FANDOM


Originally I wanted to put the proof that uparrow notation is strictly increasing, but it turned into an extremely long proof. I cut it off and will probably put it in another post.

\(a\uparrow^m(b\uparrow^mc)\) vs. \((a\uparrow^mb)\uparrow^mc\)

For all a,b,c > 2

When m = 1, \(a\uparrow(b\uparrow c) > (a\uparrow b)\uparrow c = a\uparrow(b\times c) \) holds.

Assume that : \(a\uparrow^m(b\uparrow^mc)> (a\uparrow^mb)\uparrow^mc\) for \(m\geq1\)

\((a\uparrow^{m+1} b)\uparrow^{m+1} c\)

\(=\underbrace{\underbrace{(a\uparrow^ma...a)}_{b}\uparrow^m...\uparrow^m\underbrace{(a\uparrow^ma...a)}_{b}}_{c}\)

\(<\underbrace{a\uparrow^m...a}_{b\times c}\)

\(=a\uparrow^{m+1}(b\times c)\)

\(<a\uparrow^{m+1}(b\uparrow^{m+1}c)\)

And further,

Assume that : \(a\uparrow^m(b\times c)> (a\uparrow^mb)\uparrow^mc\) for \(m\geq2\)

\((a\uparrow^{m+1} b)\uparrow^{m+1} c\)

\(=\underbrace{\underbrace{(a\uparrow^ma...a)}_{b}\uparrow^m...\uparrow^m\underbrace{(a\uparrow^ma...a)}_{b}}_{c}\)

\(<\underbrace{a\uparrow^m...a}_{b\times c}\)

\(=a\uparrow^{m+1}(b\times c)\)


In different cases :

\(a+b=a\underbrace{+1+1...+1}_{b}\)

if \((a+1)+1=a+(1+1)\)

\((a+b)+c=(a\underbrace{+1+1...+1}_{b})\underbrace{+1+1...+1}_{c}\)

\(=a\underbrace{+1+1...+1}_{b+c}\)

\(=a+(b+c)\)

\(<a+(b\times c)\)


if \((a+b)+c=a+(b+c)\)

\((a\times b)\times c=\underbrace{(\underbrace{a+a+...a}_{b})+...\underbrace{a+a+...a}_{b}}_{c}\)

\(=\underbrace{a+a+...a}_{b\times c}\)

\(=a\times(b\times c)\)

\(=a\times(b\times c)\)


if \((a\times b)\times c=a\times(b\times c)\)

\((a\uparrow b)\uparrow c=\underbrace{(\underbrace{a\times a\times ...a}_{b})\times ...\underbrace{a\times a\times ...a}_{b}}_{c}\)

\(=\underbrace{a\times a\times...a}_{b\times c}\)

\(=a\uparrow(b\times c)\)

\(<a\uparrow(b\uparrow c)\)


if  \((a\uparrow^mb)\uparrow^mc<a\uparrow^m(b\uparrow^mc) \) for \(m\geq2\)

\((a\uparrow^{m+1} b)\uparrow^{m+1} c\)

\(=\underbrace{\underbrace{(a\uparrow^ma...a)}_{b}\uparrow^m...\uparrow^m\underbrace{(a\uparrow^ma...a)}_{b}}_{c}\)

\(<\underbrace{a\uparrow^m...a}_{b\times c}\)

\(=a\uparrow^{m+1}(b\times c)\)

\(<a\uparrow^{m+1}(b\uparrow^{m+1}c)\)

\(+\) \(\times\) \(\uparrow\) \(\uparrow\uparrow\)&beyond

(AoB)oC

vs.

Ao(BoC)

\(=\) \(=\) \(<\) \(<\)

(AoB)oC

vs.

Ao(B\(\times\)C)

\(<\) \(=\) \(=\) \(<\)

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