Multiplication and Division

I have been doing multiply and divide but never really tried to understand it.

E.g. (4 x 12 / 4) could be written as: (12, 4 x 3, 48 / 4) but yet the result is always same, can anyone prove or explain why?

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I'm pretty sure it can't be written as 60/4. – Zev Chonoles Mar 12 '11 at 10:55
I suggest that you try reading en.wikipedia.org/wiki/Fraction_%28mathematics%29 and then ask more specifically on what you don't understand. – Rasmus Mar 12 '11 at 11:02
Corrected to 48, thanks. – user8149 Mar 12 '11 at 11:11
I am not really sure what you are asking. The same thing can go under different names and ways of writing, true enough. You, user3231, are you, your father's child, your neighbour's neighbour, the person with such-and-such social security number and so on. The same is true for number 12 (and lots of other things). – DaG Mar 12 '11 at 11:29
@user3231: Dividing by 4 is by definition the inverse operation to multiplying by 4. – Rasmus Mar 12 '11 at 12:15

2 Answers

It may help to think of the numbers as products of smaller numbers. It may make it easier to understand what's going on. $$4\times \frac{12}{4} = 2\times 2 \times \frac{2\times 2\times 3}{2\times 2}=2\times 2\times 3 = 4 \times 3 = 4\times 3 \times \frac{4}{4}=12$$ Notice how you can always multiply a number by a fraction $\frac{a}{a}$ for any $a \neq 0$ and get the same number (because $\frac{a}{a}=1$): $$b = 1 \times b= \frac{a}{a}\times b = \frac{a\times b}{a}=a\times \frac{b}{a}$$ We can for instance use it on your example: $$12 = 12 \times \frac{4}{4} = \frac{12\times 4}{4} = \frac{48}{4}$$

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Yes, this helped me realizing about what i was thinking, thanks. – user8149 Mar 12 '11 at 17:16

You want to interpret the expression $$a\times b\,\,\, /\,\,\, c$$ We could first multiply $a$ and $b$, and then divide the result by $c$, i.e. $$(a\times b)\,\,\, /\,\,\, c\hskip0.5in (1)$$ or we could multiply $a$ by the quantity $b/c$, i.e. $$a\times(b/c)\hskip0.5in (2)$$ Here is the reason expressions (1) and (2) are equal: dividing by a number $c$ is the same thing as multiplying by its reciprocal $1/c$. This is because, for any number $x$, the number $x/c$ is the unique solution to the equation $$\_\_\_\times c = x\hskip0.5in (\ast)$$ Anything that can go in the spot $\_\_\_$ is equal to $x/c$. But we also have $$\left(x\times (1/c)\right)\times c = x\times\left((1/c)\times c\right)=x\times 1 =x$$ where the first equality is justified by the associative law of multiplication. This means that $x\times (1/c)$ is also the solution to the equation $(\ast)$, and hence must be equal to $x/c$.

So, how does this help? We can rewrite (1) and (2) as $$(a\times b)\times (1/c)\hskip0.5in (1)$$ and $$a\times (b\times (1/c))\hskip0.5in (2)$$ respectively, and the associative law of multiplication tells us that these two quantities must be equal.

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It was good enough as I understand now at smaller level, thanks. – user8149 Mar 12 '11 at 17:18
This really should be the proper answer to this question. :-) – Victor Sep 13 '11 at 20:06