Question

Hey can someone help me understand the process of showing that binary structures are isomorphic?

Answer 1

So I get the steps,
step 1: Define \(\Phi\)
step 2: Show \(\Phi\) is 1-1
step 3: show that \(\Phi\) is onto
step 4: show that it's homomorphic

Answer 2

but my question is how do you show steps 2,3?

Answer 3

1-1 implies that f(a) = f(b) if a=b

Answer 4

onto means that for some y in the range, there is a x in the domain

Answer 5

I get that, I guess I should say step 3

Answer 6

I understand the meaning, but how do I show/prove that?

Answer 7

what did you get form step 1?

Answer 8

in my case I'm doing it where phi is given as 2n for the integers and addition to the integers and addition

Answer 9

so:\[\phi(x)=2x\
?]

Answer 10

yup

Answer 11

code flop

Answer 12

I could still read it haha

Answer 13

then for some y=2x, can we find an suitable x for all ys?

Answer 14

no, there are no fractions in the integers for the odd numbers

Answer 15

Is that all I would have to say?

Answer 16

\[y=\phi(x)=2x\]
\[\frac{y}{2}=x\]
\[y=\phi(\frac{y}{2})\]
then you need a different phi right?

Answer 17

or is y/2 defined in your domain?

Answer 18

well, it wanted me to test that particular phi

Answer 19

x= 0,1,2,3,4,5,6,7,8

2x/2 = x is definantly in x

Answer 20

So I could say that using phi=2n, the integers and addition to the integers and addition are not isomorphic?

Answer 21

right, but the other way around it is not, correct? It fails onto right?

Answer 22

phi is a function, the produces an out put

if we can define phi(ab) and phi(a) and phi(b) then we can determine of phi(ab)=phi(a)phi(b) ... the homomorphic property

Answer 23

every even number can be mapped back to some integer

Answer 24

wait, but we are doing it for addition not multiplication?

Answer 25

so it'd be the sum of 2(a+b)=2a+2b for the homomorphic part

Answer 26

first step, define some function: f(x) = 2x is the function you defined

is it 1-1? for some a,b Integer; 2a = 2b imples that a=b by cancelation

is it onto? y = f(x) = 2x, x=y/2; for some x=y/2: y = f(y/2), therefore its onto

Answer 27

is it homomorphic? thats when the operation becomes important

Answer 28

ok, so for onto, that is where I am confused

Answer 29

y=f(x)=2x I get
x=y/2 I get
^ how that is always in the integers, I don't understand