# Two Alternate Proofs that $x \neq 0 \wedge xy = xz \implies y = z$.

I believe I have been able to construct in two ways, using the field axioms, that if $x \neq 0$ and $xy = xz$, then $y = z$. However, I've seen similar proofs like this assume that we can perform arithmetic operations, such as multiplying both sides by an inverse--which mirrors in some sense some proofs I've written in an abstract-algebra context--whereas others are more 'purist' in this sense. The similar proof in Rudin, for example, does not assume that we can use simple arithmetic.

My question, then, is which of these is 'more' standard in a first-year analysis course?

Proof 1: Assuming I can use arithmetic .

Since $x \neq 0$, $\exists x^{-1}$ s.t. $xx^{-1} = x^{-1} x = 1$ by the field axioms. Therefore, \begin{align*} xy = xz & & \text{By assumption} \\ x^{-1} (xy) = x^{-1} (xz) & & \text{Multiply on left by $x^{-1}$} \\ \left(x^{-1} x\right)y = \left(x^{-1} x\right)z & & \text{Associativity} \\ 1y = 1z & & \text{Inverse properties} \\ y = z \end{align*}

Example 2: Without assuming arithmetic, and mirroring Rudin.

\begin{align*} y & = 1 \cdot y & & \text{Multiplicative identity} \\ & = \left(x \cdot \frac{1}{x}\right) y & & \text{Mult inverse axiom with $x \neq 0$} \\ & = \left(\frac{1}{x} \cdot x\right)y & & \text{Commutativity of multiplication} \\ & = \frac{1}{x} \left(x \cdot y\right) & & \text{Associativity of multiplication} \\ & = \frac{1}{x} \left(xz\right) & & \text{Assumption that $xy = xz$} \\ & = \left(\frac{1}{x} \cdot x\right) z & & \text{Associativity of multiplication} \\ & = 1z & & \text{Inverse properties} \\ & = z \end{align*} Thanks in advance.

• If $K$ is a field, then $G=(K^{\times},\cdot)$ is an abelian group, so that the cancellation law holds. For $x,y\in G$ we have that $xy=xz$ implies that $y=z$. – Dietrich Burde Aug 17 '18 at 18:20
• In your "mirroring Rudin" example the proof is just one long chain of equal quantities: You want to show $y=z$ so you start with $y$ and write down expressions you know are equal to it using axioms and assumptions until you have a $z$. In your "arithmetic" proof, you have a list of equalities, and you use your axioms to transform them to get the claim you want: that $y=z$. I don't really think these proofs are different, and I expect people looking at your work would agree with me. But, if you want to make sure, I would suggest asking your grader/professor. – James Aug 17 '18 at 18:28
• Also, I think your worry about "using arithmetic" is illfounded. You have a claim such as $xy = xz$ in some structure you are reasoning about. You also have a binary function on that structure: multiplication. Therefore the quanties $x^{-1}(xy)$ and $x^{-1}(xz)$ are both defined because you are just plugging in elements of the domain into your function. That $y=z$ follows from the assumed properties of multiplication and the existance of inverses. – James Aug 17 '18 at 18:31
• One more thing. The only thing you must avoid in a proof of $y=z$ is starting with $y=z$ and deriving $0=0$ or $1=1$. As long as your proof starts with assumptions you are given, follows logically valid steps, and ends up with what you want, then the proof is good. Many of my students try to show $x=y$ and argue "$x=y$ ... <operations> ... $0=0$, QED". What is most frustrating is that often if they just turned the proof up-side-down, then it would be valid, i.e, the operations they effect on the equation can be done backwards to start with $0=0$ and derive $x=y$. – James Aug 17 '18 at 18:35
• Those two proofs are exactly the same as far as I can tell. Or aren't significantly different. "Assuming arithmatic" is a meaningless thing to say. To prove this we must have a well defined set of axioms. "Assuming arithmetic" is simply referring to them. – fleablood Aug 17 '18 at 18:37

The two proofs are essentially the same and the first doesn't use arithmetic, but rather field axioms. I wouldn't use $\frac{1}{x}$, but that's more cosmetic than substantial.
More substantial is that you don't need to appeal to commutativity: \begin{align} y &=1y &&\text{(multiplicative identity)} \\ &=(x^{-1}x)y &&\text{($x\ne0$ has an inverse)} \\ &=x^{-1}(xy) &&\text{(associativity)} \\ &=x^{-1}(xz) &&\text{(hypothesis)} \\ &=(x^{-1}x)z &&\text{(associativity)}\\ &=1z &&\text{(property of the inverse)} \\ &=z &&\text{(multiplicative identity)} \end{align}
On the other hand, the other proof seems shorter \begin{align} & xy=xz &&\text{(hypothesis)} \\ & x^{-1}(xy)=x^{-1}(xz) && \text{($x\ne0$ has an inverse)} \\ & (x^{-1}x)y=(x^{-1}x)z && \text{(associativity)} \\ & 1y=1z && \text{(property of the inverse)} \\ & y=z && \text{(multiplicative identity)} \end{align} and less “rabbit out of a top hat”.