Exercise 2.3.4 (Binomial formula)

Hints. Use Proposition 2.3.4 and Lemma 2.3.2.

How to think about the exercise. This exercise is basically just an application of all of the propositions and lemmas that appeared in this section (Section 2.3). Exercises like this one test your ability to rigorously apply definitions and results. For each step, you want to cite the result which justifies it.

Model solution. Let $a,b$ be natural numbers. By definition of exponentiation (Definition 2.3.11), ${(a+b)}^2={(a+b)}^1\times (a+b)=(a+b)(a+b)$. By the distributive law (Proposition 2.3.4), we have $(a+b)(a+b)=(a+b)a+(a+b)b$. By the distributive law again, $(a+b)a=\mathit{aa}+\mathit{ba}$; by the definition of exponentiation and commutativity of multiplication (Lemma 2.3.2) this is just $a^2+\mathit{ab}$. Similarly, by the distributive law, $(a+b)b=\mathit{ab}+\mathit{bb}=\mathit{ab}+b^2$. Summarizing everything so far, we have ${(a+b)}^2=a^2+\mathit{ab}+\mathit{ab}+b^2$. It now suffices to show that $\mathit{ab}+\mathit{ab}=2\mathit{ab}$. By the associativity of multiplication and definition of multiplication, $2\mathit{ab}=2\times (\mathit{ab})=(1\times (\mathit{ab}))+\mathit{ab}=\mathit{ab}+\mathit{ab}$.

Source: Issa Rice’s blog.