# Negatives and positives

thanks for the help, I'll check it out.

matt grime
Homework Helper
Ok, step by step again:

you agree that 0*x=0 for any x, in particular

0*1=0
is that ok?

you agree that

0=1+(-1)

and that if is substitute any equivalent ways of writing a number into a statement it remains unchanged so that i put these together and get

0=0*1=(1+(-1))*1

now we also agree that we can distribute multiplication, ie that a(b+c)=ab+ac?

so now we have 0=1*1+(-1)*1

(notice the brackets to avoid confusion)

so we have

0=1+(-1)*1

subtract that one on the right and we get

-1=(-1)*1

thus this fact is deducible from the properties of addition and multiplication behaving as we declare they do.

Well put matt grime :)

Jonny_trigonometry, I also have a question. Dont take it the wrong way if it may sound offensive lol

Do you honestly not understand how negatives and positives work?
Or are you just trying to figure out why it works that way?
And/Or your tring to figure out why another system (dubed &) will or won't work?

heh I still think my english grammer analogy is not a bad example. Oh, that must mean its a good example. :lol:

I'm trying to figure out why people chose to do things that particular way. I've followed Matt grimes arguments the whole time, but I still see that they would work with either multiplication rules.

because if 1*1 is defined to be -1, then 0=1*1+(-1)*1 yields 0=-1+(-1)*1
and then 1=(-1)*1 which shows that the multiplication rules I propose would still work
-1*-1=-1
1*1=-1
-1*1=1
1*-1=1

see what I mean?

I think the reason why we do it this way
-1*-1=1
1*1=1
-1*1=-1
1*-1=-1
is because people started using natural numbers before they started using negatives, so when negatives were introduced, they still wanted to keep thier previous multiplication rules (that a positive times a positive is a positive). Thats it. Thats why I think it's more of an anthropology question.

matt grime
Homework Helper
But, and this is something you really really must accept, the unique element that satifsies e*x=x for all x *is* 1 (in the reals). it is the meaning of the symbol 1 in mathematical terminology.

if you want to think of someother system where 1 is not the multplicative identity then feel free to do so but you *have* to accept that you aren't talking abuot the ordinary multiplication on the reals as the rest of the world knows it. No one is denying that there are other binary opertions one may define on the the reals but we are explaining why in the usual one we can show certain things follow from other certain things.

shall we do a quick "what maths is about" lesson? it's only a roguh guide.

maths is essentially about proving (if possible) why fact X follows from Y. As such we often want to know the minimal set of rules that can be used to deduce the results we want. remember, there is nothing such as absolute truth in mathematics, it is all relative to our axioms.

So, we may accept that the integers are a Ring, and then it follows that (-1)*(-1)=1 where 1 is the mult identity and -1 is the additive inverse of 1, and can be easily shown.

Even if not we may show that considering the natural numbers and formally adding inverses (negatives) yields the same result.

in any case 1 is simply the label for the identity under *

Last edited:
yes, I understand that I'm not talking about ordinary multiplication on the reals as the rest of the world knows it.

hmmm... ok, I'm startin to get it... but! what if -1 is the multiplicative identity?

how far back do the rules go? does it stop at the multiplicative identity? or is there a reason that the MI must be positive 1?

Last edited:
matt grime
Homework Helper
There is a reason that 1 is the mult ident and that is becaue the operation is * the multiplcation operator as we know it where n*m means add m up n times (n, m are positive integers) and which is extend to the rest of the integers as we invented them.

The element that is the identity with respect to some opereation is dependent on the operation.

Take Z the integers with the usual operations of addition denoted as + , then defnie a new opertaion & where

x&y=x+y-1

then -1 is the identity with respect to this "addition". See, it can be done, but you are attempting to think of our declaration of identities (an inverses) as independent of an operation.

Zurtex
Homework Helper
Jonny_trigonometry said:
yes, I understand that I'm not talking about ordinary multiplication on the reals as the rest of the world knows it.

hmmm... ok, I'm startin to get it... but! what if -1 is the multiplicative identity?

how far back do the rules go? does it stop at the multiplicative identity? or is there a reason that the MI must be positive 1?
The multiplicative identity is defined such that if we let e be the multiplicative identity:

e*x = x = x*e

Just as the additive identity, say f, is defined as:

f+x = x = x+f