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 October 3rd, 2019, 07:45 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Comparing large numbers without calculus A curious logical consequence of the principle of non-contradiction is that a contradiction implies any statement; if a contradiction is accepted as true, any proposition (or its negation) can be proved from it. Let's try (*)$\displaystyle 66^{77} > 77^{66}$ . I will prove (*) using sgn function(analytic one) . By this equallity : $\displaystyle sgn(x)=-1+2\lim_{s\rightarrow \infty} (1+e^{-2sx})^{-1}$. Let $\displaystyle x=66^{77}-77^{66}$ and suppose $\displaystyle 77^{66}>66^{77}$ or $\displaystyle x<0 \equiv sgn(x)<0$. $\displaystyle sgn(x)=-1+2\lim_{s\rightarrow \infty} \frac{1}{1+e^{-2s\cdot sgn(x)(77^{66}-66^{77})}}\: \equiv \: -1+2\lim_{(s,x(s) )\rightarrow \infty} \frac{1}{1+e^{-2s|x(s)|}}.$ $\displaystyle sgn(x)=-1+2\cdot \frac{1}{1+e^{-2\cdot \infty |\infty| }}=-1+2=1>0$ which is a contradiction of $\displaystyle sgn(x)<0 \;$, therefore (*) is proven . Last edited by idontknow; October 3rd, 2019 at 07:49 AM.
 October 3rd, 2019, 09:22 AM #2 Senior Member   Joined: Jun 2014 From: USA Posts: 620 Thanks: 52 If I understand correctly, the principle of explosion... https://en.m.wikipedia.org/wiki/Principle_of_explosion ...is the reason for the principle of noncontradiction: https://en.m.wikipedia.org/wiki/Law_of_noncontradiction It is true that $66^{77} > 77^{66}$ isn’t it? Is there an issue? I guess I’m confused. Thanks from idontknow
 October 3rd, 2019, 10:21 AM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,588 Thanks: 1432 are you seriously leaving $\dfrac{1}{1+e^{-2\infty|\infty|}}$ as a directly evaluable term? I mean it's pretty clear that it's 1 but you can't just leave it like that. Thanks from idontknow
October 3rd, 2019, 10:53 AM   #4
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Joined: Dec 2015
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Quote:
 Originally Posted by romsek are you seriously leaving $\dfrac{1}{1+e^{-2\infty|\infty|}}$ as a directly evaluable term? I mean it's pretty clear that it's 1 but you can't just leave it like that.
I agree , I evaluated it in short-terms.
In this case $\displaystyle \lim_{a\rightarrow \infty } \lim_{b\rightarrow \infty} e^{-ab} = \lim_{b\rightarrow \infty } \lim_{a\rightarrow \infty} e^{-ab}$ holds true .
The limit is with two variables .

Last edited by idontknow; October 3rd, 2019 at 10:56 AM.

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