Difference between revisions of "Fundamental Theorem of Sato"
(Method 4.) |
m (added link to Sato) |
||
Line 22: | Line 22: | ||
Gmass is Ssto in cat form. Since Gmass is amazing, and since if <math>a=b</math> and <math>b=c</math>, <math>a=c</math>, Sato is amazing. | Gmass is Ssto in cat form. Since Gmass is amazing, and since if <math>a=b</math> and <math>b=c</math>, <math>a=c</math>, Sato is amazing. | ||
+ | |||
+ | |||
+ | ------ | ||
+ | Learn about Sato: | ||
+ | https://artofproblemsolving.com/wiki/index.php/Naoki_Sato |
Revision as of 21:53, 20 August 2019
The Fundamental Theorem of Sato states the following:
Sato is amazing.
Proof:
Method 1: Proof by Contradiction
Assume, for contradiction, that Sato is not amazing.
This is absurd. Therefore, Sato is amazing. (Proved)
Method 2: Proof by Authority
Whatever AoPS says is correct, and AoPS says that Mr. Sato is amazing. Thus, Mr. Sato is amazing. (Proved)
Method 3: Proof by Pigeonhole
There is only one Sato. By Pigeonhole, either Sato is amazing or he isn't (in this case, Sato goes in the unamazing pigeonhole). Fortunately, Sato cannot fit in a pigeonhole; hence, Sato is amazing. (Proved)
Method 4: Proof by Gmass
Gmass is Ssto in cat form. Since Gmass is amazing, and since if and , , Sato is amazing.
Learn about Sato: https://artofproblemsolving.com/wiki/index.php/Naoki_Sato