# Proof of a very basic identity. a=b --> xa = xb

• Nov 10th 2012, 09:39 AM
publicvoid
Proof of a very basic identity. a=b --> xa = xb
Hi, I need to prove a very basic equality which states that
if a=b , then x*a = x*b
note that a,b, and x can be scalars, matrices or vectors.

I guess I have to prove it using the ring (R, +, *), any help would be appreciated. Thanks!
• Nov 10th 2012, 11:33 AM
emakarov
Re: Proof of a very basic identity. a=b --> xa = xb
Quote:

Originally Posted by publicvoid
I need to prove a very basic equality which states that
if a=b , then x*a = x*b

This is true for any theory with equality and a binary operation *. That is, this is a corollary of equality axioms as opposed to axioms describing any particular theory, such as theory of groups, rings or vector fields. Equality axioms are usually considered the common basis of all theories, along with such axioms as "A and (A -> B) imply B."

Quote:

Originally Posted by publicvoid
note that a,b, and x can be scalars, matrices or vectors.

I guess I have to prove it using the ring (R, +, *)

Note that scalars (e.g., real numbers) and matrices do indeed form a ring, but vectors do not because of multiplication. The dot product maps two vectors into a number, not a vector, and the cross product is not associative, nor does it have a multiplicative identity. Vectors are usually considered to be a vector space over a field or its generalization, a module over a ring. In this case, multiplication takes two objects of different type: a scalar and a vector.
• Nov 10th 2012, 12:09 PM
publicvoid
Re: Proof of a very basic identity. a=b --> xa = xb
Hi, emakarov.
As I understand, you are saying that this is an axiom, which needs no proof. That is exactly what I think. But apparently my professor does not think so, so he gave it as a homework :)
So, any ideas how to prove it?
• Nov 10th 2012, 12:30 PM
emakarov
Re: Proof of a very basic identity. a=b --> xa = xb
I don't know how to prove it except for the following. Equality axioms may differ from source to source, but one approach is as follows. There are two axioms about equality: (1) for all x, x = x and (2) for all x, y, A[x] and x = y imply A[y] for any formula A. So, we have x * a = x * a by (1) and then replace one occurrence of a with b by (2) to get x * a = x * b.
• Nov 10th 2012, 12:33 PM
Plato
Re: Proof of a very basic identity. a=b --> xa = xb
Quote:

Originally Posted by publicvoid
. But apparently my professor does not think so, so he gave it as a homework . So, any ideas how to prove it?

Without knowing the exact set of ring axioms, this is a mere guess.
\begin{align*}0&=a-b\\ 0&=x(a-b)\\0 &=xa-xb\\0 &= ax-bx\\ax&=bx \end{align*}
• Nov 10th 2012, 12:42 PM
publicvoid
Re: Proof of a very basic identity. a=b --> xa = xb
Quote:

Originally Posted by Plato
Without knowing the exact set of ring axioms, this is a mere guess.
\begin{align*}0&=a-b\\ 0&=x(a-b)\\0 &=xa-xb\\0 &= ax-bx\\ax&=bx \end{align*}

yes this is how I did it, I hope it's true :)