15
$\begingroup$

Consider real numeric expressions build only from integers, operators $+,-,\times,/$ and taking a positive expression to a power (no variables involved), e.g. $$\frac{2}{7},\ 2^{1/2},\ \sqrt[5]{2+\sqrt{11}},\ 2^{\sqrt{3}},\ ...$$ Now we can write identities between such expressions that are either true, e.g. $\sqrt{3+\sqrt{8}}=1+\sqrt{2}$, or false, e.g. $\frac{3}{5}=\frac{2}{7}$.

Is it possible to prove all such true identities and disprove all false identities using only usual "high-school" algebra rules?

$\endgroup$
  • 7
    $\begingroup$ en.wikipedia.org/wiki/Tarski's_high_school_algebra_problem $\endgroup$ – Lord Soth Nov 1 '13 at 23:26
  • $\begingroup$ @LordSoth If I understand correctly, Tarski's problem asks about identities that may involve variables. $\endgroup$ – X.C. Nov 1 '13 at 23:32
  • 1
    $\begingroup$ How could taking an irrational power even be expressed "using only usual 'high-school' algebra rules"$\hspace{.02 in}$? $\;\;\;\;$ $\endgroup$ – user57159 Nov 1 '13 at 23:50
  • 2
    $\begingroup$ @RickyDemer $2^{\sqrt{3}}=2^{3^{1/2}}$ $\endgroup$ – X.C. Nov 1 '13 at 23:52
  • 8
    $\begingroup$ I am not even sure that $ZFC$ axioms are sufficient to settle all such questions. We do not yet know if $2^{2^{2^{2^{2^{2^{1/2}}}}}}$ is an integer. $\endgroup$ – Vladimir Reshetnikov Nov 2 '13 at 0:26
5
+50
$\begingroup$

As far as I know, this is an open problem. It might be the case that even $ZFC$ axioms are not sufficient to settle all such questions, or that the problem is even undecidable (although it seems unlikely). The main issue that we know very little about behavior of repeated exponentiation. For example, it is unknown if the following number is an integer: $$2^{2^{2^{2^{2^{2^{1/2}}}}}}$$ Personally, I expect that there are no surprises to be discovered in the repeated exponentiation, that this number will eventually be proved to be transcendental, and that there are no unexpected identities in your class of expressions that could not be explained using only "high-school" algebra rules. But these are only conjectures.

$\endgroup$
  • 3
    $\begingroup$ I like your conjectures. $\endgroup$ – Betty Mock Nov 8 '13 at 20:59
-5
$\begingroup$

I'll answer no.

Suppose you could prove all identities using such rules. Then there could get written some infinite list of all such identities (you could enumerate all such identities by a characteristic function which returns "true" for each true identity and "false" for each false identity). But, that would imply you could have a bijection between the set of all natural numbers and all of those identities. So, you could encode all of those identities via the natural numbers. Included in this list identities is all identities involving real numbers including pi=pi, e=e, sqrt(2)=sqrt(2). That set of identities clearly has the same cardinality as the set of all real numbers. But, there does not exist any bijection between the set of real numbers and the set of natural numbers. So there does not exist any bijection between the set of all natural numbers and that of all such "high-school" algebra identities. Therefore, you cannot prove "all such true identities and disprove all false identities using only usual "high-school" algebra rules".

$\endgroup$
  • 8
    $\begingroup$ The (hypothetical) list does not include all identities involving real numbers. e.g. $\pi=\pi$ is not identity involving real numeric expressions build only from integers, operators +,−,×,/ and taking a positive expression to a power. $\endgroup$ – Hurkyl Nov 2 '13 at 4:45
  • 2
    $\begingroup$ The set of identities in question is countable and recursively enumerable. $\endgroup$ – OlegK Nov 2 '13 at 16:42
  • 1
    $\begingroup$ @Hurkyl Strictly speaking, we do no know if $\pi$ is expressible in this way. But you are right, it does not include all real numbers. $\endgroup$ – Vladimir Reshetnikov Nov 2 '13 at 17:03
  • $\begingroup$ @Vladimir: To be picky, "$\pi$" is not an expression of the form, although there may be other forms expressible in this way that have the same value as $\pi$. $\endgroup$ – Hurkyl Nov 3 '13 at 0:24
  • $\begingroup$ @Hurkyl You're right, but the list can get used to form sequences of pairs of identities like the following: [(2=2, 3=3), (2.7=2.7, 2.8=2.8), (2.71=2.71, 2.72=2.72), (2.718=2.718, 2.719=2.719), (2.7182=2.7182, 2.7183=2.7183), (2.71828=2.71828, 2.71829=2.71829), ...], and in general sequences of pairs of identities that get closer and closer to any real number can get found. So, how does it matter that the (hypothetical) list doesn't include all identities involving real numbers? $\endgroup$ – Doug Spoonwood Nov 3 '13 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.