Chapter 2: An ancient theorem and a modern question: Difference between revisions

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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 [[Recommended| recommended]] section if you are interested but are not strictly necessary for understanding this chapter.
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 \( a^2 + b^2 = c^2 \) ===