# Spivak calculus chapter 1 consequence of property 9

Spivak state's that one of the consequence of the distributive property is being able to solve multiplication of Arabic numerals.

    1 3
x 2 4
_______
3 1 2


is arranged as:

$$13 . 24$$

$$=13 . (2 . 10 + 4)$$

$$= 13 . 2 . 10 + 13 . 4$$

$$= 26 . 10 + 52$$

$$= 312$$

and

  1 3
x 4
_____
5 2


is arranged as:

$$=13 . 4$$

$$=(1 . 10 + 3) . 4$$

$$=1 . 10 . 4 + 3 . 4$$

$$=4 . 10 + 12$$

$$=4 . 10 + 1 . 10 + 2$$

$$=(4 + 1) . 10 + 2$$

$$=5 . 10 + 2$$

$$=52$$

so why isn't $$4 . 10 + 12$$ directly written as $$52$$ like in the first problem?

• To save additional writing effort? With the first illustration, Spivak clearly outlines how the distributive property operates- and that is the entire point. May 23 '20 at 7:23
• @Manan how does writing additional lines save effort? it increases it May 23 '20 at 8:21
• Call it a typo, or an overlooked error. I think the author has made his point about the distributive property clear. May 23 '20 at 8:59

He is showing us how did he obtain $$13.4=52$$ in the first working using $$P9$$ property explicitly.
\begin{align}13\cdot 24 &= 13 \cdot (2\cdot 10 + 4) \\ &=13 \cdot 2 \cdot 10 + \color{blue}{13 \cdot 4} \\ &= 26 \cdot 10 + \color{blue}{52} \end{align} The multiplication $$13\cdot 4=52$$ uses $$P9$$ also: