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I want to show that $0=x^x$ has no solution for $x > 0$ in $x \in \mathbb{R}$.

I know that there isn't a solution, but I don't know how to show it mathematically.

EDIT: What I have finally written in my exercise as proof:

$x^x = 0$

$\Leftrightarrow x = \dfrac{ln(0)}{ln(x)}$

$\Rightarrow ln(0)$ is not defined, therefore no solution exists.

Basti Funck
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    Hint: What is the definition of $x^x$? – Git Gud Jun 15 '14 at 16:46
  • $x>0$, but can i assume of that, that there is no solution at all? – Basti Funck Jun 15 '14 at 16:49
  • $a^b = \exp(b \ln a)$? – Myself Jun 15 '14 at 16:56
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    @Myself if that were really the definition, then exponentials wouldn't be defined for $a < 0$. This is a somewhat more demanding question than it seems at first glance. – DanielV Jun 15 '14 at 18:56
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    @DanielV, The precise definition of taking power is what "Myself" has just stated. Negative numbers are NOT used in taking power. Just for integer powers we define, Manually, taking power for negative numbers. For example $(-1)^\sqrt2$ means nothing :) – Fardad Pouran Jun 15 '14 at 20:34
  • @DanielV that is really 'a' definition, and I think the most common one in most contemporary mathematical writing. You can provide an ad hoc definition by first defining rational powers (and showing that they actually exist) then defining real powers by a limiting process (showing that the limit exists) and passing to complex numbers by demanding that the correct properties hold; but in the long run this approach will spare you the trouble and reduce everything to basis properties of the exponential function and the logarithm. – Myself Jun 15 '14 at 21:20

3 Answers3

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Hint What's the range of the exponential function? Write $x^x$ with exponential.

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    you may want to address the fact that $(x,y)\mapsto (x^y)$ has a funny-looking domain, with the portion where $x\le0$ being all messed up. – dfeuer Jun 15 '14 at 19:03
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If Supposed that there is $t^t=0$, with $t>0$, if $t>0$ then $$\ln{t}\in \mathbb{R}$$ $$t\cdot\ln{t}\in \mathbb{R}$$ $$\ln{t^t}\in \mathbb{R}$$ but $$\ln{t^t}=\ln{0}\in \mathbb{R}$$ is a contradiction

AsdrubalBeltran
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  • Ah thanks! I think I currently have a little blackout but this clarified it. ln(0) is forbidden so no solution at all. Will take a little break and then try to comprehend all the answers. – Basti Funck Jun 15 '14 at 17:12
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    To be pedantic, the key here is that $\ln$ is undefined for zero and a bijection over reals. Just the fact that it is undefined is not sufficient to claim contradiction. – DanielV Jun 16 '14 at 15:24
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Since no one has addressed the $x<0$ case yet, I'll suggest:

For an arbitrary complex number $C = Me^{i\theta}$ and arbitrary real number $R$, $$C^R = M^Re^{iR\theta}$$ As $M>0$, $M^R > 0$ and thus $C^R \ne 0$.

DanielV
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  • There is no case $x<0$. See also the comment by Fardad Pouran beneath the original post. Yes you can 'define' it for rational exponents (essentially by choosing a non standard branch cut for the natural logarithm) but that requires a level of sophistication not present in the question. – Myself Jun 15 '14 at 21:15
  • Interesting to take the thought a step further, but we won't introduce complex numbers until the next weeks. – Basti Funck Jun 16 '14 at 14:26