Transcendental question?

First off, it is typical to say "algebraic" instead of "non-transcendental", especially sense the transcendental numbers are the strangers to intuition while the algebraic ones make "sense".

And on to your questions:

How do you prove that an irrational number in general is transcendental?

Assume it is algebraic and find a contradiction. In general, this is quite hard, and some questions surrounding this topic are still unknown.

Or does someone know how to prove that pi is transcendental?

I've seen the proof before... search online for it, but it's definitely not simple...

Is e transcendental?

Yes, so is 2^(2^(1/2)) and many others (in fact, given a randomly chosen real number, the probability that it is transcendental is 100%).

Can we prove that for any transcendental number z, for all "transcendentally conservative" functions f, f(x) is also transcendental?

Nope... in general, we'd also have f(x) = 0, or nontrivially, we might have a function that that maps all of the real numbers into the algebraic numbers.

For some classes of functions, your conjecture will be true (for instance, the nonconstant polynomials with algebraic coefficients).

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Sketch of how to show the transcendental numbers are uncountable (in fact, co-countable).

Step 1: Show that the countable union of countable sets is countable.

Step 2: Show that the algebraic numbers are countable.

Step 3: Show that the real numbers are uncountable.

Step 4: Use contradiction to show that the transcendental numbers must be uncountable. In particular, they are "most" of the reals, since we showed in step 2 that the algebraic numbers are countable.

I'll elaborate a little on step 2:

Consider a general n-degree polynomial with integer coefficients. For a given one, there is at most n zeros. Since the integers are countable, the number of such polynomials is countable, so that the number of zeros for n-degree polynomials is at most countable.

Now, there are a countable number of possibilities for the degrees, so this means there are at most countable number of zeros of polynomials with integer coefficients. This is precisely the statement that the algebraic numbers are at most countable.

(this is enough, but to show they are countable, note that the integers are all zeros of linear polynomials, so they must all be algebraic, forcing the algebraic numbers to be countable)

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Sorry about that, I didn't notice you said a bijection. If f is a bijection as well as continuous, then I believe it may be true... I may look into it further.

Actually, I believe it still won't be true, although I don't have a counterexample - only intuition - behind my stating this, so I would be open to being wrong here.

Let me add to the discussion... I may not really answer your question forthright.

The first number shown to be constant is known as Liouville's constant (1st reference). This number was shown to be transcendental as a result of Liouville's theorem (which is listed as a lemma in the 1st reference and is pretty easy to prove). Basically, Liouville's theorem says that algebraic number cannot be approximated very well by rational numbers in some technical sense. If you can construct a number which is approximated too well by rationals, then it's a transcendental number. Such numbers are easy to construct and are generally called Liouville numbers. However, there aren't that many of them--the set of Liouville numbers has measure 0. So most transcendental numbers don't look like this, and you have to resort to other methods.

The Gelfond-Schneider theorem (2nd reference) was a big breakthrough in constructing transcendental numbers. Given any algebraic number a ≠ 0, 1 and any irrational algebraic number b, a^b is transcendental (where a^b = e^(b log(a)) for any value of log(a)). However, this only gives countably many examples.

However, both the Gelfond-Schneider theorem and the Lindemann-Weierstrass theorem are results of a more general statement about algebraic independence of analytic functions. See the third reference for complete details. I believe that this statement can be used to show the transcendence of e and π in addition to a while lot of other numbers, like e^π. The general arguments that are used in the classical proofs of the transcendence of e and π are generalized here.

I wish I could remember the arguments a little better, but the idea is something like... if f(z) is an analytic function so that f'(z) is algebraically related to f(z) over the rationals, then there cannot be too many algebraic numbers a so that f(a) is also algebraic. Something like that. So for e, you use that (d/dz)(exp(z)) = exp(z), and for π, you use that (d/dz)(sin(z))^2 + sin^2(z) = 1.

I don't really have any thoughts about your conjecture.... but I'll think about it.

Not all irrational numbers are transcendental. However, it is true that all transcendental numbers are irrational, since all rational numbers are algebraic. For instance, √2 is algabraic as it is a solution to the function x² - 2 = 0.

pi is the number 4/1 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + 4/13 - 4/15 + ....

e is the number 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + ....

These numbers cannot possibly be the solutions to any polynomial. So they are transcendental

e transcendental:

See top two sources (two very different proofs).

π transcendental:

Follows from the fact about e and the Lindemann-Weierstrass Theorem, see third source (this can also be used to show e is transcendental). As you said, if π was algebraic, so would be iπ, which contradicts the L-W theorem and e^(iπ) = -1.

The reason you gave, however, is incorrect.

Let x = 2^(√2)

y = √2

Then x is transcendental (Gelfand-Schneider theorem, see fourth source) and y is algebraic and irrational. However,

x^y = 2^2 = 4 is not transcendental.

As for your function, I will continue to think about it.

Steve

I think I can answer all these with the following:

Such questions are *usually* (as far as I know) treated with Mathematical Induction, which I'm not going to go into, and if that disqualifies me from the points...oh well..... Frankly I don't care, cuz I don't spend time here to gain points....but, anyway in order to prove the behavior of an equation is ongoing, or does this or that, the preferred approach is that of induction.

Perhaps others will either deliver all the math to your doorstep, or offer other ways to show the behavior you are asking about.

As for "pi", there are tons of algorithms for enumerating the decimal places. I know this sounds like a cop-out, but do a search or two on Google (pi algorithms) and look thru the different methodologies. Pi is one of the most studied mathematical objects in the mathematical sciences, and therefore is heavily and thoroughly documented.

Hope this helps.