Is -4 multiplied by -4 equal to 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 I've seen this before somewhere

 

[dextercioby]

what a de ja vu... i swear I've seen this before somewhere

And that's actually déjà vu as in nonadapted orginal French... :tongue2:

Daniel.

 

[WORLD-HEN]

If I understand correctly, you want me to prove that negative times negative is positive, right?

$$-a(a-a) = 0$$

This statement is true since $$(a-a) = 0$$

Using the distributive law :

$$-a(a-a)=0$$

$$-a^2 + (-a X -a) = 0$$

$$-a X -a = a^2$$

So, substituting 4 for a gives
$$-4 X -4 = 4^2$$
$$-4 X -4 = 16$$

 

[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" $X$ was actually $\times$ or $\cdot$ all along... Ingenious... :rofl:

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]

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

 

[WORLD-HEN]

Sorry, haven't used latex much before

 

[Raza]

If I understand correctly, you want me to prove that negative times negative is positive, right?

$$-a(a-a) = 0$$

This statement is true since $$(a-a) = 0$$

Using the distributive law :

$$-a(a-a)=0$$

$$-a^2 + (-a \times -a) = 0$$

$$-a \times -a = a^2$$

So, substituting 4 for a gives
$$-4 \times -4 = 4^2$$
$$-4 \times -4 = 16$$

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]

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

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=-4*-4&btnG=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]

http://www.google.com/search?hl=en&q=-4*-4&btnG=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.

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.

 

[eNathan]

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 We could travel back in time! But of course, there is the hole thing about the fact that negative numbers don't 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

 

[Zurtex]

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.

 

[mathwonk]

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".

 

[eNathan]

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.

Nah, I already know what those are. what I mean when I said that negative roots are not REAL numbers, but imaginary, is this.

Sqr(-25)=5i Right? 5*5=25? -5*5=25? There is not mathematical root, unless you get into imaginary numbers.

And the point I was trying to get at is the concept of negative numbers. They are not like positive numbers, in the sense that they are the exact oposite, or inverse of it. Nevermind tho...No need to start a huge argument, as I sense it might happen.

 

[Zurtex]

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".

Sorry, my fault entirely, big mistake there using the word negative number. I of course meant additive inverse and that the additive inverse of any positive number is negative.

Nah, I already know what those are. what I mean when I said that negative roots are not REAL numbers, but imaginary, is this.

Sqr(-25)=5i Right? 5*5=25? -5*5=25? There is not mathematical root, unless you get into imaginary numbers.

And the point I was trying to get at is the concept of negative numbers. They are not like positive numbers, in the sense that they are the exact oposite, or inverse of it. Nevermind tho...No need to start a huge argument, as I sense it might happen.

Imaginary are mathematical though, the word real means a well known and defined set in mathematics it does not mean the English word 'real' as in something that exists.

Just trying to clarify not argue.

 

[eNathan]

Yes, your right, I was using real as an English word, and you interpeted it as a mathematical word.

 

[josh6541]

the question is really funny. isn't it?

 

[mathwonk]

the reason for my objection is that i am a teacher and every year, every day almost, i have students who think that -a is always negative, independent of what a is.

many people find it almost impossible to understand the definition: |x| = x when x is positive, and |x| = -x, when x is negative, for example.

they cannot grasp that -x in that case is actually positive because they are in the habit of saying "negative x" for -x.

this apparently trivial difference actually gives rise to some genuine misunderstandings.

whatever.