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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 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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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. Tags calculus, comparing, large, numbers Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post kmkv6dl Advanced Statistics 3 July 16th, 2019 08:48 PM Sequoia Algebra 3 August 16th, 2017 11:22 AM jk12 Math 0 May 13th, 2017 04:45 PM Tomi Advanced Statistics 0 February 12th, 2014 12:00 PM sachinrajsharma Algebra 1 February 21st, 2013 08:42 PM

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