Walking Randomly

[Cross-posted at Crossed Streams]

How’s this for rationale for the influence of negative factors on the sign of a product (for the integers at least)?

If -a = 0 – a, and

if a * b = 0 + a + a + a + … + a wherein a appears b times as an addend,

then -a * b = (0 – a) * b = 0 + (0 – a) + (0 – a) + (0 – a) + … (0 – a) wherein 0 – a appears b times as an addend.

Hence -a * b = 0 – a – a – a – … – a wherein a appears b times as a subtrahend,

and thus -a * b = -(ab).

However if a * -b = 0 – a – a – a – … – a wherein a appears b times as a subtrahend,

then -a * -b = 0 – (0 – a) – (0 – a) – (0 – a) – … (0 – a) wherein 0 – a appears b times as a subtrahend,

then -a * -b = 0 + a + a + a + … + a wherein a appears b times as an addend,

and thus -a * -b = ab.

I think the distributive example is a good one, because the real, deep reason that negatives multiply to positives is that our very nice rules about positive numbers wouldn’t work with negatives unless it is the case that (-3)x(-4)=12, say.

Here’s another way to argue it though:

(-3)x(4)

is equal to

0-4-4-4 = -12.

That is, if multiplying by a positive integer is repeated addition, then multiplying by a negative integer is repeated subtraction. So

(-3)x(-4) = 0 – (-4) – (-4) – (-4) = 0+4+4+4 = 12

I would illustrate the distributive law argument with different numbers though:

(-5)x(10-10) = -50 + 50 = 0

A modification of the vector explanation should probably hold…

All numbers can be represented using Euler notation. Let’s start with a number, a*e^(j*0), that sits on the right hand side of the x-axis in the complex plane. If this number is now multiplied by -1, i.e. e^(j*pi), we end up with a number,

b= a*e^(j*0)*e^(j*pi)=a*e^(j*pi)

‘b’ then sits on the left hand side of the x-axis, being, as it is, the result of rotating ‘a’ by 180 degress. Now consider a further multiplicationn by -1 resulting in,

c=b*e^(j*pi)=a*e^(j*2pi)

This further rotation by 180 degrees causes ‘c’ to coincide with the location of ‘a’, thus proving that two minuses make a plus.

I think like everything in math it is just a convention with no deeper meaning. I am (almost) sure that the whole edifice could also be build on e.g. (+)*(+)=(-) and (-)*(-)=(-)

It just comes natural to us because we are used to it – but that means nothing – like e.g. the decimal system ( digital system) or standard analysis ( non-standard analysis).

The only thing you have to explain is “-1 times -1 = 1″ and the general rule follows easily also for a non-mathematician.

Explain that multiplication of a number by -1 is defined as reflection around zero (in the real number line if you don’t want to mention complex numbers). It is then obvious that if you do this twice to the number 1 (i.e., if you compute -1 times -1 times 1) you get back to where you started, i.e., to 1.

The fact that this definition makes the distributive rules work can then be taken as a lucky coincidence by non-mathematicians.

My high school algebra teacher had a pretty intuitive answer for this. According to him, multiplication was adding the number a to itself b times. Multiplying a negative a by a positive b was like adding the negative number a to itself b times. Alternatively, it was like adding b to itself a times, and then putting a negative out front. So, from the alternate view, adding another negative sign simply canceled out the original, resulting in a positive.

The distributive law is of course _why_ we require a negative times a negative to be positive (and it’s what I use to explain it, but I’m not at all convinced that it’s an accessible explanation).

I’ve heard the “debt” explanation used. Say I am given 2 tenners. My net worth changes by 2*10 (=20). Say I have 2 tenners and I burn them. My net worth changes by -2*10 (=-20). Say I take on two debts of 10 pounds each (example, someone buys me a pint in london, twice, and I owe them back for it): my net worth changes by 2*-10 (=-20). A debt is like burning money. Say I cancel my two debts. My net worth changes by -2*-10 (=20): I get richer!

I have to say I’m not a great fan of this explanation either, but I’ve seen it used.

If your audience accepts that multiplication is modeled by rectangles, the side lengths being the factors and the area being the product, then it is fairly easy to draw a picture. Set the side lengths to something like (a-b) and (c-d). You will end up drawing several rectangles (e.g. ac, a(c-d), etc). The area of the (-b)(-d) rectangle will be forced to be positive for the resulting equation relating the area of (a-b)(c-d) and ac to make sense. This is another visual demonstration of the distributive law, but might appeal better to intuition.

When you tire of this on, you’ll want to tackle the “why is dividing by a fraction the same as multiplying by its reciprocal?” question…

On the way to the chippy I was thinking: to work out “(minus something) times (something)” do the “times” without the minus, then minus the result, so it’s the same as “minus (something times something)”. We use that rule _again_ when both somethings have a minus. So “(minus something) times (minus something)” is “minus (something times (minus something))” which is (same rule again) “minus minus (something times something)”. “minus minus thing” is the same as “thing”. Why? (actually I think most people capable of posing the original question are probably happy to accept that minus minus thing is thing) Because of the rule we learnt when doing subtraction of negative numbers: “minus minus thing” is “0 – minus thing” which is “0 + thing” which is “thing”.

Algebraically: (-a)*(-b) == -(a*(-b)) == -((-b)*a) == -(-(b*a)) = b*a == a*b

Basically, it is an arbitrary convention. I read a whole book on this subject a few years ago. The author demonstrated that you could construct a coherent mathematics without this convention. He sketched out the beginnings of a system in which minus times minus is still minus.

If I can recall the name of the book, I’ll post it. But I read it a few years ago and didn’t keep it.

Thanks to the rather long memory of amazon dot com, I was able to find the book I was thinking of:

“Negative Math” by Alberto A. Martinez

http://www.goodreads.com/book/show/547416

Math teacher here.
There’s no such thing as minus. Only “opposites”

Say these out loud:

I have 8 dollars.
I have the opposite of 8 dollars. (what do you have?)
I have the opposite of the opposite of 8 dollars. (what do you have?)

1) 8
2) – 8
3) – - 8, which we know as “8″

The negative numbers are the concept of an opposite to the positives. Their very origin is in the question: “what number can I add to 8 to get zero?”

Your initial question is doomed by vocabulary to begin with: take “Minus times a minus is a plus”.

do the following replacements:
minus -> opposite
times -> of
plus -> normal

Surely we can agree that the opposite of an opposite is back to normal.

(and technically the opposite is called the “additive inverse” since there are other kinds of opposites)

Other kinds like — multiplicative inverses!

Divide 100 tiles into two piles. (100 / 2)
Didn’t you just take “half of 100?” (1/2 * 100) remember, of = multiply.

Divide 100 tiles into four piles. (quarter of 100?)

Divide 100 tiles into 1 pile.

Divide 100 into … 1/2 pile. (how many more tiles do you need?)

Each time we are taking our 100 tiles and making them into a newly sized pile. We are given the number of piles to make, and we make them. When we need to make half a pile, what would the other half “be” ?

A last one to think about: divide 100 into 2/3 of a pile.

I think you could just see a – (minus) as a sign-change, and for multiplication we need to add the number of sign changes: 2 sign-changes off 1*1 makes 1*1….

Hey, i always approach this from the angle of direction, and specifically with videos of people walking forwards and backwards. I’ve uploaded the videos i use to my blog, john cleese and michael jackson. The kids in my class really seem to grasp the concept pretty much intuitively from spending fifteen minutes or so mucking around with these clips (we tend to finish on powerpoint being presented with them all simultaneously).
Hope this helps?

Ryan

There are several good explanations here: http://mathforum.org/dr.math/faq/faq.negxneg.html Ask Dr. Math is a great resource that I often refer to if I need to explain something mathematical to someone else.

I wouldn’t say that it’s a proof. It’s just verifying that the axioms hold.

Ray

I think that, despite our nice rules for the simple arithmetic operations, there are more to their personalities than we are willing to admit. In particular, although multiplication is commutative — and there’s no doubt of that — what’s being meant there is the binary or two-argument form of multiplication.

If, instead, a*b is seen through the eyes of Frege and Curry (logicians, but no matter), it might be parsed as [*(a)](b). Interpreted, the “*(a)” generates a function given the value of “a” which multiplies its argument by the value of “a”. Fed the value of “b” it produces the value of “a*b”. But “*(a)” might be considered the class of functions which “times by the value of ‘a’”.

Now, consider “*(-1)”. What does it do? Well, one thing it does is to map the positive real line into its negative, and the negative into the positive, leaving the exact zero where it belongs. If the value of “a” is positive, then “*(-a)” reflects and scales positive numbers into negatives, and negatives into positives.

So, one of my answers would be the reason why if the values of “a” and “b” are both positive, then “(-a)*(-b)” is positive is because of this latent characteristic of the binary operation of multiplication as a signed, scaling transformer.

There’s not really anything to explain. It’s a convention, just as it’s a convention that we accept Euclid’s fifth postulate. It’s a necessary convention so that the accepted rules of the numbering system follow what we would take as physical intuition (see many of the above arguments).

In the words of Abstract Algebra, multiplication by a negative reciprocal gives unity (the existence of an inverse). This is just a fancy way of stating that we want the real numbers, along with the operation of multiplication gives a group.

But no, if you’re a formalist, then there’s no real reason why -1*-1 = 1. It’s a matter of convention and a matter of making everything symmetric and pretty.

I think the following is a rigour explanation using only the axioms of a field:

Every element, a, in a field F has a additive inverse, denoted (-a), that is

a + (-a) = 0.

Now the additive inverse of (-a) is (-(-a)), so

(-a) + (-(-a)) = 0.

By adding a on both sides we get

a + (-a) + (-(-a)) = a
0 + (-(-a)) = a
(-(-a)) = a

Hence by the commutative and associative axioms we get

(-a)(-b) = (-1*a)(-1*b) = (-(-(a*b)))= a*b.

Best regards
Jonathan

Just realized that this of course doesn’t work for integers, since they do not form a field. An easy solution is of course to consider integers as rational numbers with 1 in the denominator. If we don’t want to consider rational numbers, you must notice that integers under addition is a abelian group and under multiplication a commutative monoid.

1 * a = a, – – ”here ‘a’ is any number”
hence, 1 * (-1) = (-1)
also -1 * a = -a, – -”here ‘a’ is any number”
hence, -1 * (-1) = – (-1)
Negative of +1 = -1
negative of (-1) = 1, because a number plus the negative of the
number adds up to zero. i.e. -1 + x = 0, that means x=1
From above, (-1) * (-1) = – (-1) = negative of (-1) = 1
That shows (minus * minus ) yields plus

Another Soln
This is not a proof, however a demonstration for a curious kid why
negative times negative is positive. It may be helpful to think that
when dealing with the numbers ‘-’ or ‘+’ signs are identical to ‘-1
times’ or ‘+1 times’. Here is an example:
-5 = ‘-1 times’ 5
A number plus a negative number can result zero. Lets take the
number ‘1’ that is ‘+1’, then take ‘-1’. It could have been any other
number. The MAIN assumption is:
1 -1 = 0
Without losing the effect we can times the equation by ‘-’ or ‘-1’ then:
-1 -(-1) = 0
Since both equations are the same, that means:
-1 -(-1) = 1 -1
that means:
-(-1) = 1
it shows:
‘-’ times ‘-’ is plus
—————————————————-
This approach is also helpful to show why ‘+’ times ‘-’ is ‘-’
Let us safely assume ‘1’, which is ‘+1’, times a number results in the
number, i.e. ‘1 times’ 10 is 10. Then ‘1 times’ ‘-1’ is ‘-1’ . That is:
‘+1 times’ -1 = -1
That shows: ‘-’ times ‘+’ results ‘-’

It’s definitely not arbitrary and not just a convention. That is to say that changing it would be like stating 3 = 4.

To demonstrate it numerically rather than conceptually (ie through some real world application) I would first talk about why a negative divided by a negative equals a positive. I think most people more easily understand that any non zero value divided by itself equals 1. That is, -x / -x = 1 where x > 0 and thus a negative divided by a negative is a positive. If from that they can’t see that a negative times a negative is also a positive then an example may be in order.

I would offer the following from the above:
1 = (-1 / -1)

Knowing that a positive divided or multiplied by a negative is a negative also establishes that:
-1 = (1 / -1)
-1 = (1 * -1)

With these it is easy to show that a negative times a negative must be positive:
-1 = -1
-1 = (1 / -1)
-1 * (-1) = -1 * (1 / -1)
-1 * -1 = ((-1 * 1) / -1)
-1 * -1 = (-1 / -1)
-1 * -1 = 1

This isn’t a proof of course, but it should be enough to illustrate why our mathematical system requires that a negative times a negative equals a positive. The point here being that this isn’t dependent on any particular real world application, but is an inherent truth in mathematics.

Alternately, it might be better to demonstrate why it can’t be otherwise. For example:

-5 + 5 = 0
-1 * (-5 + 5) = -1 * (0)
(-1 * -5) + (-1 * 5) = 0

Now, if a negative times a negative is a negative then (-1 * -5) would equal -5 so:
(-1 * -5) + (-1 * 5) = 0
(-5) + (-5) = 0
-10 = 0
-10 is not 0 obviously so that isn’t possible mathematically.

However, if a negative times a negative is a positive then (-1 * -5) equals 5 so:
(-1 * -5) + (-1 * 5) = 0
(5) + (-5) = 0
5 – 5 = 0
0 = 0

Consequently, the fact that a negative times a negative is a positive isn’t just some convention for the sake of convenience, it’s absolutely necessary and true.

I also wanted to point out that the original example given doesn’t constitute a formal proof. It states that for the equation, a*(b+c) = a*b + a*c, the values a = 1, b = -1, c = 1 satisfy it such that 1 = -1 * -1. Just plugging in a specific set of values to an equation won’t prove anything except the behavior of the equation with those specific values.

I’ll demonstrate with an example. Suppose I want to prove that the sum of any number with itself is equal to the product of that number by itself. That is to say:

x + x = x * x

Using the above technique I could say let x = 2 so:

(2) + (2) = (2) * (2)
4 = 4

So showing that the equation is true for x = 2 is misleading. It doesn’t indicate anything about other values of x so it does not prove that the equation is true for all values of x.

Now back to the original problem. Stating that a negative times a negative equals a positive could be expressed as:

-x * -y = z where x, y and z are real numbers greater than 0

It’s not enough to provide a single set of values that satisfy the equation. A formal proof has to prove that for all positive values of x and y, there exists a positive value of z for which the equation is true. I suspect someone has developed a formal proof for this, but I don’t have an example to show and it’s certainly beyond my skills.

That’s not to say that the example in the original post isn’t good for explanations, but it’s just not a proof.