Why exactly does the distributive property work?

Question

Suppose I have this expression that needs to be simplified:

$$4(2x + 4)$$

It can be simplified down to this:

$$8x + 16$$

In this case, this expression has been simplified down using the distributive property. My question is, how exactly does the number, in this case, $4$, get distributed to the two numbers in the parentheses? Why exactly does the distributive property work?

Answer 1

Check out this page.

I think the picture there is worth a thousand words.

Answer 2

Note that $$\begin{align}4(2x+4)&=(2x+4)+(2x+4)+(2x+4)+(2x+4)\\&=(2x+2x+2x+2x)+(4+4+4+4)\\&=4\times (2x)+4\times 4.\end{align}$$

Answer 3

For most people, the fact that the $\text{LHS}$ and $\text{RHS}$ weigh the same is proof enough.
I'm glad that you're not that people.

Let's see an example of distribution of multiplication over addition of integers: $$ \begin{align} 3\times\left(\color{red}{1} + \color{blue}{2}\right) &= \color{red}{3} + \color{blue}{6}\\ 3 \times \left(\color{red}{\star}\space\color{blue}{\star\star}\right) &= \begin{array}{cc} \color{red}{\star} &\color{blue}{\star\star} \\ \color{red}{\star} &\color{blue}{\star\star} \\ \color{red}{\star} &\color{blue}{\star\star} \end{array} \end{align} $$

Multiplication can elementarilly defined as:

$$a\times b = \sum_{i=1}^{a} b $$

So, even when we have $n$ variables, this still applies. $$k\times (x_1 + x_2 + \dots + x_n)= k\times\sum_{i = 1}^{n}x_i = \sum_{j = 1}^{k} \sum_{i = 1}^{n}x_i$$

Hence giving the illusion of distribution:

All that's happening here is this:

$$ \begin{align} a\times(b+c) &= (b+c) + (b+c) + (b+c) + \dots \text{a times}\\ &= (b+b+b+\dots \text{a times}) + (c+c+c+\dots \text{a times})\\ &= (a\times b) + (a\times c) \end{align}$$

@MathLove has used this fact to note the result of the example you've given.

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Note: The definition I gave of multiplication gets shaky for non-whole numbers. In which case, you turn to the geometrical interpretation of multiplication as the Area of a Rectangle

Answer 4

I would say "the reason" is works is that, at least when you look at just the natural numbers, multiplication is repeated addition. See mathlove's answer which uses this definition for an idea of how it works. You could generalize this to a proof for $n(a+b)=na+nb$ using induction on $n$.

To expand to the integers, the rationals, the reals, etc. you would want to look at how you construct them from the natural numbers, and see that the distributive law carries over through the construction.

Answer 5

a(b+c)=ab+ac: This is the distributive property.

a(b+c)=ab+ac

a(b+c)= a(b+c): All I did here is factor out the "a"

You are left with a true statement. a(b+c) will always equal a(b+c) because something will always equal itself. If a(b+c)=ab+bc is the distributive property, and I just simplified the distributive property down to a true statement, then the distributive property must equal the true statement because simplifying something does not change its value. So since the true statement equals the distributive property, then the distributive property must always be true because the true statement will always be true.

Answer 6

$$a(b+c)$$ $$=a(\frac {ba}{a} +\frac {ca}{a})$$ $$=a(\frac {ba+ca}{a})$$ $$=ba+ca$$

$$\therefore a(b+c)=ba+ca$$