Negatives and positives

  • #1
Jonny_trigonometry
451
0
this is the dumbest question I've ever come up with, but I can't answer it, so that would make me horribly stupid.

Why is:
-1*-1=1
1*1=1
-1*1=-1
1*-1=-1

why can't:
-1*-1=-1
1*1=-1
-1*1=1
1*-1=1
?

Is there any reason why we've defined things in the latter way rather than the former?

p.s. sorry folks, I'm sure there is an obvious and simple reason, but I can't see it right now.
 

Answers and Replies

  • #2
honestrosewater
Gold Member
2,136
5
Well,
1x = x​
is an axiom. :smile: The general case probably has another proof, but I can't think of it just now. Right, here's a start.
Prove: (-x)y = -(xy) = x(-y).

xy + (-x)y = (x + -x)y = 0y = 0​
Say you already know that if w + v = 0, then v = -w. That gives you the first equality, (-x)y = -(xy). The other, -(xy) = x(-y), is proven the same way.
The next theorem you want is
Prove: (-x)(-y) = xy.
Use the previous theorem to prove this.
 
Last edited:
  • #3
HallsofIvy
Science Advisor
Homework Helper
43,021
970
In other words, if you define
"-1*-1=-1
1*1=-1
-1*1=1
1*-1=1 "

Then the distributive law would give -1*(1+ -1)= -1*(1)+(-1)*(-1)= -1-1= -2

But 1+ -1= 0 (by definition of -1) so it is really (-1)*0= 0. The definitions you give would mean that the distributive law (a very important algebraic law) would not be true.
 
  • #4
Jonny_trigonometry
451
0
ahhh... I see! thanks, I was really only looking at commutivity i guess.

well, are you sure?

based on
-1*-1=-1
1*1=-1
-1*1=1
1*-1=1

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

becasue -1*1=1, then the distributive law would still work.
You just messed up in your example.

hmm...

"Well,

1x = x

is an axiom. The general case probably has another proof, but I can't think of it just now. Right, here's a start.

Prove: (-x)y = -(xy) = x(-y).

xy + (-x)y = (x + -x)y = 0y = 0

Say you already know that if w + v = 0, then v = -w. That gives you the first equality, (-x)y = -(xy). The other, -(xy) = x(-y), is proven the same way.
The next theorem you want is

Prove: (-x)(-y) = xy.

Use the previous theorem to prove this." - Honestrosewater

I can't, based on this:
-1*-1=-1
1*1=-1
-1*1=1
1*-1=1
I don't think I can... But I can prove (-y)(-x)=-(xy), and (yx)=-(xy) with it.
Almost all the current theorems would be bassackwards, but wouldn't math still work?

or is there some case in my logic that would end up with -1=1?
hmm...
-1*-1=-1=1*1
multiply -1 on both sides
(-1)(-1)(-1)=1*1(-1)
(-1)(-1)=(-1)(-1)
-1=-1

how about
multiply 1 on both sides
1*(-1)(-1)=1*1*1
1*-1=-1*1
1=1

I think the logic still works, but every function would be flipped along the y-axis, so for example the zeta function's solutions would be 2,4,6,... and it's non-trivial ones would all lie on real part -1/2, the line y=x would have a negative slope of 1, but symmetrical functions like even polynomials (y=x^n, where n=2,4,6,...) and the cosine function would all be the same. Am I interpreting this wrong, or is there a flaw, and therefore a reason why we've defined things to be the way they are?
 
Last edited:
  • #5
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,967
19
"1*1 = -1"

Note that this contradicts the fact that 1 is defined to be the multiplicative identity. :smile:

Well, not exactly: it says that 1 = -1, which is a perfectly fine thing to say when working mod 2.
 
  • #7
eNathan
352
1
Jonny_trigonometry said:
this is the dumbest question I've ever come up with, but I can't answer it, so that would make me horribly stupid.

Why is:
-1*-1=1
1*1=1
-1*1=-1
1*-1=-1

why can't:
-1*-1=-1
1*1=-1
-1*1=1
1*-1=1
?
Is there any reason why we've defined things in the latter way rather than the former?

p.s. sorry folks, I'm sure there is an obvious and simple reason, but I can't see it right now.

Well that kinda gets into the very logic of what a negative is. Think of it as logic "not" operations. Think about using these in sentences and you will get the idea.

-1*-1=1 ; No + No = Yes
1*1=1 ; Yes + Yes = Yes
-1*1=-1 ; No + Yes = No
1*-1=-1 ; Yes + No = No

For instance, if I said...
I did not not go to the store. That really means, you DID go to the store. -1 * -1 = 1 ; No + No = Yes ... and so forth.

You kinda see how there is a relationship between mathematics and grammer? lol I am not sure if this is an official way to present it, but I just though it up a while back.
 
  • #8
Jonny_trigonometry
451
0
so are all words considered to be multiplied together in a sentance? I've always thought they were a sum.

hmm...

-1*-1=1 ; No + No = Yes => if someone says no twice, they are unsure of what they really want because they feel the need to back up their first answer, so therefore they really mean yes... eek! sounds like the guy that never took no for an answer!

1*1=1 ; Yes + Yes = Yes => if someone says yes twice, they really mean it!

-1*1=-1 ; No + Yes = No => if someone says both no and yes to something, then let's look on the pessimistic side and assume they meant no. where's our sense of adventure?!

1*-1=-1 ; Yes + No = No => same reason as above
 
Last edited:
  • #9
Jonny_trigonometry
451
0
Hurkyl said:
"1*1 = -1"

Note that this contradicts the fact that 1 is defined to be the multiplicative identity. :smile:

Well, not exactly: it says that 1 = -1, which is a perfectly fine thing to say when working mod 2.


well, could there be a different multiplicitive identy?

hmm... I'll have to study "multiplicitive identy" for a while to fully follow your argument. For now I'll say that you may have the reason why it won't work
 
  • #10
eNathan
352
1
Jonny_trigonometry said:
so are all words considered to be multiplied together in a sentance? I've always they were a sum.

Yes, they would most definately have to be multiplied in a sentence. Let me give an example.

John Kerry didnt not lose the election. == John Kerry lost the election.

In the first sentence, the words "Didnt", "not", and "lose" all count as logical "not" operations.

In the second sentence, the word "lost" counts as the only "not" operator. So notice that when you multiply them, they equal the same thing

-1 * -1 * -1 == -1

If you got sum, you would get -3 == -1 :surprised what would that mean lol

Edit :: I say that "lost" is a not operator, and therefore equals -1 because "lost" would be the same as "didnt win".
 
Last edited:
  • #11
Jonny_trigonometry
451
0
poor Kerry...

but it's worse for the nation...

poor us...

but who doesn't not know how not bad Kerry would not have not not been?
 
  • #12
eNathan
352
1
but who doesn't not know how not bad Kerry would not have not not been?
But who -1 * -1 know how -1 * -1 Kerry would have -1 * -1 * -1 been?
Just multiply them, cancel out the 1's, and w00t

But who knows how good Kerry would not have been? :wtf:

Yea I was going for Kerry too. Too bad...
 
Last edited:
  • #13
matt grime
Science Advisor
Homework Helper
9,426
4
Jonny_trigonometry said:
well, could there be a different multiplicitive identy?


yes and no. there may be a different multiplicative identity but it would be for a different multiplicatve structure. all of the algebraic properties of the numbers you are thinking about follow from extending those on the natural numbers and are simple and natural consequences.
 
  • #14
Jonny_trigonometry
451
0
so, would the consequences be that all functions are a transform of all known functions? ie. a mirror image about the y-axis?

y=x would have a -1 slope?
imaginary numbers would deal with the sqrt of 1 rather than -1?
the left side of the y-axis would be the right and vice versa?

would the linear transformation in [tex]\Re^2[/tex] then be:

[tex] \left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right)[/tex]
?
 
Last edited:
  • #15
matt grime
Science Advisor
Homework Helper
9,426
4
what on earth?
 
  • #16
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,967
19
I think he's saying that the multiplication you get by using -1 as the multiplicative identity is isomorphic to the ordinary reals through the map x → -x.
 
  • #17
matt grime
Science Advisor
Homework Helper
9,426
4
one more that i can add to the list of questions that i cannot even begin to answer.
 
  • #18
Jonny_trigonometry
451
0
well, I'm just trying to understand why we've defined multiplication rules between negatives and positives, negatives and negatives, and positives and positives the way we did. I proposed different rules:
-1*-1=-1
1*1=-1
-1*1=1
1*-1=1
and I'm trying to find the consequences of how this change would change all the math that is built on multiplication rules, and I'm thinking that everything would be backwards, but math would still work. Negatives would be treated like how we treat positives and positives would be treated like how we treat negatives.

yes, no?

but the main question at hand, that is if math would still work but just "backwards", is... Why did people define things the way they are rather than doing it the other way? OR, if this whole idea isn't logical, then that would be the reason why, and I can put it to rest.

eh, nevermind...

this is probably more of an anthropology question. or psychology, or evolution of ideas type question or something. I sound like a bumbling fool. :rolleyes:
 
Last edited:
  • #19
matt grime
Science Advisor
Homework Helper
9,426
4
We (well, not me) already explained this to you.

0*x=0 for any number, right?

now 0=a-a for any a, right?

so put these together and it follows that

0=0*x=(a-a)*x

now we want multiplication to be distributive so that

0=a*x+(-a)*x

or

(-a)*x=-(a*x)

right?

so you want to undo some or all of the following:

the definition off additive inverses (ie x-x is no longer 0), the distributivity of multiplication over addition (or that 2=1*2=1*(1+1)= 1*1+1*1 is no longer true, or that 2 is not 1+1), or that 0*x is not 0, but this again follows from the fact that 1*0=(1*(0+0)=2*0, so you would be again denying that 1 is the identity or that multiplication is distributive, or that 0 is the additive identity or that additive inverses exist).

people defined things the way they are in order to consistently extend the operations that we can think of easily for positive whole numbers. in order to do it differently we would have to redefine most everything and create many special subcases such as a(b+c)=ab+ac if a,b and c are all positive, but not necessarily otherwise. and of cuorse you would then need to define every single possible variation if we are to have nunderlying universal rules.
 
  • #20
Jonny_trigonometry
451
0
I don't follow the reasoning from this:
(-a)*x=-(a*x)
to the idea that 1-1 is no longer equal to zero, and 1+1 is no longer equal to 2 or 0*x doesn't equal zero.
because if:
1*x=x, and -1*x=-x, then
(-a)*x=-(a*x)
0=a*x+(-a)*x
0=(1)a*(1)x + (-1)a*(1)x
0=(1*1)*a*x + (-1)*(1)*a*x
0=(-1)*a*x + (1)*a*x
which is still true, even with diferent multiplication rules of neg and pos:
-1*-1=-1
1*1=-1
-1*1=1
1*-1=1

man, I'm tired of this, are you? I give up, I'll trust that there is a logical reason that you're trying to show me, but I just don't get it because I just don't catch on. btw, I've already pretty much figured that my interpretation that all functions would be flipped across the y-axis is incorrect, y=x would have a slope of 1. I'm bettin that I just don't get it right now so pardon me for being headlong.
 
Last edited:
  • #21
matt grime
Science Advisor
Homework Helper
9,426
4
if

0=a*x +(-a)*x,

which follows from the fact that 0*x=0 and that 0=a-a and that mult is distributive, then it follows that

-(a*x)=(-a)*x

after subtracting a*x from both sides. that is how all these elementary proofs are done (eg showin 0*x=0)
 
  • #22
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,967
19
Jonny: I think what you're trying to say is that you'd like to define a new multiplication (I'll call it &) which is given by:

a & b := -(a * b)

Is that correct?
 
  • #23
Jonny_trigonometry
451
0
I'm not really trying to invent something new, I just got to wondering why is it that
-1*-1=1
1*1=1
-1*1=-1
1*-1=-1

if a&b would give rise to
-1*-1=-1
1*1=-1
-1*1=1
1*-1=1

then you can interpret my curiosity as a new multiplication. I don't know.

I understand why 1+ (-1) =0, but I don't understand why 1*(-1)=-1, besides knowing that it's true, I can't get any further than the knowledge of -1*1=-1 being true. I'll have to accept that it's true for now until I can understand why it's true.
 
  • #24
Zurtex
Science Advisor
Homework Helper
1,120
1
Are you aware of the 10 axioms of the field [itex](\mathbb{R}, +, *)[/itex]? Once you are aware of them then the rest follows quite simply.
 
  • #25
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,967
19
I agree with Zurtex... what you need are definitions!

The real numbers are, by definition, a complete ordered field. (Start with the right link and work leftwards)

(Sigh, that's not the best link for complete... but oh well)

In particular, the definition of a field says that there is a multiplicative identity which, by convention, we write as 1. To be a multiplicative identity means that, for any x, 1 * x = x = x * 1.
 
Last edited:
  • #26
Jonny_trigonometry
451
0
thanks for the help, I'll check it out. :biggrin:
 
  • #27
matt grime
Science Advisor
Homework Helper
9,426
4
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.
 
  • #28
eNathan
352
1
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:
 
  • #29
Jonny_trigonometry
451
0
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 their 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.
 
  • #30
matt grime
Science Advisor
Homework Helper
9,426
4
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:
  • #31
Jonny_trigonometry
451
0
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:
  • #32
matt grime
Science Advisor
Homework Helper
9,426
4
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.

can we talk about addition since that is simpler?

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.
 
  • #33
Zurtex
Science Advisor
Homework Helper
1,120
1
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
 

Suggested for: Negatives and positives

  • Last Post
Replies
1
Views
335
Replies
1
Views
474
  • Last Post
Replies
3
Views
426
Replies
4
Views
492
  • Last Post
Replies
4
Views
388
Replies
7
Views
657
Replies
6
Views
873
Replies
3
Views
389
  • Last Post
Replies
1
Views
549
  • Last Post
Replies
1
Views
399
Top