Recent content by tc_11

So if our ring does not contain 1... then our ring does not have unity (there is no element such that a*1=a). Then 1=0 would mean... I'm not sure.. that the only element must be 0 because 1's not in there?

I'm sorry, I'm just trying to understand... my initial question is: we want to prove that the only time 1=0 is in the trivial ring {0}. And in the proof, it is said a=a*1. And so I am trying to clarify... we can use the property a=a*1, because we are talking about a ring where 1=0, we know the...

Okay.. but if we are talking about a ring where 1=0, don't we already know 1 is in the ring?

Yeahh okay that's what I was thinking. We know 1 is in R... and there is no other way for the number one to behave... 1*x = x always. And so since 1 is in R, we must have unity. Thanks!

Hi, I found a couple of proofs proving that 1=0 only in the trivial ring {0}. They say Suppose 1 = 0. Let a be any element in R; then a = a ⋅ 1 = a ⋅ 0 = 0. But what I don't understand is that they say a = a ⋅ 1. But that is only true if a ring has unity (x*1=1*x=x), and it is possible to...

But an integral domain has unity by definition. So since R has unity, 1 is in R, so since aR contains all n element of R, R=aR, and 1 must be in aR too right? So why do you have to use yx_i_0 = (x_ia)x_i_0 = x_i(ax_i_0) = x_ia = y ?

That's what I'm trying to prove though, that the additive identity v + 0 = v does in fact hold, and if not it's not a vector space, but we have to test the axioms anyway to see which ones do hold. Addition of 2 vectors in this problem translates to: vector v := C (where C is a 2x2 matrix)...

Hi, I am new with linear algebra, and I'm having a hard time wrapping my mind around the 0 vector and the additive identity v + 0 = v, where 0 is the 0 vector. If I had a 2x2 matrix, and v + w = C + (C^T)*D ... (where (C^T) is the transpose, v & w are vectors, and C & D are matrices)...