Can someone prove that -4 x -4 =16?
Lets say that 4 and 16 are members of a ring such that 4*4 = 16
In any associative ring:
1) a*0 = a*0 + 0 = a*(b + -b) + ab + -(ab) = a(b + -b + b) + -(ab) = ab + -(ab) = 0
2) 0*a = 0*a + 0 = (b + -b)*a + ba + -(ba) = (b+ -b + b)a + -(ba) = ba + -(ba) = 0
3) a*-a = a*-a + 0 = a*-a + (a*a + -(aa)) = a(-a + a) + -(aa) = a*0 + -(aa) = -(aa)
4) -a*-a = -a*-a + 0 = -a*-a + (a*-a + aa) = (-a + a)*-a + aa = 0*-a + aa = aa
edited to add: here are the axioms of a ring http://mathworld.wolfram.com/Ring.html
I haven't shown which axioms is use din each step but i hope you can see thta each step does follow on directly from the ring axioms and what was shown before it.
It's a simple extension of the properties of real number to prove that -1*a = -a (where -a is the additive inverse), also that -1*-1 = 1 and also that (-a)*b = -(a*b). I think the proof is fairly elementary once you have properties like these.
Edit: Well it appears the person above me had done a lot more general proof with full details so nevermind.
what a de ja vu... i swear ive seen this before somewhere
And that's actually déjà vu as in nonadapted orginal French... :tongue2:
If I understand correctly, you want me to prove that negative times negative is positive, right?
[tex] -a(a-a) = 0 [/tex]
This statement is true since [tex] (a-a) = 0 [/tex]
Using the distributive law :
[tex] -a^2 + (-a X -a) = 0[/tex]
[tex] -a X -a = a^2[/tex]
So, substituting 4 for a gives
[tex] -4 X -4 = 4^2 [/tex]
[tex] -4 X -4 = 16[/tex]
WORLD-HEN we then to use a * when absolutely needed to denote multiplication. But (-4)(-4) = 16 might be easier.
So that big "ex" [itex] X [/itex] was actually [itex] \times [/itex] or [itex] \cdot [/itex] all along... Ingenious... :rofl:
Thank you all, I finally get this.
A substitute teacher was asking me this question and I couldn't answer her at all
The last time a simalir question was asked someone quoted W. H. Auden:
"Minus times minus is plus, the reasons for this we need not discuss"
If you don't want to bother with all the ring theoretic stuff just tell your teacher that
Sorry, havent used latex much before
your probability won't work if they are two different numbers, let say -5 and -3. How will you work that out?
what does -1 mean? it means the number which added to 1 gives zero. and hence -(-1) is the number which added to -1 gives zero, so -(-1) = 1.
morover (0+0) = 0, so 0a = (0+0)a = 0a + 0a, so subtracting gives 0a = 0.
hence 0 = 0a = [1+(-1)]a = a + (-1)a.
hence (-1)a =-a.
thus (-a)(-a) = (-1)(-1)a^2 = -(-1)a^2 = a^2.
gee, this is as long and tedious as alkl the oithers.
obviously this kind of nonsense was never meant to be fun.
The exact same way.
(-5)(3) + (-3)(-5) = 0
-15 + (-3)(-5) = 0
-3 x -5 = 15
-a(b-b) = 0
-ab + (-b)(-a) = 0
(-b)(-a) = ba
world hen, you are assuming that (-a)(b) = -ab, which i proved above.
that is why your post looks shorter.
Is that simple enought. No, it probally not. But its fairly simple. The concept of a negative number is totally different from that of real numbers. A negative number is not only the asbsense of a real number, but its used to make up for the non-existance of the real numbers. This is why a negative times a negative is a positive.
Negative numbers ARE real numbers.
There's a simple amendmnt you make to the proof I made earlier so that it proves (-a)(-b) = ab rather than (-a)(-a) = aa.
Yes, your right on that one. I tend to get lost with all the terms :yuck:
but my concept of what negative numbers are correct, its like saying "I am not not going to the store", which really means you ARE going to the store. Its sort of hard to explain, and for even me to understand (...) but a negative a negative amount of times is really a positive, because like I said, a negative number is the absense of positive numbers. If our mathamatical system was not like that :surprised We could travel back in time! But of course, there is the hole thing about the fact that negative numbers dont have true roots, but imaginary roots. I guess, mathematicly, to get the square root of p, when p is -negative, you would do 0-(Sqr(0-p)) That should work
eNathan I can't help thinking when reading your posts that you don't really understand the number system that well.
A negative number: -a is defined such that: a + (-a) = 0. That is it, there is no more to it, however from this and the other rules of real numbers we can prove that -1*(a) = -a, -1*-1=1, (-a)*b = -(a*b) and many other useful identities.
Just because the square root of a negative number is imaginary it does not mean it is not the 'true' square root of the number.
I think you would find it rather helpful to look up the mathematical definitions of the words 'real', 'imaginary' and 'complex'. As well as looking up what Natural Number, Integers and Quotients are.
this has nothing to do with negative numbers, -a is not a negative number, it is the additive inverse of a.
a need not be a real number at all, a could belong to any additive group.
In any case even if a were a real number, -a is mereoly the opposite of a, if a is zero, -a is zero not negative, and if a is negative then -a is positive.
-a is not read "negative a", but "minus a" or "additive inverse of a", or "the number you add to a to get zero".
the word negative means "less than zero". this am,kes no sense in amny situations where -a makes perfect sense, for example complex numbers where there are no negative complex numbers, since "greater than" and "less than" are not defined for complex numbers, for the same reason you cannot say which horse on a merry go round is in "front".
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