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,

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

We now use commutivity of multiplication of Lemma 2.3.2,

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

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 $(a + b)^2 = a^2 + 2ab + b^2$. $\square$