Constructing $\mathbb Q$ from $\mathbb Z$

Question

Let $X = \{(a,b)|a,b \in \mathbb Z, b \neq 0 \}$. Define a relation $\sim$ on $X$ by $(a,b) \sim (c,d)$ iff $ad = bc $.

a) I'm trying to show that ~ is an equivalence relation.

So, is it reflexive?

$(a,b) \sim (a,b) $ iff $ab = ab$, with $b$ not equal to zero

so its reflexive

Is it transitive?

$(a,b) \sim (c,d) $ iff ad = bc and if $(c,d) \sim (e,f)$ then $cf = de$ so, $c = ad/b$ and $d = bc/a$

and we have then $ad/b *f = bc/a *e$

Is it symmetric?

$(a,b) \sim (a,b)$ iff $ab = ab$, same as $ab = ba$, (why?) with $b$ not equal to zero

b) Let $\mathbb Q$ be the set of equivalence classes and denote the equivalence class of $(a,b)$ by $[a,b]$. I'm needing help in showing that the "addition and Multiplication" of $\mathbb Q$ are well defined:

(1) $[a,b] + [c,d] = [ad+bc,bd]$

and

(2)$ [a,b][c,d] = [ac,bd]$

What I have so far:

(1) $[a,b] + ([c,d] +[e,f]) = [a,b] + ([cf,de]) = [ade,bcf]$

but this is true for when ?

(2) If $c =b$ and $d = a$

then, $[a,b][b,a] = [a^2,b^2]$ iff $a^2 = b^2$, which is not well defined, since $a = -5$ and $b$ could be equal to $5$, with $25 = 25$

This is the standard construction of the fraction field of a commutative ring (In particular this case is for the commutative ring $\mathbb{Z}$). For the symmetric bit, the reason why $ab = ba$ is because $\mathbb{Z}$ is a commutative ring — , Jan 22 '12 at 01:09
Note that you're not applying the formula for addition, in what you have so far for (1). Try a little more carefully. As for (2), why is there a problem for a=5, b=-5? — Niel de Beaudrap, Jan 22 '12 at 01:12
Note that one should not use division in proving that this relation is an equivalence realtion, which is what you are doing when proving transitivity. Division is not defined for arbitrary integers, and since you are constructing the rationals, you should be introducing rationals by doing things like "$c=ad/b$". That quotient may not be defined! — Arturo Magidin, Jan 22 '12 at 04:02

Henry · Accepted Answer · 2012-01-22T01:00:57.780

4

Looking at what you have said for (a):

Transitive: I do not like you saying "c = ad/b and d = bc/a and we have then ad/b *f = bc/a *e" because you seem to be using division a little freely. Better to say that if $(a,b) \sim (c,d)$ and $(c,d) \sim (e,f)$ means $ad = bc$ and $cf = de$ then $bcf = bde$ so $adf = bde$ so (since $d \not = 0$) $af = be$ meaning $(a,b) \sim (e,f)$
Symmetric: you have confused this with reflexive. What you want is $(a,b) \sim (c,d)$ iff $ad = bc$ iff $bc = ad$ iff $(c,d) \sim (a,b)$

Now for part (b):

For addition you need to show that if $(a_1,b_1) \sim (a_2,b_2)$ then $(a_1 d +b_1 c, b_1 d) \sim (a_2 d + b_2 c , b_2 d)$
For multiplication you need to show that if $(a_1,b_1) \sim (a_2,b_2)$ then $(a_1 c, b_1 d) \sim (a_2 c , b_2 d)$

It is worth noting that in fact $[a,b][b,a] = [ab,ba] = [1,1]$

edited Jan 22 '12 at 01:00

answered Jan 22 '12 at 00:46

Henry

157,058

Thanks, does part (b) look good? – Buddy Holly Jan 22 '12 at 00:48
No - part (b) needs work too. I will add something – Henry Jan 22 '12 at 00:54
It does not make sense to me that I found a counterexample to something that I'm supposed to prove, above in part b – Buddy Holly Jan 22 '12 at 00:55
@Buddy: Your counter-example is built on a misapprehension. $[a,b][b,a]$ is usually not $[a^2,b^2]$ – Henry Jan 22 '12 at 01:04
Henry: All I can understand for part b, what you've shown is:
a1b2 = a2b2 but where does c and d fit in the context? Is it a typo?
– Buddy Holly Jan 22 '12 at 01:04
@Buddy: I have not shown it, I have assumed it and shown what you need to prove. – Henry Jan 22 '12 at 01:06

score 3 · Answer 2 · edited Jan 22 '12 at 01:09

HINT $\ $ To show that addition is well-defined, you need to show that the definition doesn't depend on the choice of equivalence class representatives of the summands. Since addition is commutative it suffices to do this for one summand, i.e. it suffices to show that

$$\rm [A,B]\ =\ [a,b]\ \Rightarrow\ [A,B]\ +\ [c,d]\ =\ [a,b]\ +\ [c,d] $$

$$\rm [A,B]\ =\ [a,b]\ \Rightarrow\ [Ad+Bc,Bd]\ =\ [ad+bc,bd] $$

Unwinding the definition of fraction equality, the implication reduces to simple algebra.

Applied twice, this immediately implies the sought result for both summands, namely

$$\rm [A,B]\ =\ [a,b],\ [C,D]\ =\ [c,d]\ \ \Rightarrow\ \ [A,B]\ +\ [C,D]\ =\ [a,b]\ +\ [c,d] $$

which says precisely that addition doesn't depend on the choice of argument representatives.

Holdsworth88 · Answer 3 · 2012-01-22T00:57:55.130

1

Reflexivity: $(a,b)\sim (a,b)$ since $ab=ab$.

Symmetry: Assume $(a,b) \sim (c,d)$. This implies $ad=bc$, which implies $cb=da$ by the symmetry of equality of integers. The condition $(c,d) \sim (a,b)$ is equivalent to $cb=da$, which we have already shown.

Transitivity: Assume $(a,b) \sim (c,d)$ and $(c,d)\sim (e,f)$. Then $ad=bc$ and $cf=ed$. We want to show $af=eb$. Multiplying the two statements together yields $adcf=bced$. By canceling $c$ and $d$ from both sides we see that $af=eb$, which implies $(a,b) \sim (e,f)$.

In your attempt to show rational addition is well defined, you mixed up the definition of addition and multiplication.

To prove rational addition is well-defined, it is enough to show that if $(a,b)=(a',b')$, then $(a,b)+(c,d)=(a',b')+(c,d)$.

edited Jan 22 '12 at 00:57

answered Jan 22 '12 at 00:44

Holdsworth88

8,818

We can't cancel d since with (a,b) ~ (c,d) we have asserted that b, d are not equal to zero – Buddy Holly Jan 22 '12 at 00:46
2

@Buddy: No - you can cancel $d$ precisely because $d$ is not $0$. What you cannot do is divide both sides by $0$ – Henry Jan 22 '12 at 00:48
and you can assume $c\neq 0$, for if it was the statement would be trivial. – Holdsworth88 Jan 22 '12 at 00:55

Constructing $\mathbb Q$ from $\mathbb Z$

3 Answers3