Question

Define a product of pairs setting \((a,b) \cdot (c,d) = (ac+bd,ad+bc)\) . Show that this operation is well defined for equivalence classes.

Answer 1

My attempt: By definition 6.2.9, let \(R\) be an equivalence relation on a set \(S\). For each element \(x \in S\)

\([x]=[y \in S: (x,y) \in R]\)

is the equivalence relation and partition.

Suppose \((a,b)\) or (a',b')\( and \)(c,d)\( or \)(c',d')\(. Then we need to show that \)((ac+bd+a'd'+b'c'=a'c'+b'd'+ad+bc))\(

\)ac+bd+a'd'+b'c'=a'c'+b'd'+ad+bc\(

Adding \)bc'+a'c'\( to both sides, we have


\)ac+bd+a'd'+b'c'=a'c'+b'd'+ad+bc\(.

Answer 2

so what I don't get is where does this \[(a,b) ~ (a'b') \rightarrow a+b'=b+a'\] come into play... I feel like the reflexivity definition is involved

Answer 3

Latex spaghetti

Answer 4

hold on

Answer 5

My attempt: By definition 6.2.9, let \(R\) be an equivalence relation on a set \(S\). For each element \(x \in S\)

\([x]=[y \in S: (x,y) \in R]\)

is the equivalence relation and partition.

Suppose \((a,b)\) or \((a',b')\) and \((c,d)\) or \((c',d')\).

Then we need to show that \(((ac+bd+a'd'+b'c'=a'c'+b'd'+ad+bc))\)

\(ac+bd+a'd'+b'c'=a'c'+b'd'+ad+bc\)

Adding \(bc'+a'c'\) to both sides, we have


\(ac+bd+a'd'+b'c'=a'c'+b'd'+ad+bc\).

Answer 6

but before the adding part suppose ... there's this part.. \[(a+b')c' = (b+a')c'\] which becomes

\[ac'+b'c'=bc'+a'c'\]

so where does that come from because apparently I need to add that to both sides and cancel out b'c' and a'c' from\[ac+bd+a'd'+b'c'+bc'+a'c'=a'c'+b'd'+ad+bc+ac'+b'c'\]

Answer 7

so after cancelling...

we have \[ac+bd+a'd'+bc'=b'd'+ad+bc+ac'\]

From \[(a+b')d'=(b+a)d'\]
we have\[ad'+b'd'=bd'+a'd'\]

so adding bd'+a'd' to both sides yields,

\[ac+bd+a'd'+bc'+ad'+b'd'=b'd'+ad+bc+ac'+bd'+a'd'\]

Answer 8

That definition is confusing

Answer 9

Can you just restate the definition again?

Answer 10

cancelling, a'd' and b'd' we have

\[ac+ad'+bc'+bd=ad+bc+ac'+bd'\]

factoring out a and b , we have

\[a(c+d')+b(c'+d) = a(d+c')+b(c+d')\]

there is an equivalence relation on the left and the operation is well defined on the right.

Answer 11

By definition 6.2.9, let $R$ be an equivalence relation on a set $S$. For each element $x \in S$

$[x]=[y \in S: (x,y) \in R]$

is the equivalence relation and partition.

Answer 12

grtrr

Answer 13

By definition 6.2.9, let \(R\) be an equivalence relation on a set \(S\). For each element \(x \in S\)

\([x]=[y \in S: (x,y) \in R]\)

Answer 14

So \([x]\) is an equivalence class, right?

Answer 15

This is the definition of an equivalence class?

Answer 16

all that letter gibberish stuff is what my prof wrote... and that stuff is correct... but what I don't get is how the heck does this happen \[(a,b) ~ (a',b') \rightarrow a+b'=b+a'\]

(a,b) ~(a'b')

Answer 17

feels like reflexivity

Answer 18

I cannot help, because I don't understand the problem.

Answer 19

What does it mean for an operation to be well defined for equivalence classes?

Answer 20

same goes for \[(a+b')c'=(b+a')c' \rightarrow ac+b'c'=bc'+a'c'\]
\[(a+b')d'=(b+a')d' \rightarrow ad+b'd'=bd'+a'd'\]

where dd that come from ?!

Answer 21

an expression is well-defined if it is unambiguous and its objects are independent of their representative. :/

Answer 22

ok

Answer 23

That is not a very formal definition

Answer 24

How would you prove something is unambiguous?

Answer 25

???

Answer 26

You gave a definition for products of pairs...

Answer 27

Yet you don't have any pairs anywhere....

Answer 28

I still don't know the formal definition, of even any examples and counter examples, of what a well defined operation are.

Answer 29

Maybe they mean it is closed?

Answer 30

\((a,b)\in R\) and \((c,d)\in R\)
We do: \[
(a,b)\cdot (c,d)=(ac+bd,ad+bc)
\]Is \( (ac+bd,ad+bc) \in R\)?

Answer 31

I guess we have to explore:
reflexivity
symmetry
transitivity

Answer 32

well defined \(a=b\implies f(a) = f(b)\)

Answer 33

Hmmmmm, well defined applies to operations? What would be not well defined?

Answer 34

operations are functions

Answer 35

XxX -> X

Answer 36

But functions seems to be well defined by their definition as is based on your definition of well defined.

Answer 37

definition of function includes well defined and defined everywhere

Answer 38

\(f:\mathbb{Q}\rightarrow R, f(\frac{a}{b})=b\) is not well defined relation

Answer 39

\(\frac{2}{4}=\frac{1}{2}\) but \(4\ne2\)

Answer 40

anyway i didn't read what you guys where doing, I just thought you wanted exact definition of well defined.

The only time you even have a problem with something being or not being well defined is when dealing with equivalence classes.

Answer 41

I think your definition helps quite a bit.

Answer 42

so you need to show that if \((a,b) = (c,d)\) and \((a',b')=(c',d')\)
then \((a,b)\cdot(a',b')=(c,d)\cdot(c',d')\)

Answer 43

I see.

Answer 44

\[
(a,b)\cdot (a',b') = (aa'+bb',ab'+ba')
\]

Answer 45

I just had a similar topic in class today trying to show that coset multiplication is well defined operation on the set of equivalence classes. its fresh:)

Answer 46

\[
(c,d)\cdot (c',d') =(cc'+dd',cd'+dc')
\]

Answer 47

We need to show that \(aa'+bb' = cc'+dd'\) then? And also that \(ab'+ba' = cd'+dc'\)?

Answer 48

yeah, using the fact that (a,b) = (c,d), and (a',b')=(c',d') and what that is defined to mean

Answer 49

im on my phone so some of the code is showing weird where he defined things. Ill go get the wife's pc.

Answer 50

we still need to figure this out?