Proof of Cancellation Law for Multiplication

New to Linear Algebra and trying to get decent at proofs. Any help is appreciated.

Is this a valid way to prove the Cancellation Law of Multiplication? I've only seen it done some alternative way that doesn't make much intuitive sense to me.

Given: For all real numbers $x,y,k$ where $k\neq 0$ .

If $kx = ky$, then $x = y$

*First, $ky, kx$ are also real numbers by the Closure property of Multiplication

$kx = ky$ : Implication

$k(1/k)x = k(1/k)y$: Substitution

$(k(1/k))x = (k(1/k))y$: Associative Property of Multiplication

$(1)x = (1)y$ : Multiplicative Inverse

$x = y$ : Multiplicative Identity

• Not sure that the second step is "substitution". Placing the 1/k in the middle rather than the end makes me want to see an reasoning by comutivity. Seems good though. – fleablood Sep 4 '16 at 0:40

I could be wrong but I think the second step is rushed.

I'd do something like.

$\frac 1k(kx )= \frac 1k(kx )$: identity;binary operation

$\frac1k (kx)=\frac 1k (ky)$:substitution

But maybe I don't understand your second line. The rest is perfect.

You have the right idea for the proof. I'd change some of the wording.

$kx = ky$ : Hypothesis (not Implication)

In the next line you have implicitly used the commutativity of multiplication. Instead write

$(1/k)(kx) = (1/k)(ky)$: Multiply both sides by the $1/k$, which exists because we've assumed $k \ne 0$.

$((1/k)k)x = ( (1/k)k)y$: Associative Property of Multiplication

The rest is fine as is

$(1)x = (1)y$ : Multiplicative Inverse

$x = y$ : Multiplicative Identity

This theorem and its proof really have nothing to do with linear algebra - the right context is the real number system (more abstractly, a field) although you encountered it in a linear algebra course, which is often where students first have to write careful proofs.