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.

Answer 1

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 $$\frac ba\cdot\frac dc=\frac{bd}{ac}.$$

Answer 2

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.

1. 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 $a \cdot \frac{1}{b}\,$.

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

Answer 3

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.

More on that later...

Answer 4

According to you $w = \frac{(xy)(bd)}{bd}$ if and only if $(bd)w = (xy)(bd)$. If $\dfrac ab$ and $\dfrac cd$ are to be considered valid fractions, then we must assume that $bd \ne 0$. Hence we can divide both sides by $bd$ and get $w=xy$ So $xy = \frac{(xy)(bd)}{bd}$