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
$\endgroup$ 12 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$
