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

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<math> 2^a \cdot 2^b = 2^{a+b} </math>
<math> 2^a \cdot 2^b = 2^{a+b} </math>


Now, you may notice that this doesn't help if we are interested in numbers like <math> 2^{\frac{1}{2}}</math> or <math>2^{-1}</math>. These cases are covered in the [[Recommended| the recommended section]] but 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 <math> 2^{\frac{1}{2}}</math> or <math>2^{-1}</math>. These cases are covered in the [[Recommended| recommended]] section but if you are interested but are not strictly necessary for understanding this chapter.


== Preliminaries ==
== Preliminaries ==
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