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Pythagorean theorem (view source)
Revision as of 09:25, February 15, 2021
, 09:25, February 15, 2021→Conceptualization (cont'd): typo
2. And since ''l''<sub>1</sub> and ''l''<sub>2</sub> are whole numbers, that which results when they are multiplied by themselves, namely ''l''<sub>1</sub><sup>2</sup> and ''l''<sub>2</sub><sup>2</sup>, certainly are as well.
2. And since ''l''<sub>1</sub> and ''l''<sub>2</sub> are whole numbers, that which results when they are multiplied by themselves, namely ''l''<sub>1</sub><sup>2</sup> and ''l''<sub>2</sub><sup>2</sup>, certainly are as well.
3. So let's assign ''o''<sub>1</sub> as being equal to ''l''<sub>1</sub><sup>2</sup> and ''o''<sub>2</sub> as being equal to ''l''<sub>2</sub><sup>2</sup>, ''o''<sub>1</sub> and ''o''<sub>2</sub>, to limit them to showing they are natural numbers like the ''n'''s in the "simple assumption":
3. So let's assign ''o''<sub>1</sub> as being equal to ''l''<sub>1</sub><sup>2</sup> and ''o''<sub>2</sub> as being equal to ''l''<sub>2</sub><sup>2</sup>, to limit them to showing they are natural numbers like the ''n'''s in the "simple assumption":
:(''o''<sub>1</sub> / ''h'') + (''o''<sub>2</sub> / ''h'') = ''h''.
:(''o''<sub>1</sub> / ''h'') + (''o''<sub>2</sub> / ''h'') = ''h''.