as an example on wiki]

Assume log2 3 is rational. For some positive integers m and n, we have

log2 3 = m/n

2^(log 2 3) = 2^(m/n)

3 = 2^(m/n)

3^n = 2^m

This is a contradiction since 3^n for a positive n is an odd number and 2^m for a positive m is an even number.

The only assumption is that log2 3 was rational, so this assumption must be wrong.

Therefore, log2 3 is irrational.

√2 ^ (2 log2 3) = 3 proves a^b can be rational even if a and b are irrational.

May 24th, 2012 at 2:27 pm

Hi Cazayoux. Use powers rather than logs.

May 24th, 2012 at 3:36 pm

Posts crossed. Another way is that 3^n = 2^m isn’t possible by the fundamental theorem of arithmetic.

I’ll post my stuff in 10 mins or so.

May 24th, 2012 at 3:46 pm

I remember this from some distant past (but using √2

^{√2}).Let x = √a

^{√b}where a and b are non-square. As I showed, in post 4, √a and √b are irrational. It is possible (a priori) that x is rational in which case we’re done. If not, consider y = x^{2√b}= a^{b}and that is rational (in fact integral).Kel9902 suggested √2

^{x}= 3. Squaring => 2^{x}= 3^{2}. If x is rational, then we may set x = p/q and assume gcd(p,q) = 1. Then raising to the power of q => 2^{p}= 3^{2q}and that isn’t possible according to the fundamental theorem of arithmetic. So x is irrational and Kel9902’s assertion is proven.May 25th, 2012 at 3:52 am

Cazayoux’s proof that 2log2 3 is irrational in post 6 is very nice. When I proposed the post I had in mind using x = √2^√2, and to used the development proposed by Chris in post 9.

What I like about this answer is that the answer to the question is yes, but the proof does not need to distinguish if it is x or x^√2 which is irrational.

May 25th, 2012 at 4:39 am

Hi Zorglub. I like the sneaky existence only aspect too.

Kel9902’s idea can be extended: √b

^{x}= n, where b is non-square and n is integral. It is obvious that if b or n have at least one non-shared prime factor, that alone guarantees that x must be irrational. So Kel9902’s idea can be used to construct x.June 19th, 2012 at 11:36 pm

Let x = 2^sqrt(2), which is irrational.

Let y = sqrt(2), which is also irrational.

Now, x^y = 4, by simple application of Law of Indices!

Hence, Irrational^Irrational = Rational (Q.E.D.)

June 20th, 2012 at 4:58 am

Hi Gerami. I accept that it is well known that sqrt(2) is irrational. You haven’t justified your assertion that 2^sqrt(2) is irrational.

June 20th, 2012 at 10:05 pm

Hi Chris,

2^sqrt(2) is not only irrational but also transcendental.

This follows from the solution to Hilbert’s 7th problem (of the 23 problems he formulated in 1900).

Please refer:

http://en.wikipedia.org/wiki/Hilbert%27s_problems

June 21st, 2012 at 3:52 am

Hi Gerami. Thank you

I had unsuccessfully tried to establish the case for 2^sqrt(2) after my last post.

I’d completely forgotten about the Gelfond-Schneider theorem.

I also note that 2^sqrt(2) is known as the Gelfond-Schneider constant.

PS I’ve just found a supposed proof of the Gelfond-Schneider theorem.

June 21st, 2012 at 5:06 am

On this site, over the years, transcendental numbers have been mentioned a few times. So I thought I’d write a few comments about why they are important. The following is off the top of my head and uses memories from a long time ago.

First a digression. For a long time, the geometers thought that geometry was the real mathematics and the algebra was a lesser topic. This was because in geometry, for each theorem that stated a relationship of points to lines, there was a dual theorem that related lines to points. i.e. you simply swapped the words line and point.

However, some bright spark changed all that. He noticed that e.g. the (point) equation of the plane: ax + by + cz = p was usually understood to mean that a, b, c, p were all constant, and that x, y, z were variables. But if you regard x, y, z, p to be constants, and a, b, c to be variables, you get the plane equation of the point. i.e. the equation can be interpreted as describing all the planes that pass through a given point. Voila! – by exchanging the words variable and constant, the algebraists have duals too.

Back to the topic. It turns out that if a geometrical object is constructible with compass and straightedge alone (in a finite number of steps), then that construction can be described in terms of the roots of a finite order polynomial equation with integer coefficients. Such roots are called algebraic numbers. The converse of algebraic is transcendental. i.e. a transcendenal number cannot be a root of a finite order integer coefficient polynomial and cannot be constructed by geometrical means.

Put another way, algebraic numbers can be constructed by geometric means. It turns out that e.g. sqrt(2) is algebraic. It can be constructed as the hypoteneuse of a 90° triangle (the other two sides be of length 1) and as the root of the polynomial x

^{2}– 2 = 0. However, π and e are transcendental, and so cannot be constructed by geometric means. Because π is transcendental, it is not possible to square the circle – see http://en.wikipedia.org/wiki/Squaring_the_circleThese considerations also “explain” why integer coefficient polynomials are so important.

PS. Integer coefficients can be replaced with rational coefficients. The latter can be converted to the former simply by multiplication by the LCM of the denominators of the rationals.