# Proof of -m=(-1)m

1. Sep 18, 2009

### JoshSmith

1. The problem statement, all variables and given/known data

Prove $$\forall m \in \mathbb{Z}$$ that $$-m=(-1)m$$

2. Relevant equations

The axioms of $$\mathbb{Z}$$ are assumed, as are some of its propositions and corollaries. They would be too many to retype here, so assume that I have referenced the correct axioms and propositions if those are not stated outright.

3. The attempt at a solution

I believe, after working through it, that I got the solution. This has taken me hours, though, so please tell me if this works.

Proof. We'll show -m=(-1)m.

m=m by definition.
m+m=m+m by replacement.
m+m(1)=m+m by the multiplicative identity.
m+((-m)(-1))=m+m by some proposition not stated here.
(-m)(-1)=m by some proposition not stated here.
(-m)(-1)=m(1) by the multiplicative identity.
(-m)(-1)=m(-1)(-1) by some corollary not stated here.
-m=m(-1) by cancellation.
-m=(-1)m by associativity. Q.E.D.

2. Sep 18, 2009

### lanedance

1 + (-1) = 0-----------------by def'n (addditive inverse)
(1+(-1))*m = 0*m = 0--------multiply by m
1*m + (-1)*m = 0-----------distributive multiplication
m + (-1)*m = 0 -------------1*m = m
---> -m = (-1)*m by def'n of -m

may need to show it is also unique, not sure

Last edited: Sep 18, 2009
3. Sep 18, 2009

### JoshSmith

At the end I think you mean -m = (-1)m. You're missing a sign.

I don't think that works because we haven't defined -m, so you can't say it's by definition of -m. In fact, we're actually trying to define -m via this proof. Also, in the first step I think you'd need to be clear that 1+(-1)=0 is by the additive inverse, not by definition.

The whole thing is all very basic, I know. But does the proof I gave seem to work?

4. Sep 18, 2009

### lanedance

i find alot of the basic stuff tricky, beacuse you're so used to skipping the steps you're so familiar with...

you're right about 1 + (-1) = 0 being the additive inverse property, and about the missing -

however the definition of -m is that
m + (-m) = 0

so as soon as you are at the step
m + (-1)*m = 0
you should be able to see it they are equivalent, i skipped the last steps

but to be a tiny bit more explicit...
m + (-1)*m = 0
m + (-1)*m + (-m) = (-m).............add (-m) to both sides
m + (-m) + (-1)*m = (-m)............commutative
(-1)*m = (-m).............................defn of (-m)
QED

I like this a bit better as, i think it only relies on the basic properties:
- multiplicative distribution
- multiplicative identity & zero
- existence of (-m)

there are quite a few corollaries in yours, that i can't see where they came from
eg. in your 4th step you assume:
m*1 = (-m)*(-1) by some proposition not stated here

which is pretty close to what you're trying to prove...?

Last edited: Sep 18, 2009
5. Sep 18, 2009

### JoshSmith

I find the basic stuff pretty tricky, too, which is why this was so difficult.

The 4th step is close, but not quite it. If you had the book I have (which lays out all these axioms and propositions) you'd see it. I'm not going to retype it all.

I do like yours better; it uses less steps and only axioms.

So, question for you: how did you know how to approach this? My question throughout has been whether an approach to proof can come from some general method, or whether such mathematical intuition is developed over time. I was only able to solve this by working backwards, but I feel there must have been an easier way.

6. Sep 18, 2009

### lanedance

not sure, i think its just an intuition thing you build up after working some examples, and some trial and error

i guess i thought, whats in what we want to prove.. m, (-m), and (-1), whilst 1 is always inherent... can get most of them involved by multyiplying 1+(-1) = 0 by m.... Usually starting from zero can relate alot of things as well

And actually i think for correctness, though the idea is the same, the first step should more be like
0 = 0*m = (1+(-1))*m = 1*m + (-1)*m = m + (-1)*m
and so on from there

7. Sep 18, 2009

### JoshSmith

I see what you mean. You just kind of pulled apart the individual components and worked your way up from there. Nice.

I don't think you actually have to lay out the steps nearly as much as you're saying in your reply. I'm pretty sure you had it down right the first (revised) time.