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This question already has an answer here:

I know it may sound stupid but I'm just genuinely wondering about it....

$$\frac ab\times\frac1c=\frac a{bc}$$ where $b,c\ne0$.

How can we multiply numerators by numerators and denominators by denominators?

Is it just a rule? Or can it be proved?

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marked as duplicate by YuiTo Cheng, Shogun, Robert Soupe, postmortes, Wouter Jun 19 at 7:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This question says what I said in my answer. I didn't see any comment from OP which says I want this. All of people say their thoughts without what the OP really wanted. $\endgroup$ – Nosrati Aug 4 '18 at 10:53
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    $\begingroup$ @user108128 I feel you leave the wrong impression to the reader. You say it's simply an axiom and that's all as if to say axioms can't be questioned, that's how it is, get used to it etc. Yes, axioms are something we regard to be true, therefore they don't require proof, but we can certainly question whether some notions or operations are well-defined. I think that is the question to be asked in this topic even if the OP doesn't realise it. I mean no offense. $\endgroup$ – Alvin Lepik Aug 4 '18 at 16:58
  • $\begingroup$ This question, as many other questions of the same complexity on the Stack Exchange network proves that the model of knowledge sharing promoted by the network does not improve the global knowledge. On the contrary, it promotes a world where people don't bother learning and understanding how the things work (not even things so simple as the multiplication of fractions) because they can always ask somebody and get a (possibly wrong) answer. $\endgroup$ – axiac Aug 5 '18 at 9:54
  • $\begingroup$ @axiac You have 40k rep on stackoverflow, surely you yourself have gained many insights by reading answers on the stackexchange network. I know I do all the time, it's been invaluable to me. $\endgroup$ – littleO Aug 5 '18 at 15:31
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You can think of multiplication as meaning "of". So what is $2/5$ of $3/7$ (for example)?

Draw a picture of a cake (a rectangular cake) sliced into 7 equal vertical slices, with $3$ of those slices having red frosting. That's $3/7$ of the cake.

Take that 3/7 of the cake and slice it horizontally into 5 equal pieces, and pour sprinkles on 2 of those 5 pieces. (When you're doing the horizontal slicing, slice the entire cake horizontally while you're at it.)

The portion of the cake with sprinkles is 2/5 of 3/7. But if you draw the picture, you see that the cake has been chopped into 35 equal pieces (5 groups of 7), and 6 of those 35 pieces have sprinkles. So, $$ \frac{2}{5} \text{ of } \frac{3}{7} = \frac{2 \times 3}{5 \times 7}. $$

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There are three steps to this process

  1. Define what a fraction actually is
  2. Define what multiplying two numbers actually does
  3. Prove that multiplying fractions is done by multiplying numerator with numerator and denominator by denominator

Steps 1 and 2 can be done many different ways, and for each combination, step 3 will be done differently.

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  • $\begingroup$ But defining an arbitrary fraction $a\div b$ is letting it be equal to $a(1\div b) = (a\div 1)(1\div b)$, but now we are back to the question again: why are we multiplying? I guess you might have to reverse steps $1$ and $2$. $\endgroup$ – Mr Pie Aug 4 '18 at 8:59
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    $\begingroup$ @user477343: That's no definition. I think Arthur had in mind something like in this answer, for example. $\endgroup$ – Hans Lundmark Aug 4 '18 at 9:07
  • $\begingroup$ @HansLundmark Or like in the article "Teaching fractions according to the common core standards" a bit down on this site. But yes. $\endgroup$ – Arthur Aug 4 '18 at 9:18
  • $\begingroup$ @HansLundmark well, I mean, that is how I define it, with some thorough explanation of course. But a key essential is definitely multiplication; i.e., $a\div b=a\cdot b^{-1}$. Nonetheless, the answer in the link serves a top-notch explanation :) $\endgroup$ – Mr Pie Aug 4 '18 at 9:20
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Great question! The short answer to this: it works because we defined it like this. (I assume we are talking about multiplication of rational numbers)

We are worried, however, whether the operation is well-defined. It means that the result $$\frac{a}{b} \cdot \frac{x}{y} = \frac{ax}{by} $$ must not depend on the choice of fractions. It can't be that this equality holds for some fractions, but not for some other fractions. That would make the operation ill-defined.

The process of verification is quite an involved one, however, especially for multiplication.

On a quick search I did find this which covers all one needs.


To give a slightly different spin on this problem. Intuition paves the way for how we want to define certain operations. Other answers give an intuitive explanation why multiplying two fractions produces a certain fraction. We used these intuitions to define how multiplication of two fractions behaves. But to be absolutely sure we didn't make a mistake, we must also verify the operation is well-defined and that is beyond reach for intuition.

This idea of well-definedness is very important in mathematics not just as a failsafe for addition and multiplication of (rational) numbers to be bulletproof.

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As @Arthur points out, understanding why fractions multiply as they do depends on understanding what a fraction is. That's a subtle question.

There are ways to answer your particular question if you choose to think of fractions as what you get when you cut up pies, but I think the best way starts with defining (thinking about) $1/x$ as the number $?$ that solves the equation $$ ? \times x = 1 . $$ Then you can use the ordinary rules of arithmetic to show that the left side of your equation is a solution to the equation $$ ? \times bc = a $$ and so must equal $a/(bc)$.

Related How to make sense of fractions?

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This is just how we've (in most cases, at any rate) chosen to define multiplication of rational numbers $\mathbf Q.$ And it can be proven to work (by this I mean that it is a well-defined binary operation on $\mathbf Q$). Of course, this definition has some intuition behind it about how rational numbers should behave under multiplication. However, all this can be seen quite neatly by a development of the system $\left(\mathbf Q, \times \right)$ from the natural numbers $\mathbf N:=\{0,1,2,3,\ldots\}$ and the operation $\times$ defined on them in the usual recursive manner which eventually boils down to the primitive successor function. There may be other ways to effect this development, though, but I think this accords most with intuition.

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The reciprocal of a nonzero number $x$ is usually denoted by $\dfrac 1x$ or by $x^{-1}$. Both of these notations are useful because they imply a lot of things that can later be show to be true. In other words, the notation coerces you into treating the multiplicative inverse of $x$ correctly when you learn more stuff.

Particularly in your case, the notation is confusing things instead of simplifying things.

So, just for this discussion, lets assume this postulate.

1. POSTULATE For every non zero real number $x$ there exists a non zero real number $x^*$, called the multiplicative inverse of $x$, such that $x x^* = x^* x = 1$.

We start with an important detail.

2. THEOREM. Let $x$ be a non zero number. If $xy=1$ for some real number $y$, then $y = x^*$.

In other words, the multiplicative inverse of $x$ is unique.

PROOF. Suppose $xy=1$. Then $x^* = x^*(1) = x^*(xy) = (x^*x)y = 1y = y$.

3. NOTATION. Let $x$ and $y$ be real number with $y$ non zero. Then we define $x \div y = \dfrac xy = xy^*$

4. THEOREM. Let $x$ and $y$ be non zero real numbers. Then $x^*y^* = (xy)^*$.

PROOF. $(xy)(x^*y^*) = (xx^*)(yy^*) = 1 \cdot 1 = 1$. By theorem (2.), $x^*y^* = (xy)^*$.

5. THEOREM. Let $a,b,c,d$ be real numbers with $b,d$ non zero. Then $\dfrac ab \cdot \dfrac cd = \dfrac{ac}{bd}$.

PROOF. $\dfrac ab \cdot \dfrac cd = (ab^*)(cd^*) = (ac)(b^*d^*) = (ac)(bd)^* = \dfrac{ac}{bd}$.

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