3

I'm sure these proofs i'm doing are trivial to most of you, but I'm going to persist with these from this book because the aim is to improve the technique of proof writing and mathematical communication. I'm also finding out some properties of numbers I did not realise.

If $d\in \mathbb{Z}$ and $d|d$, then $\exists k \in \mathbb{Z}$ such that $d\cdot k = d$
The number k exists and is 1.

Would that last line be an acceptable ending of the proof?

Bucephalus
  • 1,386
  • 4
    This is the right idea, but: "If $d\in\mathbb Z$ and $d\mid d$ then..." Is a confusing way of writing the proof. You don't want $d\mid d$ implying something, you want something to imply $d\mid d$. So what you should be saying is roughly the reverse: "Since $d\cdot 1=d$ then there is a $k$ such that $d\cdot k=d$, and thus $d\mid d$." – Thomas Andrews Jun 16 '17 at 01:27
  • Oh that's nice @Thomas. Thanks for that. – Bucephalus Jun 16 '17 at 01:29
  • 1
    Indeed. Bucephalus, you had structured a proof along the lines of ${A\to B~,~ B}~\vdash A$. This is not valid. Thomas gives the correct form: ${B~,~ B\to A}~\vdash A$. (Modus Ponens) – Graham Kemp Jun 16 '17 at 01:37
  • But couldn't I be stating "If $d\in \mathbb{Z}$ and $d|d$, then $\exists k \in \mathbb{Z}$ such that $d\cdot k = d$" as the axiom I will be using to perform the proof. Like a precondition. But then say, what Thomas said, "since $d \cdot 1 = d$ then there is a $k$ such that $d \cdot k =d$, and thus $d|d$." Would you do this, or it's just accepted in the maths community. – Bucephalus Jun 16 '17 at 02:03

1 Answers1

4

This is the right idea, but: "If $d\in\mathbb Z$ and $d\mid d$ then..." is a confusing way of writing the proof.

You don't want $d\mid d$ implying something, you want something to imply $d\mid d$. So what you should be saying is roughly the reverse:

Since $d\cdot 1=d$ then there is a $k$ such that $d\cdot k=d$, and thus $d\mid d$.

What you've written is a common error in beginning proof-writing. You are reflecting the order of your reasoning: "I want to show $d\mid d$, which means that $d\cdot k=d$, but then $k=1$...."

But that is not the order of the proof, that is the order of your reasoning. The result you want should always be the end of the proof.

Thomas Andrews
  • 177,126