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Why (-1)(-1)=1

  1. Jun 8, 2006 #1
    So I'm teaching a college algebra course this summer and I'm trying to go nice and slow and give good reasons for most things. One thing I can't seem to justify is why (-1)(-1)=1. I can do it with equivalence class (pairs of natural numbers) or the following little proof but I'd like to have a good grounded explanation (like you can use holes and piles of dirt for adding negative and positive numbers together).

    1-1=0
    -1(1-1)=0
    (-1)(1)+(-1)(-1)=0
    (-1)(-1)=1

    So does anyone have a nice meat and potato way of explaining it?
     
    Last edited: Jun 8, 2006
  2. jcsd
  3. Jun 8, 2006 #2

    matt grime

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    But that is a meat and potatoes way of doing it. This is college algebra, if they can't handle simple deductive reasoning and implication then something is very wrong.
     
  4. Jun 8, 2006 #3
    Actually, I think that's the status quo - at least here in the States.
     
  5. Jun 8, 2006 #4

    arildno

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    I would present this wholly without the operation of subtraction, and a bit more fully:

    1+(-1)=0
    (-1)*1+(-1)*(-1)=(-1)*0
    (-1)+(-1)*(-1)=0
    1+(-1)+(-1)*(-1)=1+0
    0+(-1)*(-1)=1
    (-1)*(-1)=1
     
  6. Jun 8, 2006 #5
    Here is the way I look at it (while not necessarily rigorously mathematical, it makes sense):

    The friend of my friend is my friend (+)(+) = +
    The friend of my enemy is my enemy (+)(-) = -
    The enemy of my friend is my enemy (-)(+) = -
    The enemy of my enemy is my friend (-)(-) = +

    (I'm not sure it can get any more meat and potatoes than this)
     
  7. Jun 9, 2006 #6
    Ha, I like Motai's version. MattGrime, I didn't say they couldn't handle it, but its not very satisfying. Sometimes a proof isn't proof in the sense that it may prove the theorem valid but give no real insight into why. With adding numbers of different signs you can illustrate the idea with holes and piles, which makes good sense. Mutliplication of a positive number by a negative number you can write in terms of adding up a lot of holes. But the product of two negative numbers reduces to the question of (-1)(-1) and except for Motai's cute post I've yet to find a reason as to why. I mean, at somepoint when human beings possesed multiplication and negative numbers they must have thought about what (-1)(-1) was and came up with an answer. Given that a great deal of math historically is well motivated and not particularly formal, I bet that answer made use of some grounded concept, I'm curious as to what it was.

    But of course, thanks for all your replies.
     
  8. Jun 9, 2006 #7

    matt grime

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    These people are at college; they don't need their hand holding. The sooner they meet formal reasoning, and the sooner we stop treating them like kids, the better.

    Incidentally, why must 'they' have wanted to multiply -1 by -1? They might well have treated it as something one can't do. Just like they treated -1 as something you can't square root. An idea that still pervades students today and causes all kinds of problems, exactly because we choose to overly ground things in 'real' terms. It is amazing the number people who will no accept i as the square root of -1 simply because they were told that you can't square root negative numbers by their high school teacher. And then when they think about dividing by zero they come up against a whole new problem of 'well I was told you can't square root -1 and you can, now do I believe my teacher who said I can't divide by zero?'

    Anyway, yes, maths was very informal up until the mid 19th C. Which makes it entirely plausible that they did not have a consistent definition of multiplication of negative numbers at all, just as there were inconsistent uses of radicals of negative numbers, just as there was no such thing as a 'function' in the modern sense.
     
  9. Jun 9, 2006 #8

    shmoe

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    A minor change of view from the standard 'meat and potatoes', showing (-1)*x+x=0 tells you (-1)*x is the additive inverse of x. Then (-1)*(-1) is the additive inverse of (-1), the additive inverse of -1 is 1 (need to show inverses are unique). In other words, multiplying by -1 gets you the additive inverse, multiply by -1 again gets you another additive inverse again, additive inverse of an additive inverse is what you started with.

    If someone understands complex multiplication amounts to adding the arguments of the terms, multiplying by -1 is a rotation by pi degrees. Rotate by pi degrees twice.

    enemy of my enemy is not always my friend, some people are just jerks and enemies of everyone. Using vague colloquial sayings to try to justify maths to college students is horrid (as humor no problem)
     
  10. Jun 9, 2006 #9

    DaveC426913

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    The effect of multiplying by -1 is that you are left with the original number, except the direction (sign) is reversed.

    So, what happens if you reverse direction twice? You're back where you started: 1 (unity).
     
  11. Jun 9, 2006 #10
    1 is the multiplicative identity. Anything multiplied by 1 remains the same.
    1(-1) = -1
     
  12. Jun 9, 2006 #11

    0rthodontist

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    One way to explain the algebra from an intuitive perspective is by viewing multiplication of integers as repeated addition, so when you multiply -1 by 5, you view it as adding up -1 five times. When you multiply -1 by -1, you want to add up -1 negative one times. This means that if you add -1 ONE MORE time, you will have added up -1 a total of zero times to get 0. In other words, that says (-1 added up -1 times) + (-1 added up once) = (-1 + 1)*(-1), so (-1 * -1) (looking on the left) is the number that when added to -1 equals 0, so it is 1.
     
  13. Jun 9, 2006 #12

    es

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    Personally, I think if they don't get DaveC's argument then you're going to have much bigger problems on your hands.

    But if it were me, I would present Arildno's demonstration and conclude with if a negative multiplied by a negative is not defined to be positive then a multiplication operation can break the addition operator in a similar way that a divide by zero can. Since we would like to avoid creating special cases (math is complicated enough right?) and because there is a sound geometrical interpretation, any other definition seems to create a lot of extra work with no benfit. But, if they would like to create thier own algebra which does so, then, of course, they are free to. That particular algebra is not what you will be studing in this class though. :)
     
  14. Jun 11, 2006 #13
    Courant simply feels that such things can not really be proved because they are built into the system. Courant says that (-1)(-1) can be only -1 or 1. The first choice is unsuccessful. (Remember absolute value of (ab)=absolute value (a) times absolute value (b).)

    (Richard Courant founded the Courant Institue at New York University in 1935. From the start, as I understand it, Courant was very interested in practical, applied mathematics, and somewhat critical of the direction of pure mathematics. As a result the institue was heavily subsidized by the government which wanted answers to practical questions.)
     
    Last edited: Jun 11, 2006
  15. Jun 12, 2006 #14
    Thanks for all your replies. I think math can be made relatively grounded for the majority of students. Formal reasoning at this level obscures a lot of the points. As math students we put up with stuff like universal mapping problems and the like because we know that in the end thats the better way. The formalism and succintness of the definitions makes sense once you work on it. These kids are not going to take another math class. They're not going to "work through it" to see why certain things work and I don't blame them. So while I do show them formalities I also like to illustrate them with motivation from the world around us. Once again thanks for all your replies.
     
  16. Jun 12, 2006 #15

    arildno

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    I cannot agree, as you do, that hand-waving and teaching nonsense (like motai's enemy "argument"*) is to be considered a virtue.
    Math CAN be hard, that only means the students must discipline themselves in understanding it.


    *Why, for example, does the operation of multiplication provide a model for the analysis of friendships??
    A somewhat less worse analogy would be in connection with the proposition(-(-1))=1, which is deducible solely from the properties of addition.
     
    Last edited: Jun 12, 2006
  17. Jun 12, 2006 #16
    I like Courant's approach; any other is superficial in comparison. (-1)(-1) = 1 by definition. We define it such to preserve the distributive law. Following Courant, consider -1*(1-1). Now, if we apply the law, (-1)(1)+(-1)(-1) = -1*(1-1). (-1)(1) and -1*(1-1) can be calculated ordinarily, so: -1 + (-1)(-1) = 0. Then (-1)(-1) must equal 1. That is, by imposing the distributive law on negative integers, we are forced to this conclusion.
     
    Last edited: Jun 12, 2006
  18. Jun 12, 2006 #17

    matt grime

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    But you see you haven't really defined it to be 1. You have deduced that it is 1 in order for the integers to satisfy the definition of a ring.
     
  19. Jun 13, 2006 #18

    t!m

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    Ian Stewart presents two suggestions to explain this to students, the first is similar to yours/arildno's explanation. The other is more 'meat and potatoes:'

    He says, think of numbers as money in a bank. A positive number is money you have, a negative number is money you owe. So if you owe, for example, 3 payments of $5, then this is 3 (-5) = -15, which is a correct $15 owed to the bank. Now suppose the bank 'forgives' 3 owed payments of $5, then this is (-3)(-5), which only makes sense to be positive 15, the equivalent of a $15 gain. If it were -15, then you would still owe the bank $15 which hardly makes sense.

    I also think it is worth making a distinction that 'college algebra,' at least how it is often used in the states, is a very basic math class, often intended for non-science/math majors.
     
  20. Jun 13, 2006 #19
    I have another explanation, one that I've worked out from several discussions on this topic with (who else!) my mother.

    I'm going to adopt a strange notation for numbers:

    (r, [tex]\Theta[/tex])

    where r is the modulus of the number n, and where [tex]\Theta[/tex] is how much we have to rotate ourselves about on the real number line about the origin, and as measured from the positive direction, i.e. a positive number has [tex]\Theta = 0[/tex] and negative numbers have [tex]\Theta = 180 \degree[/tex]. However, instead of writing [tex]\Theta = 180 \degree[/tex], from now on I'll measure angles as "revolutions" to avoid ugly numbers, so a negative number has [tex]\Theta = 0.5[/tex].

    Now there are an infinity of ways to express, say, 9:
    (9, 0), (9, 1), (9, 42), (9, -27)

    or -7:
    (7, 0.5), (7, 8.5), (7, -2.5)

    which all get us to the same place on the number line.

    This was actually how the Greeks thought about numbers, in terms of their direction and their magnitude. (OMG VECTOR!) Not only that, but the Greeks defined multiplication too. So I'll use the Greek method of multiplying the numbers (p,q) and (r,s):
    1. Start with the unit number, (1, 0).
    2. "Scale" the magnitude of this number up p times, and rotate it by q.
    3. "Scale" this magnitude of this number up r times, and then rotate it by s.

    For example, multiply -4 and 3, which in our system is (4, 0.5) and (3, 0). Under our system, we take the unit number (1, 0), scale it up 4 times, and rotate it half a turn. Now scale this up 3 times, and rotate it zero turns. We get (12, 0.5), which is of course the expected result: -12.

    It's now obvious under this definition of numbers and multiplication why -1*-1 = 1, because it's equivalent to doing half a turn to the negative side of the number line, and then half a turn again, so I end up on the positive side! Our ideas of numbers and multiplication came from the Greeks, so I'm sure you'll find that these definitions are consistent with what we normally think of as multiplication. So the answer to your question is: because the Greeks said so! (And for the numerous reasons mentioned above.)

    This idea is similar to the idea to the multiplication of vectors in the Argand plane, but there, we don't limit ourselfs to multiples of one-half for the rotation.

    This is a very weird explanation, I know.

    [edit]Of course, shmoe's second argument is a lot like this one.[/edit]
     
    Last edited: Jun 13, 2006
  21. Jun 16, 2006 #20
    I stopped reading this thread up until I saw this. Well said man, now even I understand why (-1)(-1)=1 :biggrin:
     
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