Prove that -4 x -4 = 16

Raza

Can someone prove that -4 x -4 =16?

jcsd

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.

Zurtex

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.

cronxeh

what a de ja vu... i swear ive seen this before somewhere

dextercioby

And that's actually déjà vu as in nonadapted orginal French...

Daniel.

WORLD-HEN

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(a-a)=0[/tex]

[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]

Zurtex

WORLD-HEN we then to use a * when absolutely needed to denote multiplication. But (-4)(-4) = 16 might be easier.

dextercioby

So that big "ex" [itex] X [/itex] was actually [itex] \times [/itex] or [itex] \cdot [/itex] all along... Ingenious...

Daniel.

Raza

Thank you all, I finally get this.
A substitute teacher was asking me this question and I couldn't answer her at all

jcsd

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

WORLD-HEN

Sorry, havent used latex much before

Raza

your probability won't work if they are two different numbers, let say -5 and -3. How will you work that out?

mathwonk

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.

WORLD-HEN

The exact same way.

-5(3-3)=0

(-5)(3) + (-3)(-5) = 0

-15 + (-3)(-5) = 0

-3 x -5 = 15

More generally

-a(b-b) = 0
-ab + (-b)(-a) = 0

(-b)(-a) = ba

mathwonk

world hen, you are assuming that (-a)(b) = -ab, which i proved above.

that is why your post looks shorter.

eNathan

http://www.google.com/search?hl=en&q...=Google+Search

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.

jcsd

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.