The question states:

Simplify $2^{\sqrt2}$

I tried doing: $$2^{\sqrt2}=2^{(2^{\frac{1}{2}})}$$ $$=(2^2)^{\frac{1}{2}}$$ $$=4^{\frac{1}{2}}$$ $$=2$$ $$=2^1$$

Clearly this is ridiculous as I've just "proved" $2^{\sqrt2}=2^1$. Yet my proof seems quite convincing to me. Where did I go wrong? I tried debugging by trying a similar technique on $2^4$ and it worked fine.

$$2^{4}=2^{2^{2}}$$ $$=(2^2)^{2}$$ $$=4^{2}$$ $$=16$$ $$=2^4$$

  • 1
    $\begingroup$ Your second equality is false. RHS is $2^{1}$ and LHS is $2^{\sqrt 2}$. $\endgroup$ Sep 28 at 9:42
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    $\begingroup$ $a^{b^c}\neq(a^b)^c$. $\endgroup$
    – Gonçalo
    Sep 28 at 9:43
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    $\begingroup$ People tend to overlook the fact that iterated exponentiation is not associative. As a matter of definition, $a^{b^c}$ means $a^{(b^c)}$. $\endgroup$
    – lulu
    Sep 28 at 9:44
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    $\begingroup$ Anyway, I think $2^{\sqrt2}$ is already as simple as it's going to get. I don't know what someone had in mind when they asked for it to be simplified. $\endgroup$ Sep 28 at 9:45
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    $\begingroup$ The second example works because of $2^2=2\cdot 2=4$ $\endgroup$
    – Peter
    Sep 28 at 10:59

1 Answer 1


Because $x^{(y^z)}$ (like you did in the first example) does not mean $(x^y)^z$. It works in your second example because by chance $2^2=2\times 2$.

We say that the exponentiation does not have the associative property.


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