To get an integer to become a different integer, you have to add or subtract another integer, e.g. $1+2=3$

To get a rational number to become an integer, you have to multiply by the denominator, e.g. $\frac{3}{4}(4) = 3$

To get an irrational number to become an integer, you have to raise it to some power, e.g. $(\sqrt{3})^2 = 3$

From my understanding, transcendental numbers are numbers where there is no operation (done in one step) that can bring a transcendental number back into the integer realm; I know it is easy to get an integer from a transcendental number in two operations/steps, e.g. $e+e=2e\quad \frac{2e}{e}=2$

Is this because (1) no operation exists in mathematics (because 'mathematics transcends the human mind' - pun intended) or (2) because no operation known to man exists?

I know that there is some ambiguity in that question so I will try my best clarify: by option 1, I mean is there a proof that says that such an operation cannot exist and by option 2, I mean no such proof exists.

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    $\begingroup$ Well, if we are allowed the function $\ln$, then $\ln(e^{47})=47$. And $\ln$ is one button press on my calculator! $\endgroup$ – André Nicolas Aug 8 '13 at 1:25
  • $\begingroup$ Can you give a precise definition of operation? For example, if I take $\alpha = \ln 2$, then I have $\exp(\alpha) = 2$. Is this not one operation? $\endgroup$ – Brandon Carter Aug 8 '13 at 1:26
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    $\begingroup$ the "floor" and "ceiling" functions will take any of your examples as input and output an integer. So will "f(x) = x / x" $\endgroup$ – scott_fakename Aug 8 '13 at 1:26
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    $\begingroup$ "To get an irrational [but not-transcendental] number to become an integer, you have to raise it to a power." Um ... $\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1} = 2$. No powers of either factor used. $\endgroup$ – Blue Aug 8 '13 at 1:57
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    $\begingroup$ It's also worth pointing out that there are non-transcendental irrationals that aren't expressible via radicals; for instance, "most" roots of "most" polynomials of degree greater than $4$. Your "done in one step" criterion (as I understand it) for converting to integers wouldn't apply to those. $\endgroup$ – Blue Aug 8 '13 at 1:59

You're a little bit off with your definition of transcendental numbers. The "algebraic numbers" (i.e. the "non-transcendental" numbers) can be thought of as follows. We begin by considering the numbers that can be brought to zero under the following operations:

  • addition by an integer
  • multiplication by a (nonzero) integer
  • (positive) integer powers

Any such number and any sum of finitely many of these numbers is algebraic. If a number isn't algebraic, it's transcendental.

$3$ is algebraic because $3-3=0$.

$\frac34$ is algebraic because $4\times \frac34 = 3$, and $3-3=0$.

$\sqrt3$ is algebraic because $(\sqrt3)^2=3$, and $3-3=0$.

$e$ is transcendental because no amount or order of these operations will bring $e$ to $0$. We can't simply divide $e$ by $e$ or do anything like that because $e$ is not a rational number, and we can't call that one step because $e$ certainly is not an integer.

The equivalent (and more concise) statement of this definition, by the way, is that an algebraic number is one that is the root of some polynomial $a_0+a_1x+a_2x^2+\dots+a_nx^n$, where each coefficient $a_k$ is an integer. So for example, $\sqrt2-1$ is algebraic because $$ (\sqrt2-1) + 1=\sqrt 2\\ (\sqrt 2)^2=2\\ 2-2=0 $$ Equivalently, $\sqrt 2-1$ is a root of the equation $$ (x+1)^2-2=0\implies\\ x^2+2x-1=0 $$

Note: You could (as some do) add to the list of allowable operations division by an integer, which would allow you to consolidate some steps by using rational numbers. However, this does not expand the set of numbers you can bring to zero because any root of a polynomial with rational coefficients is the root of some polynomial with integer coefficients.

Along the same lines, you could allow negative/rational powers with out changing the outcome.

Second note: if you're trying to understand all this, I highly recommend this video (if you aren't already a numberphile fan), which does a good job of explaining what I've said here.

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    $\begingroup$ @Omnomnomnom Thanks for the clarification and video! $\endgroup$ – zerosofthezeta Aug 8 '13 at 1:58
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    $\begingroup$ I think the definition you give can only produce algebraic polynomials whose minimal polynomial has a solvable Galois group: specifically, you forgot to allow adding together numbers produced by the operations you listed, so you only get those algebraic numbers that can be constructed with radicals. (with the caveat you have to do so in a nontrivial way; e.g. not simply add $x$ to $(-1) \cdot x$) $\endgroup$ – user14972 Dec 29 '14 at 12:26
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    $\begingroup$ The video is really funny and good! $\endgroup$ – Bman72 May 11 '15 at 12:50

Your question is a little vague so I am going to give you step by step answers. Firstly here is an operation which gives you an integer:

On input $x$ output 1.

Now you may consider this cheating, but if you want to rule this out you have to be more explicit by what you mean by "operation". Lets say you allow addition, subtraction, multiplication, division, exponentiation and logs. Here is a new operation:

On input $x(\neq 0)$ output $x/x$.

If you don't like this, then you have to say why. Maybe I can only divide by a number I specify before you tell me what $x$ is or something. The restrictions which apply to the transendentals are to do with polynomials with rational coefficients. So if you take as your operation "evaluate a polynomial in rational coefficients on input $x$" then you have your proof that transcendentals can't be mapped to integers by an "operation". Why?

Because a real is transcendental precisely if it is not the root of a polynomial with integer coefficients. If there was such a polynomial $p(x)$ so that your transcendental $y$ satisfied $p(y)=n$ for some $n\in\mathbb{N}$ then $y$ would be a root of the polynomial $p(x)-5$ which has integer coefficients, which contradicts that $y$ is transcendental.


What about e and then you do $e^{i\pi}$ ? I feel like that is a good counterexample.

  • $\begingroup$ to clarify e^i(pi) = -1 and it's a single operation. $\endgroup$ – Neil Mar 14 '15 at 18:32

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