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Difference between revisions of "Chapter 2: An ancient theorem and a modern question"
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Chapter 2: An ancient theorem and a modern question (view source)
Revision as of 00:34, 20 October 2022
, 00:34, 20 October 2022no edit summary
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$$ 2^a \cdot 2^b = 2^{a+b} $$ | $$ 2^a \cdot 2^b = 2^{a+b} $$ | ||
Now, you may notice that this doesn't help if we are interested in numbers like \( 2^{\frac{1}{2}}\) or \(2^{-1}\). These cases are covered in the | Now, you may notice that this doesn't help if we are interested in numbers like \( 2^{\frac{1}{2}}\) or \(2^{-1}\). These cases are covered in the recommended section if you are interested but are not strictly necessary for understanding this chapter. | ||
=== Pythagorean Theorem \( a^2 + b^2 = c^2 \) === | === Pythagorean Theorem \( a^2 + b^2 = c^2 \) === | ||