Tao Analysis I - 2.3.4

Exercise 2.3.4

Prove the identity $(a+b{)}^2=a^2+2ab+b^2$ for all natural numbers $a$, $b$.

We want to show the following.

$$((a++)+b{)}^2=(a++{)}^2+(2\times (a++)\times b)+b^2$$

Let's expand the LHS using the Definition 2.3.11 of exponentiation that $x^2=x^1\times x$,  and the distributive law of Proposition 2.3.4 twice,

$$\begin{array}{rl}((a++)+b{)}^2& =((a++)+b)\times ((a++)+b)\\ \\ & =[((a++)+b)\times (a++)]+[((a++)+b)\times b)]\\ \\ & =(a++{)}^2+(b\times (a++))+((a++)\times b)+b^2\end{array}$$

We now use commutivity of multiplication of Lemma 2.3.2,

$$\begin{array}{rl}((a++)+b{)}^2& =(a++{)}^2+(b\times (a++))+((a++)\times b)+b^2\\ \\ & =(a++{)}^2+(b\times (a++))+(b\times (a++))+b^2\\ \\ & =(a++{)}^2+(2\times (a++)\times b)+b^2\end{array}$$

The last step uses Definition 2.3.1 that $2\times m=0+m+m$.

The final expression is the RHS, and so we have shown that for all natural numbers $(a+b{)}^2=a^2+2ab+b^2$. $◻$