Rule for multiplying rational numbers (follow-up)

Question

I was thinking about the rule where $\space\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$ and why it worked, and I found this question: Rule for multiplying rational numbers.

The best answer was:

Did:

One knows that $a/b$ may be defined as the number $x$ such that $bx=a$, and that $c/d$ may be defined as the number $y$ such that $dy=c$.

If one wants the multiplication on these objects to be associative and commutative as it is on the integers, one should ask that $ac=(bx)(dy)=(bd)(xy)$ hence that the object $xy$ fits the definition of $(ac)/(bd)$.

And I was thinking, if in the end we get $\frac{(xy)(bd)}{bd}=\frac{ac}{bd}$, how do we know how to simplify the two $(bd)$s here? Don't we need the property itself $\space\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$ to know how to do that?

$$\frac{(xy)(bd)}{bd}=\frac{(xy)\cdot (bd)}{1\cdot (bd)}=\frac{xy}{1}\cdot\frac{bd}{bd}=\frac{xy}{1}\cdot 1=\frac{xy}{1}=xy$$

So do we just assume the property is true inductively (not mathematically inductively), or am I missing something?

PS: As a beginner, I may say some ridiculous things, excuse me for that.

Check and see if this follows from the "$a/b$ is defined by the solution $x$ to $bx = a$" definition. — pjs36, Feb 25 '17 at 22:44
You're asking "why it works". The definition of the multiplication of two rational numbers is arbitrary. Perhaps you really mean to ask "why it is useful". One justification that the usual definition is the only sensible one is that it is the only binary operation $\star$ on the set of rationals numbers such that for all rational numbers $q$, $r$ and $s$ the properties $1\star q=q$ and $(q+r)\star s=q\star s + r\star s$ are satisfied. Surely those are two properties you want your multiplication to satisfy! — NeedForHelp, Feb 25 '17 at 22:54

score 6 · Answer 1 · answered Feb 25 '17 at 23:18

6

A completely formal proof depends on the particular axioms and definitions you work with, but as far as the intuition of it goes, consider the following points.

The fraction $\frac{1}{b}$ denotes the multiplicative inverse of $b \ne 0\,$, which is to say that $x=\frac{1}{b}$ is the unique solution to $b \cdot x = 1\,$. The fraction $\frac{a}{b}$ can be thought of as shorthand for $a \cdot \frac{1}{b}\,$.
Multiplying together $b \cdot \frac{1}{b}=1$ and $d \cdot \frac{1}{d}=1$ gives $b \cdot \frac{1}{b} \cdot d \cdot \frac{1}{d}=1\,$. Since multiplication is commutative, this can be rewritten as $(b \cdot d) \cdot (\frac{1}{b} \cdot \frac{1}{d})=1\,$. But the latter means that $\frac{1}{b} \cdot \frac{1}{d}$ is the multiplicative inverse of $b\cdot d\,$, which proves that $\frac{1}{b} \cdot \frac{1}{d} = \frac{1}{b \,\cdot\, d}\,$.

Therefore $\;\frac{a}{b}\cdot \frac{c}{d} \stackrel{\;(1)\;}{=} a \cdot \frac{1}{b} \cdot c \cdot \frac{1}{d} = (a \cdot c) \cdot (\frac{1}{b} \cdot \frac{1}{d}) \stackrel{\;(2)\;}{=} (a \cdot c) \cdot \frac{1}{b \,\cdot\, d} \stackrel{\;(1)\;}{=} \frac{a \,\cdot c}{b \,\cdot\, d}\,$.

answered Feb 25 '17 at 23:18

dxiv

76,497

This is in my opinion the right way to make both intuition and rigour meet. Namely, this ties to the definition of a fraction field of an arbitrary ring such that the product of nonzero elements is nonzero. Intuitively, we have natural numbers that can be used for counting or for scaling, and obviously for scaling we want to be able to undo. So letting "$\times \frac1b$" denote the inverse of the scaling "$\times b$", we find that all the scalings composed of integer scalings and inverse of nonzero scalings ought to have all the multiplicative properties of rational numbers. – user21820 Feb 26 '17 at 05:57
Next, if we look at the way the scalings move the point $1$ from the origin, we find as described here that multiplication ought to distribute over addition. From this we know exactly how we should define addition and multiplication of rational numbers. It now remains for us to prove that those operations defined this way actually make rationals satisfy the field axioms. – user21820 Feb 26 '17 at 06:01
In short, intuition about scalings in physical space leads to field axioms and tells us precisely how we can construct rationals from integers if rationals are actually an abstract reflection of certain scalings in the physical world. This is not mathematics, so we prove separately and completely mathematically that this construction in fact gives us the desired field independent of the meaning of rationals in the physical world. The difference from the usual approach is that we didn't pluck the construction out from the sky, but was guided by intuition. – user21820 Feb 26 '17 at 06:07
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@user21820 Thank you for the comments. When answering such broad open-ended questions, it's never easy to hit the right balance between correct but overboard vs. easy to sell but misleading. What I tried with my answer above is outline a credible path towards the full answer. It's not complete as posted, of course, but it could be fully formalized in most systems I am familiar with. Best hope is that somewhere down the road the OP would think "this makes more sense now" rather than "that's a hopelessly wrong idea that some idiot on MSE planted into my head" ;-) – dxiv Feb 26 '17 at 06:28
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Yes I try to do the same! I like to have students know precisely what is intuition and what is proven, and where they match. – user21820 Feb 26 '17 at 06:35

score 3 · Answer 2 · answered Feb 25 '17 at 22:52

A long as we only have the integers, wewe obserev that equations of the form $ax=b$ sometimes have a solution, and sometimes not (and in the special case $a=b=0$ there are infinitely many solutions). We define the rational numbers as a set of numbers where an equation $ax=b$ with integers $a\ne 0$ and $b$ always has a unique solution. We could call the unique solution of $3x=5$ simply "the unique solution of $3x=5$", but that is inconvenient. To simplify this, we introduce the symbolic notation $\frac ba$ (i.e., the number $b$ atop the number $a$, with a horizontal line between them) for the unique solution of $ax=b$.

Note that we can easily show, for example that $\frac 32=\frac 64$: The unique solution of $2x=3$ must also be a (hence the) solution of $4x=6$ because we obtain the latter equation from the former by multiplying both sides with $2$. In particular, we see that $\frac n1=n$, i.e., some of our new rational numbers are our good old integers.

Next we wonder if we can extend addition and multiplication to rational numbers. Well, for addition, we ask: If $x,y$ are rationals with defining equations $ax=b$ and $cy=d$, can we find an equation that should hold for $x+y$? If we multiply the first equation by $c$ and the second by $a$, we obtain $acx=bc$ and $acy=ad$; if we add these, we formally obtain $ac(x+y)=bc+ad$ (formally, because we used several laws such as associativity and distributivity to arrive at the last equations, even though we can hardly claim that these laws hold for rational numbers when we have not even defined these operations). Thus $x+y$ is a solution of $ac\cdot z=bc+ad$, i.e., by definition $x+y=\frac{bc+ad}{ac}$. Hence there is only one reasonable (because it allows us to keep associativity etc.) way to define addition for rational numbers: $$\frac ba+\frac dc =\frac{bc+ad}{ac}.$$

Similarly, if $ax=b$ anc $cy=d$, then formally $acxy=bd$, i.e., $xy$ is the unique solution of $ac\cdot z=bd$, that is we are "forced" to say $$\frac ba\cdot\frac dc=\frac{bd}{ac}.$$

fleablood · Answer 3 · 2017-02-26T01:17:06.100

Using Did's definition that $[a|b]= x:b\times x = a$ and the implied axiom, such an $x$ exist and is in rational number while the rationals are commutative and associative: we have

$[a|b] = x: b\times x = a$ and we have $[c|d] = y: d\times y = c$

It follows that $ac = (b\times x)(c\times y)$. Now $b,c,x,y$ are just rationals and multiplication is commutative.

So $ac = (bd)(xy)$

So $[ac|bd] = z|bd \times z = ac$. And we can see that as $ac = (bd)(xy)$ that $z$ is $xy = [a|b]\times [c|d]$.

So $[ac|bd] = z = xy = [a|b]\times [c|d]$.

However it is a bit of an assumption that for every $a, b: b\ne 0$ that there does exist a rational $x$ so that $a = bx$ and that such a rational is unique. But then, that's usually an axiom.

And it's a fair axiom, I think... maybe I'll add more on that later.

==== okay, more on that later =====

The thing is that we are taught arithmetic in elementary school and if we are taught well, it is consistent and so Did's argument would hold and be valid. But when it comes to learning how to do formally to rigorous proofs we usually toss all the arithmetic out, and start over with abstract Field definitions.

In "arithmetic" we start with counting and the numbers themselves are very concrete concepts as is grouping them which addition and multiplication are defined to describe, as is taking away an splitting things which subtraction and division is meant to define.

But in formal math we just say: "An ordered field is any system that obeys these sets of rules and we have minimal ordered field called The Rationals". The Rationals obey all the rules because we say they do.

Among the rules (listed as I think of them; for a complete list of axioms google "Field Axioms").

--There is a rational called $1$ so for any rational $b$ then $b*1 = 1*b = b$.

--For every rational except $0$, for $b \ne 0$ there exists a rational called $\frac 1b$ so that $b * \frac 1b = \frac 1b * b = 1$

It is NOTATION and notation ONLY that we define $\frac ab$ to be the number $a*\frac 1b$.

--The rationals are commutative and associative.

From this we can prove that $\frac 1b$ is unique to $b$. (Pf: If $b*a = 1$ and $b*c = 1$ then $b*a = b*c$ and $\frac 1b*b*a = \frac 1b*b*c$ so $1*a = a = 1*c = c$. For any $b*a = 1$ then $a$ can only be one possible value.)

And we can prove that $\frac 1{bc} = \frac 1b* \frac 1c$. (Pf: $bc*(\frac 1b \frac 1c) = bc *(\frac 1c\frac 1b)= b(c*\frac 1c)\frac 1b = b*(1)*\frac 1b = b*\frac 1b = 1$. So $\frac 1{bc} = \frac 1b * \frac 1c$.

And then it's just a bit of notation that $\frac ab *\frac c*d = a*\frac 1b*c*\frac*d = a*c*\frac 1b* \frac 1d = (ac)*(\frac 1b* \frac 1d) = ac*\frac 1{bd} = \frac {ac}{bd}$.

It's all axiom and notation. It is true because we say it is true.

So that is a proof by field definitions.

But what about in terms of "arithmetic" and elementary school.

=========

Well, we take it as a given that if we have a value $m$ and a positive integer $b$ that we can divide $m$ into $b$ equal parts notated as $\frac mb$ and that $\frac mb + ..... + \frac mb = b\times \frac mb = m$.

We can verify so a number must be distinct: If $d < e$ then $d+d+d+.... < e+e+e+...$ so for any positive integer $b$ we know $d*b < e*b$. So if $e \ne d$ it is impossible for $db = eb$. So if $b*\frac mb =m$ then $\frac mb$ is the only number that that is so.

The next question comes down to can we define $\frac mb + \frac nc$ in any meaningful way and can we define $\frac mb\times \frac nc$ in any meaningful way.

Rule for multiplying rational numbers (follow-up)

4 Answers4