Let us consider the number $$\Large\pi^{\pi^\pi}=\pi^{\pi\cdot\pi}=\pi^{\pi^2}$$
As the bases are equal, the exponents must be equal, So $$\pi=2$$
You can take any $x$ instead of $\pi$.
What is wrong in this proof?
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3What is the argument why that should somehow indicate that $\pi = 2$? – Daniel Fischer Dec 02 '13 at 10:51
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I edited the question – Kartik Dec 02 '13 at 10:52
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Maybe you are mixing up $2\pi$ and $\pi^{2}$ here, – drhab Dec 02 '13 at 10:53
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4The first equality is false. $x^{x^x}\neq x^{x\cdot x}$ – Callus - Reinstate Monica Dec 02 '13 at 10:53
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@callus Why? It was taught to us in school in 7th class that $a^{b^c}=a^{b\cdot c}$ – Kartik Dec 02 '13 at 10:54
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7@Kartik No (hopefully). What was taught was most likely $\left(a^b\right)^c = a^{(b\cdot c)}$. In general, $a^{(b^c)} \neq \left(a^b\right)^c$. – Daniel Fischer Dec 02 '13 at 10:55
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@DanielFischer So we can change the question to $(\pi^\pi)^\pi$ – Kartik Dec 02 '13 at 10:58
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So we have $\left(\pi^\pi\right)^\pi = \pi^{(\pi\cdot\pi)} = \pi^{\left(\pi^2\right)}$. All correct so far. Nothing to suggest $\pi=2$. – Daniel Fischer Dec 02 '13 at 11:01
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@DanielFischer Why? Still we can say that the bases are equal, so exponents will be equal – Kartik Dec 02 '13 at 11:02
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1@Kartik That the exponents are equal just says $\pi \cdot \pi = \pi^2$. – Daniel Fischer Dec 02 '13 at 11:09
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Oh, you just changed it. Now the first equation is wrong, $a^{b^c} = a^{\left(b^c\right)} \neq \left(a^b\right)^c = a^{b\cdot c}$. – Daniel Fischer Dec 02 '13 at 11:11
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I just changed it because ${\pi^\pi}^\pi$ looked like $\pi^{\pi\pi}$ It does not change the meaning of the question. – Kartik Dec 02 '13 at 11:12
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1@Kartik It changes the meaning of the expressions in the chain of equations. $\pi^{\pi^\pi} = \pi^{\left(\pi^\pi\right)}$ is different from $\left(\pi^{\pi}\right)^\pi = \pi^{\pi\pi}$. – Daniel Fischer Dec 02 '13 at 11:15
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I meant to say that I only changed it because the typesetting was not correct, I did not change the meaning of the question. – Kartik Dec 02 '13 at 11:17
3 Answers
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I think you are mixing up $$ \left(\pi^\pi\right)^\pi=\pi^{\pi^2} $$ with $$ \pi^{\left(\pi^\pi\right)}\neq \pi^{\pi^2}. $$
In general, $$ \left(a^b\right)^c\neq a^{\left(b^c\right)} $$ but if $a=b=c=2$ it is true since then $b\times c=b^c$.
hejseb
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6You might add that unadorned $a^{b^c}$ means $a^{(b^c)}$ by notational convention. – Marc van Leeuwen Dec 02 '13 at 11:33
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Lets write $a \uparrow b$ to mean $a^b$.
Then the following reasoning is correct: $$(\pi \uparrow \pi)\uparrow \pi = \pi \uparrow (\pi \cdot \pi) = \pi \uparrow (\pi \uparrow 2)$$
However, we cannot necessarily deduce that the RHS equals
$$(\pi \uparrow \pi) \uparrow 2$$
because exponentiation isn't associative. Indeed, Google calculator tells me that:
$\pi \uparrow (\pi \uparrow 2) \approx 80662.6659386$
$(\pi \uparrow \pi) \uparrow 2 \approx 1329.48908322$
so if the calculator is correct to even the first decimal place, then
$$(\pi \uparrow \pi) \uparrow 2 \neq \pi \uparrow (\pi \uparrow 2).$$
Moral of the story: if in doubt, find better notation!
goblin GONE
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If you can take any $x$ instead of $\pi$, why didn't you take $3$ or $10$ instead of $\pi$?$$10^{10^{10}}=10^{10000000000}$$ $$10^{10\cdot10}=10^{100}\ne10^{10000000000}$$
bof
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