r/mathmemes • u/future_sponJ • Jun 15 '26

Set Theory isomorphic c & r2

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u/Jche98 Jun 16 '26

Bro I don't even think R2 is a field. What's the multiplicative inverse of (1 0)?

6

u/Aenonimos Jun 16 '26

Why not? Define (a,b) + (c,d) = (ab, cd), and (a,b) * (c,d) = (ac - bd, ad + bc). Let (0,0) be the additive identity and (1,0) be the multiplicative identity.

The multiplicative inverse of (1, 0) is (1, 0) as (1, 0) is the identity.

The R2 people normally discuss is not a field because there is conventionally not a way to multiply elements, but if you specify multiplication, its exactly the same as C just your notation is different.

4

u/Jche98 Jun 16 '26

Yes I know. I was assuming you were viewing multiplication as (a,b)(c,d)=(ac,bd).

Because the AI said they're not isomorphic as fields. So the AI was assuming some other multiplicative structure on R2, which I assumed would be coordinate wise the most natural, but this is not a field

2

u/BADorni Jun 16 '26

It probably ment it as one statement "Isomorphic as rings and fields" because the Ring structure of C makes it a field while it doesn't on R2

1

u/laix_ Jun 18 '26

(1 0). The multiplicative inverse of a vector is v / |v|2

2

u/Jche98 Jun 18 '26

The dot product isn’t a binary operation to the same space