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Difference between revisions of "User:Objectz/Math"
 
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<math> x \downarrow \downarrow \downarrow \downarrow y = (x \uparrow \uparrow \uparrow y)^{-(x \uparrow y)}</math>. So, <math>3 \downarrow \downarrow \downarrow \downarrow 3 = (3 \uparrow \uparrow \uparrow 3)^{-(3 \uparrow 3)} = (3^{7625597484987})^{-27} = 3^{-205891132094649}</math>.
 
<math> x \downarrow \downarrow \downarrow \downarrow y = (x \uparrow \uparrow \uparrow y)^{-(x \uparrow y)}</math>. So, <math>3 \downarrow \downarrow \downarrow \downarrow 3 = (3 \uparrow \uparrow \uparrow 3)^{-(3 \uparrow 3)} = (3^{7625597484987})^{-27} = 3^{-205891132094649}</math>.
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<math>10 \downarrow \downarrow \downarrow \downarrow 10 = (10 \uparrow \uparrow \uparrow 10)^{-(10 \uparrow 10)} = (10 \uparrow \uparrow \uparrow 10)^{-10000000000} = \boxed{(10^{10^{10000000000}})^{-10000000000}}</math>.
  
 
<math>10 \underbrace{\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \downarrow \downarrow \downarrow \downarrow 10} 10 = ObjectZ_{1}</math>.
 
<math>10 \underbrace{\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \downarrow \downarrow \downarrow \downarrow 10} 10 = ObjectZ_{1}</math>.
  
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<math>10 \underbrace{\downarrow \downarrow \downarrow\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \underbrace{\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \downarrow \downarrow \downarrow \downarrow 10} 10} 10 = ObjectZ_{2}.</math>
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<math>10 \underbrace{\downarrow \downarrow \downarrow\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \underbrace{\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \downarrow \downarrow \downarrow \downarrow 10} 10} 10 = ObjectZ_{2}</math>.
  
 
<math>ObjectZ_{513298}</math> is ObjectZ's Number.
 
<math>ObjectZ_{513298}</math> is ObjectZ's Number.
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<math>n_{\Omega}(1) = \underbrace{ObjectZ_{ObjectZ_{ObjectZ_{ObjectZ_{ObjectZ_{ObjectZ_{... _ {ObjectZ_{ObjectZ's ~ Number}}}}}}}}}_{ObjectZ's ~ Number}</math>.
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<math>n_{\Omega}(2)</math> is the attachment in the post when you click on the link. https://artofproblemsolving.com/community/c1230676h2319478p18794259
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<math>n_{\Omega _ {\Omega}}(1) = \underbrace{n_{\Omega}n_{\Omega} \cdots \cdots n_{\Omega}n_{\Omega}}_{(n_{\Omega}(1))} (n_{\Omega}(1))</math>

Latest revision as of 14:38, 9 November 2020

This is my math subpage. These contain functions that I have created.

$x \downarrow y = x^{-(x \uparrow y)}$ where $x \uparrow y = x^{y}$ so $3 \downarrow 3 = 3^{-(3 \uparrow 3)} = 3^{-27}$.

$x \downarrow \downarrow y = (x \uparrow y)^{-(x \uparrow y)}$ so $3 \downarrow \downarrow 3 = (3 \uparrow 3)^{-(3 \uparrow 3)} = 27^{-27} = 3^{-81}$.

$x \downarrow \downarrow \downarrow y = (x \uparrow \uparrow y)^{-(x \uparrow y)}$. So, $3 \downarrow \downarrow \downarrow 3 = (3 \uparrow \uparrow 3)^{-(3 \uparrow 3)} = 7625597484987^{-27} = 3^{-729}$.

$x \downarrow \downarrow \downarrow \downarrow y = (x \uparrow \uparrow \uparrow y)^{-(x \uparrow y)}$. So, $3 \downarrow \downarrow \downarrow \downarrow 3 = (3 \uparrow \uparrow \uparrow 3)^{-(3 \uparrow 3)} = (3^{7625597484987})^{-27} = 3^{-205891132094649}$.

$10 \downarrow \downarrow \downarrow \downarrow 10 = (10 \uparrow \uparrow \uparrow 10)^{-(10 \uparrow 10)} = (10 \uparrow \uparrow \uparrow 10)^{-10000000000} = \boxed{(10^{10^{10000000000}})^{-10000000000}}$.

$10 \underbrace{\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \downarrow \downarrow \downarrow \downarrow 10} 10 = ObjectZ_{1}$.

$10 \underbrace{\downarrow \downarrow \downarrow\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \underbrace{\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow \cdots \downarrow \downarrow \downarrow \downarrow \downarrow \downarrow}_{10 \downarrow \downarrow \downarrow \downarrow 10} 10} 10 = ObjectZ_{2}$.

$ObjectZ_{513298}$ is ObjectZ's Number.

$n_{\Omega}(1) = \underbrace{ObjectZ_{ObjectZ_{ObjectZ_{ObjectZ_{ObjectZ_{ObjectZ_{... _ {ObjectZ_{ObjectZ's ~ Number}}}}}}}}}_{ObjectZ's ~ Number}$.

$n_{\Omega}(2)$ is the attachment in the post when you click on the link. https://artofproblemsolving.com/community/c1230676h2319478p18794259

$n_{\Omega _ {\Omega}}(1) = \underbrace{n_{\Omega}n_{\Omega} \cdots \cdots n_{\Omega}n_{\Omega}}_{(n_{\Omega}(1))} (n_{\Omega}(1))$

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