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Negative numbers

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Do you understand why taking away -1 from a number is the same as adding 1? (I remove some of your debt means you are wealthier as you owe less)

 

[latex]x-(-y)=x+y[/latex]

 

And that multiplication can be seen as adding a certain number to itself a number of times

 

[latex]3x=x+x+x[/latex]

 

multiplication by a negative number can be seen as repeated subtracting

 

[latex]-3x = -x-x-x[/latex]

 

So multiplication of two negative numbers is repeatedly subtracting a minus

 

[latex]-3(-x) = -(-x)-(-x)-(-x) = +x+x+x=3x[/latex]

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I have had some luck with the baffled, after trying the standard (as above) first, by describing the negative sign as an instruction to rotate counterclockwise 180 - to "flip" - the indicated. So multiplication by a negative 1 would be first the standard position on the line, then flipped around the zero. So multiplication buy two negatives would be two flips - back where you started.

 

This has the advantage of not using the negative sign in the explanation of the negative sign or operations involving it, agreeing with the high class development of the number system and groups etc, being intuitively generalized to handling fractions and functions and other objects not readily comprehended by subtraction, and leading naturally into the square roots of negative one (as rotations by 90) eventually.

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What helped me to remember this was that when multiplying two numbers with the same direction on a number line you treat the problem when dealing with absolute values.

 

EX:

(-3)*(-3) = |-3| * |-3| = |3| * |3| = 3*3

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Why does a negitive times a negitive equal a positive

 

The short answer is that, because a positive number of a negative amount equals a negative amount, such as (positive) three –10's equals –30, then a negative number of negative amounts must equal the opposite — a positive number — such as negative three –10's equals +30.

 

Another way of looking at it goes like this —

 

8 × 8 = 64

 

which is

 

(10 – 2) × (10 – 2)

 

which is

 

10 × 10 + 10 × –2 + –2 × 10 + –2 × –2

 

which is

 

100 + –20 + –20 + 4

 

which is

 

100 – 20 – 20 + 4

 

which is

 

64

 

(and not 100 – 20 – 20 – 4 which is 56)

 

 

 

You can do the same for 2 × 2 as being (10 – 8) × (10 – 8) and get the answer of 4.

 

 

 

I hope that one or both of these explanations helps.

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The short answer is that, because a positive number of a negative amount equals a negative amount, such as (positive) three –10's equals –30, then a negative number of negative amounts must equal the opposite — a positive number — such as negative three –10's equals +30.

 

Another way of looking at it goes like this —

 

8 × 8 = 64

 

which is

 

(10 – 2) × (10 – 2)

 

which is

 

10 × 10 + 10 × –2 + –2 × 10 + –2 × –2

 

which is

 

100 + –20 + –20 + 4

 

which is

 

100 – 20 – 20 + 4

 

which is

 

64

 

(and not 100 – 20 – 20 – 4 which is 56)

 

 

 

You can do the same for 2 × 2 as being (10 – 8) × (10 – 8) and get the answer of 4.

 

 

 

I hope that one or both of these explanations helps.

This seems a bit more complicated than other explanations. This is especially because most people don't really use the distributive property of multiplication without it being algebraic.

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The reason that a negative times a negative equals a positive is that it is a logical consequence of the basic properties of the integers, which are known as the ring axioms. You can read (way too much) about them here ...

 

http://en.wikipedia.org/wiki/Ring_(mathematics)

 

I have to apologize in advance for that link because it's way too technical and fails to provide a simplified introduction to the subject.

 

But basically we define a "ring" as any mathematical system in which we can add, subtract, and multiply; and that addition and multiplication are related by the distributive law

 

a(b + c) = ab + ac

 

valid for all objects a, b, and c, as long as they are members of the system in which we're interested. [From now on I'll use the word "numbers" to refer to the members of our system; but keep in mind that these are not necessarily the familiar integers, but rather the members of any system that obeys these rules.]

 

The ring axioms say that there's a magic number called 0 that has the property that

 

0 + n = n for all numbers n

 

and that for each number n there is an "additive inverse" called -n, whose defining property is that

 

= n + (-n) = 0.

 

There's also a magic number called 1 with the property that 1 * n = n for all numbers n.

 

All these things are true about the everyday integers, so we can use these properties to prove things about the integers. And it turns out to be the distributive property that is crucial.

 

Here's a formal derivation.

 

We want to find out what is (-1)(-1).

 

The way I'm going to do this is to evaluate the quantity (-1)(-1) - 1 and show that it must be zero. This will then prove that (-1)(-1) = 1.

 

I'll use C++ style comments ('//') to provide hopefully helpful commentary.

 

(-1)(-1) - 1 // Expression we want to evaluate.

 

= (-1)(-1) + (-1) // Because "subtraction" is actually adding the additive inverse.

 

= (-1)(-1) + (-1)(1) // Multiplying something by 1 doesn't change it.

 

= (-1)(-1 + 1) // This is the distributive law, which says that a(b + c) = ab + ac.

 

= (-1)(0) // Because -1 + 1 = 0 by definition. -1 is the additive inverse of 1.

 

= 0 // Anything times zero is zero. That's actually a consequence

// of the ring axioms and requires proof, which which we'll assume.

 

We just derived a logically certain chain of equality between (-1)(-1) - 1 and 0. If we add 1 to both sides of the equation, we get (-1)(-1) = 1.

 

Conclusion: The ring axioms logically imply that (-1)(-1) = 1. One is still free to impute metaphysical significance to all this, or to try to "figure it out." Mathematicians prefer abstraction. We write down the properties of the thing being studied; and then we derive logical consequences.

 

This methodology of abstraction often provides structural insight. We just discovered that the reason a negative times a negative is positive is simply that it's a logical consequence of the distributive law. Once you accept the distributive law, you have no other choice. That's interesting!

 

By the way we should complete the job by proving the general case (-a)(-b) = ab. That's left to the reader of course smile.png

Edited by HalfWit

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Most of the people asking the question are looking for a physical, intuitve way to comprehend the situation. The basic problem is that "negative" has been introduced to them as a direction on a number line, the degrees below zero on a thermometer, etc. That intuition does not work well for thinking about multiplication of negative numbers, neither is it well conjoined with the actual theoretical basis of that extension of the integers etc.

 

Proofs don't really meet their needs.

Edited by overtone

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Most of the people asking the question are looking for a physical, intuitve way to comprehend the situation. .............

Proofs don't really meet their needs.

 

 

In a math forum, I find this a curious comment. But leave that, here's the general case that HalfWit asked for

Consider the sum of products [math]ab+a(-b)+(-a)(-b)[/math].

 

By the associative law write this as [math]ab+[a(-b)+(-a)(-b)][/math] and by the right distributive law write this as

 

[math]ab+[a+(-a)](-b) = ab+0(-b) = ab[/math] from the multiplicative property of zero

 

On the other hand, using the exact same rules

 

[math]ab+a(-b)+(-a)(-b) = [ab+a(-b)]+(-a)(-b)][/math]

 

[math]=a[b+(-b)]+(-a)(-b) = a0+(-a)(-b)=(-a)(-b)[/math] (although you use the left distributive law here)

 

Which implies that [math]ab = (-a)(-b)[/math] as desired.

 

As a further generalization assume the above applies to the ring [math]\mathbb{Z}[/math] of integers. Let us be a bit naughty and call [math]0[/math] a proper sub-ring.

 

Then from ring theory we have that an ideal [math]I[/math] of any ring [math]R[/math] is a subring such that for all [math]x \in I[/math] and all [math]a \in R[/math] that

 

[math]xa \in I[/math] and [math]ax\in I[/math]. It is common to abuse notation somewhat and write this as [math]Ia =aI[/math] (actually equality here implies that we have a two-sided ideal). let's call [math]0[/math] a proper two-sided ideal of the integers (proper because the ring itself is also an ideal)

 

Then with very little modification, the above proof goes through for any ring if you replace [math]0[/math] by [math]I[/math]

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