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The context:

$4 - 4 = 0$

Ok, the first number is assumed as a positive and the operation here is a subtraction, so the second number is a positive also.

So I can summ it in the same way by specifying the second number as a negative: $4 + (-4) = 0$

If I did the same but making the first number a negative, $-4 + 4$, why is it equal to $0$ and not $-8$?

Why do I have to specify that the second number is a negative also? $-4 + (-4) = -8$

The question:

How can I add a number to a negative number without specifying it to be negative?

(I mean, by the same logic of specifying a negative, why don't I need to specify this: $-4 + (+4) = 0$ instead of assuming the one without a specific direction is always a positive?)

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I have no idea why so many people bother to write tag wikis and excerpts if no one is going to read them... – Asaf Karagila Aug 28 '12 at 11:25
"If I did the same but making the first number a negative" – but that wouldn't be -4+4 as you have written, it would be -4-4 (which indeed is -8). – Hans Lundmark Aug 28 '12 at 12:33
up vote 3 down vote accepted

Maybe this way of thinking about it will help you. Please correct me if this is not at all what you wanted to know. It seems to me as if you think about addition as "going in the same direction". If you have $(-4)$ and you add $4$, you seem to expect the resulting number is "more negative".

Think about it this way. Positive numbers represent balls, so a $4$ represents $4$ balls: $\bullet\bullet\bullet\bullet$

Negative numbers instead represent "holes" where the balls can fall in. Thus $-4$ is something like $\circ\circ\circ\circ$

Addition is then the operation that simply puts together everything you have. So $2+3$ would be $\bullet\bullet+\bullet\bullet\bullet=\bullet\bullet\bullet\bullet\bullet$

Similarly, $2+(-3)$ would be $\bullet\bullet+\circ\circ\circ=\bullet\bullet\circ\circ\circ$

But: Whenever a hole and a ball meet, the ball falls in the hole and clogs it, so the ball is not there anymore but the hole isn't either. Thus $\bullet\bullet\circ\circ\circ$ simplifies to $\circ$. Thus $2+(-3)=-1$. Nothing changes if you swap the numbers: $-3+2=-1$, because you add the same numbers of holes and balls.

So, if you want to have $-8$, that is $8$ holes, you would have to add $4$ holes and another $4$ holes. This means you need to sum $-4$ and $-4$.

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Nice explanation, thank you. I ended with the following logic so that it fits what I wanted: $-(4 + 4) = 8$ (determine the direction and making the operations within it). – jackJoe Aug 29 '12 at 8:08
@jackJoe I don't see how you got that. -(4+4)=-8, since you first add 4 and 4 together. According to Gergor's suggestion, I think that would mean you would first have 4 balls and 4 other balls which you added together, and then you would remove all the colored in parts, leaving you with 8 holes or -8. – Doug Spoonwood Aug 29 '12 at 11:46
@DougSpoonwood as I said, first I determine the direction (in this case negative) and make all operations there, the result will be affected by the direction specified and I don't need to tell that every value belongs to the negative side, since I already have that specification at the beginning. In this case, and using Gregor's example, you can view it as using only positive (filled) balls and in the end removing their fill (I like to see this as already working with blank balls, because I specified that at the beginning of the operation). – jackJoe Aug 29 '12 at 14:30
@jackJoe: Did you miss a minus sign in your first comment (actually meaning $-(4+4)=-8$)? If so, that may have caused the confusion. – Gregor Bruns Aug 29 '12 at 14:37
@GregorBruns spot on!, I forgot it, I meant it to be $−(4+4)=−8$. Thanks for that, I wasn't understanding why Doug commented but that answers it. Unfortunately I can't edit the first comment, but this explains it. – jackJoe Aug 29 '12 at 14:40

From the point of view of Group Theory:

Consider the Group $G$ with the set of integers and a binary operation $*$. There exists an identity element, $e$, with the property:

$$ \forall a \in G, a *e = e *a = a $$

Now, each element $a$ in G must have an unique inverse:

$$ \forall a \in G, \exists{b}, b*a = a*b = e $$

Write the inverse of an element $a$ as $-a$ (since its unique). Replace $*$ by $+$, and $e$ by $0$. We have:

$$ \forall a \in G, \exists{-a}, (-a) + a = a + (-a) = 0 $$

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If you want to add a positive number to a negative number without specifying the negative number as negative, I would suggest that you use entirely different symbols for negative numbers than positive numbers. In other words, you'll need to use something unlike Hindu-Arabic numerals, since 4 and -4 resemble each other too much (from one perspective one can even take the perspective that -4 doesn't represent a number, but rather the negative function "-" operating on 4 which equals what we call "negative four", but I digress). The following consists of one possibility along those lines, though many different symbolizations will work:

Let zero get represented by m, one by n, two by o, ..., thriteen by z. Then for fourteen we would write nm, for fifteen nn, sixteen by no, and so on.

Let negative one by l, negative two by k, ..., negative twelve by a, negative thirteen by lm, negative fourteen by ll, negative fifteen by lk, and so on.

So, then we can write q+j=0 for "four plus negative four equals zero" and p+k=n for "three plus negative two equals one".

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This is a nice idea, thanks. – jackJoe Aug 29 '12 at 8:11

There is no difference between subtracting a positive number and adding a negative number. In fact, subtraction is defined as the addition of the number's negative.

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In fact, subtraction can often be defined as the addition of the negative. (One case where it can't is when we're considering only $\mathbb N$). – Henning Makholm Aug 28 '12 at 13:29
The OP was obviously considering $\mathbb{Z}$, and to my knowledge one would define subtraction on $\mathbb{N}$ by "going through" the group inverses in $\mathbb{Z}$ first. Is there another way? – nbubis Aug 28 '12 at 13:48
One very common alternative is to define "$a-b$" to mean "the unique solution to $b+x=a$, if such a thing exists". I'm pretty sure that's how I learned it (if in less lofty language) in grammar school. It is easy enough to show that this is equivalent to adding an additive inverse (if enough such inverses exist!), but that doesn't make it the same definition. Once you're sophisticated enough to grasp abstract reasoning about inverses, your definition is more economical, but at this question's fairly basic level, the more elementary $b+x=a$ definition is more appropriate. – Henning Makholm Aug 28 '12 at 15:37

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