Why - multiplied by - gives +?
Mathematical explanations?
Quote from: Scriptavolant on July 23, 2007, 06:50:41 AM
Why - multiplied by - gives +?
Mathematical explanations?
Multiplication by a negative number means a turn of the angle by pii. When you multiply a negative number by another negative number you turn the angle by 2pii (= 360 degrees = 0 degrees) and you have a positive number.
Quote from: Scriptavolant on July 23, 2007, 06:50:41 AM
Why - multiplied by - gives +?
Mathematical explanations?
(i) Assume that a negative number is a positive number multiplied by -1, i.e. assume that -A = A(-1) where A is positive. Hence, the product of two negative numbers, -J and -K, may be written as (-1)(-1)JK
(ii) Assume that (-1)(-1) = +1. Is this assumption reasonable? The alternative is that (-1)(-1) = -1. Consider the implication of this alternative assumption by looking at the product of -1 and 0. We would get:
0= (-1)(0) = (-1)(1 + -1) = (-1)(1) + (-1)(-1) = -1 + -1 = -2
Thus, we would get 0=-2, and the "distributive" property of multiplication wouldn't work for negative numbers. Thus (-1)(-1) must = +1.
(iii). It follows that (-J)(-K) = +JK
Quote from: 71 dB on July 23, 2007, 08:08:13 AM
Multiplication by a negative number means a turn of the angle by pii. When you multiply a negative number by another negative number you turn the angle by 2pii (= 360 degrees = 0 degrees) and you have a positive number.
71 db can't figure out why multiplying two negative numbers results in a positive number, and you assume he or she will understand operations in the plane of complex numbers? Maybe an intuitive explanation involving group theory or the fully anti-symmetric tensor would be preferable. :P
Try the Sherlock Holms method, the process of elimination. There are only two possibilities, when you multiply two negative numbers the result is either positive or negative. If the result were negative then this would imply that multiplying negative number by a
positive number give the same result as multiplying a negative number by a
negative number. The result of multiplication must depend on the factors, so this can't be. The other possibility must be true, the result is positive.
Quote from: Topaz on July 23, 2007, 08:15:53 AM
(i) Assume that a negative number is a positive number multiplied by -1, i.e. assume that -A = A(-1) where A is positive. Hence, the product of two negative numbers, -J and -K, may be written as (-1)(-1)JK
(ii) Assume that (-1)(-1) = +1. Is this assumption reasonable? The alternative is that (-1)(-1) = -1. Consider the implication of this alternative assumption by looking at the product of -1 and 0. We would get:
0= (-1)(0) = (-1)(1 + -1) = (-1)(1) + (-1)(-1) = -1 + -1 = -2
Thus, we would get 0=-2, and the "distributive" property of multiplication wouldn't work for negative numbers. Thus (-1)(-1) must = +1.
(iii). It follows that (-J)(-K) = +JK
That works, why introduce the -1 thing,
0 = (-J)(0) = (-J)(K-K) = (-J)(K) + (-J)(-K)
0 = (-J)(K) + (-J)(-K)
since (-J)(K) is negative, (J)(K) is positive.
Quote from: Topaz on July 23, 2007, 08:15:53 AM
(i) Assume that a negative number is a positive number multiplied by -1, i.e. assume that -A = A(-1) where A is positive. Hence, the product of two negative numbers, -J and -K, may be written as (-1)(-1)JK
(ii) Assume that (-1)(-1) = +1. Is this assumption reasonable? The alternative is that (-1)(-1) = -1. Consider the implication of this alternative assumption by looking at the product of -1 and 0. We would get:
0= (-1)(0) = (-1)(1 + -1) = (-1)(1) + (-1)(-1) = -1 + -1 = -2
Thus, we would get 0=-2, and the "distributive" property of multiplication wouldn't work for negative numbers. Thus (-1)(-1) must = +1.
(iii). It follows that (-J)(-K) = +JK
It's getting unnecessarily complicated for the situation, but I might say that the definitions of a field (i.e., a commutative division ring) answer your question
ab initio. If you define a set,
R, with necessarily two binary operations, + and *, as the real numbers, you will see that it has all the properties of a ring. If you can show that
R has a nonzero identity element (i.e., a*b=1 and a*1=a but 1/=0), then you will have a division ring. You can show this, by the way, for
R. If, furthermore, you can show that
R is commutative, then you will have shown that
R is a field. A quick check of the properties of a field, then, will show that the binary operators are associative and commutative, and it then becomes a triviality to explain how two real numbers with a negative sign multiply to a positive.
Quote from: head-case on July 23, 2007, 08:20:20 AM
Try the Sherlock Holms method, the process of elimination.
That's exactly what I set out, viz the "Sherlock Holmes" method, in time-honoured fashion, using symbolism that many folk here should be able to understand, instead of set theory which most folk, I guess, wouldn't.
Quote from: Topaz on July 23, 2007, 09:01:09 AM
That's exactly what I set out, viz the "Sherlock Holmes" method, in time-honoured fashion, using symbolism that many folk here should be able to understand, instead of set theory which most folk, I guess, wouldn't.
What? They don't teach ring theory in high school anymore? Shock, horror!
But seriously, if you attack it in the abstract realm the problem becomes a matter of simple definitions, rearranged to yield the answer - or make the answer very easy.
Quote from: bwv 1080 on July 23, 2007, 10:11:13 AM
Other have done this to, but this is simple:
if (-1)(2) = -2
then (-2)(-2) = -1(2*2)
...which is incorrect?
Finally, a Math Question, and I'm late to the party. :'(
Quote from: head-case on July 23, 2007, 10:25:57 AM
...which is incorrect?
Sorry this is what I was trying to think of:
if 2x = x + x
and 2(-x) = (-x-x) = -2x
then -2(-x) = -(-x-x) = -(-2x) = 2x
Thank you guys for your contributions! Think I'll have to concentrate deeply on your explanations to get to the end of it at last :P
Already answered, but I like this form of the argument the most:
Let a > 0 be a real number.
We know that:
a + (-a) = 0.
Let b > 0 be another real number.
Assume the distributive property holds and multiply by (-b) :
(-b)(a) + (-b)(-a) = 0.
The left term is negative.
The magnitudes of the two terms are the same.
So the only way that this equation can be true is if the right term is positive.
So then (-b)(-a) = ab.
Then we have:
-ab + ab = 0.
Another way to get more intuition:
For any real a that is not zero:
a(1/a) = 1.
if a = -7 for example:
(-7)(1/(-7)) = 1, NOT -1 so a negative multiplied by a negative must be positive for the concept of multiplicative inverse to make sense.
Up to now I've seen a lot of most interesting formal and abstract explanations I will (try to) study soon. But what if you should explain it to your child? I mean why the hell multipling - 2 peaches by - 2 peaches you get (+) 4 peaches. There must be a conceptual way to illustrate the thing in pragmatic terms.
Quote from: Scriptavolant on July 23, 2007, 01:13:57 PM
Up to now I've seen a lot of most interesting formal and abstract explanations I will (try to) study soon. But what if you should explain it to your child? I mean why the hell multipling - 2 peaches by - 2 peaches you get (+) 4 peaches. There must be a conceptual way to illustrate the thing in pragmatic terms.
for some reason i can not conceptualize -2 peaches multiply by -2 peaches.
That is a good question, and the answer is, I think: the whole idea of negative numbers is already abstract.
The way they are introduced is by the 'elevation or temperature below 0' approach, but no one explains (and none of my classmates ever asked, either, or me until later years! :D ): "why would anyone multiply two negative temperatures or elevations"?
Quote from: Scriptavolant on July 23, 2007, 01:13:57 PM
Up to now I've seen a lot of most interesting formal and abstract explanations I will (try to) study soon. But what if you should explain it to your child? I mean why the hell multipling - 2 peaches by - 2 peaches you get (+) 4 peaches. There must be a conceptual way to illustrate the thing in pragmatic terms.
A negative quantity is one that cannot, naturally, be expressed in terms of real quantities.
Quote from: Steve on July 23, 2007, 01:35:20 PM
A negative quantity is one that cannot, naturally, be expressed in terms of real quantities.
But what if you put it this way.
John has 20 peaches.
Jack has 18 peaches (-2)
And you rehearse the condition for a x positive number of times.
Quote from: Steve on July 23, 2007, 01:35:20 PM
A negative quantity is one that cannot, naturally, be expressed in terms of real quantities.
In $ it can. If I borrow $100 worth of XYZ stock and sell it then I have a position of -$100 in the stock. If the stock's return next year is -100% then my gain is
(-100)(-1) = $100
I have made $100
Quote from: MahlerTitan on July 23, 2007, 01:25:52 PM
for some reason i can not conceptualize -2 peaches multiply by -2 peaches.
Of course not, what practical problem would have an answer in units or peaches squared.
Quote from: Scriptavolant on July 23, 2007, 01:13:57 PM
Up to now I've seen a lot of most interesting formal and abstract explanations I will (try to) study soon. But what if you should explain it to your child? I mean why the hell multipling - 2 peaches by - 2 peaches you get (+) 4 peaches. There must be a conceptual way to illustrate the thing in pragmatic terms.
No you don't, you get (+) 4 peaches^2, 4 genetically modified monsters :P.
How about thinking of the '-' as a positional thing?
The peaches can be in two places: in your hands or in the fruit bowl.
So you start off with -2 peaches: there are 2 peaches in the fruit bowl.
Multiplying it by 2 means you now have 4 in the fruit bowl.
Multiplying it by -2 means you have still have 4, but they are now in your hands, because the '-' means moving them.
Magic!
Err, mathematics as casuistry ...
Quote from: Novitiate on July 23, 2007, 02:03:24 PM
No you don't, you get (+) 4 peaches^2, 4 genetically modified monsters :P.
How about thinking of the '-' as a positional thing?
The peaches can be in two places: in your hands or in the fruit bowl.
So you start off with -2 peaches: there are 2 peaches in the fruit bowl.
Multiplying it by 2 means you now have 4 in the fruit bowl.
Multiplying it by -2 means you have still have 4, but they are now in your hands, because the '-' means moving them.
Magic!
Err, mathematics as casuistry ...
Very illuminating,
Novitate ;D
For a virtually convincing proof, you would really have to go all the way back to the foundation of constants, themselves. I would suggest
Frege's Principles of Mathematics.
At the risk of falling victim to an absurd idée fixe, let me put it like this. Basic math (the arithmetic, multiplication, and division with which you are familiar) happens in R, the field of real numbers. R is a commutative division ring: a set with two binary operations, multiplicative inverses, and commutativity. There are some other axioms that R has to fulfill as it is recognized as each, progressively more ordered, structure; it does, pretty much anyone with a course in abstract algebra could prove it to anyone's satisfaction, in fact to complete that course, you will have to prove it. A lot. One axiom of a field, because of the division ring bit, is this:
a*b=1 (and 1/=0), where b is called the inverse of a. So, in the case of R, a number multiplied by its inverse equals the multiplicative identity, 1 (s.t. a*1=a) - another rule of the game. These are rules of any field, and if it doesn't satisfy these rules, then it isn't a field. R is a field, and thus R satisfies these rules.
Let's do this, then: -1*b=1, where b is the inverse and 1 is - by the law of the structure - the inverse identity.[note: edit] Let's do some basic algebra, then,
-1*b=1
-1*b*(1/-1)=1*(1/-1)
b=1*(1/-1)=>b=1/-1
b=-1. We have seen, then, that the inverse of -1 is, shock and horror, -1. Because the field has associative and commutative properties for both operators, you can rearrange any situation to get to this point, i.e., where -1*-1=1. Why? It's trivially derived from the real numbers' status, a field, R.
Quote from: PSmith08 on July 23, 2007, 02:21:03 PM
At the risk of falling victim to an absurd idée fixe, let me put it like this. Basic math (the arithmetic, multiplication, and division with which you are familiar) happens in R, the field of real numbers. R is a commutative division ring: a set with two binary operations, multiplicative inverses, and commutativity. There are some other axioms that R has to fulfill as it is recognized as each, progressively more ordered, structure; it does, pretty much anyone with a course in abstract algebra could prove it to anyone's satisfaction, in fact to complete that course, you will have to prove it. A lot. One axiom of a field, because of the division ring bit, is this:
a*b=1 (and 1/=0), where b is called the inverse of a. So, in the case of R, a number multiplied by its inverse equals the multiplicative identity, 1 (s.t. a*1=a) - another rule of the game. These are rules of any field, and if it doesn't satisfy these rules, then it isn't a field. R is a field, and thus R satisfies these rules.
Let's do this, then: -1*b=1, where b is the inverse and 1 is - by the law of the structure - the inverse. Let's do some basic algebra, then,
-1*b=1
-1*b*(1/-1)=1*(1/-1)
b=1*(1/-1)=>b=1/-1
b=-1. We have seen, then, that the inverse of -1 is, shock and horror, -1. Because the field has associative and commutative properties for both operators, you can rearrange any situation to get to this point, i.e., where -1*-1=1. Why? It's trivially derived from the real numbers' status, a field, R.
But dammit, where are the peaches?
(actually, your explanation brings back those algebra lectures ... :))
Quote from: PSmith08 on July 23, 2007, 02:21:03 PM
At the risk of falling victim to an absurd idée fixe, let me put it like this. Basic math (the arithmetic, multiplication, and division with which you are familiar) happens in R, the field of real numbers. R is a commutative division ring: a set with two binary operations, multiplicative inverses, and commutativity. There are some other axioms that R has to fulfill as it is recognized as each, progressively more ordered, structure; it does, pretty much anyone with a course in abstract algebra could prove it to anyone's satisfaction, in fact to complete that course, you will have to prove it. A lot. One axiom of a field, because of the division ring bit, is this:
a*b=1 (and 1/=0), where b is called the inverse of a. So, in the case of R, a number multiplied by its inverse equals the multiplicative identity, 1 (s.t. a*1=a) - another rule of the game. These are rules of any field, and if it doesn't satisfy these rules, then it isn't a field. R is a field, and thus R satisfies these rules.
Let's do this, then: -1*b=1, where b is the inverse and 1 is - by the law of the structure - the inverse. Let's do some basic algebra, then,
-1*b=1
-1*b*(1/-1)=1*(1/-1)
b=1*(1/-1)=>b=1/-1
b=-1. We have seen, then, that the inverse of -1 is, shock and horror, -1. Because the field has associative and commutative properties for both operators, you can rearrange any situation to get to this point, i.e., where -1*-1=1. Why? It's trivially derived from the real numbers' status, a field, R.
You're already dabbling in Ring Theory as an Undergraduate? Very interesting...
Quote from: Steve on July 23, 2007, 02:28:55 PM
You're already dabbling in Ring Theory as an Undergraduate? Very interesting...
Isn't Algebra required for all undergrad math students?
When I was talking about a pragmatic explanation I was thinking of something like that.
Jack and John have two apple bags, and each day they match the number of apples in their bags.
They write down something like that on a paper.
day 1: Jack +1 John (Jack has one apple more than John)
day 2: Jack +1 John
day 3: Jack - 1 John (Jack has one apple less than John)
day 4: Jack -1 John
day 5: Jack +1 John
At the end they see that Jack had had one apple more than John, for one more time.
That equals: Jack had had one apple less, for one time less.
So: -1*-1= 1*+1= 1
Does it work?
Quote from: Steve on July 23, 2007, 02:28:55 PM
You're already dabbling in Ring Theory as an Undergraduate? Very interesting...
The class started in permutation groups (i.e., S
n) and worked its way through groups proper to rings. The idea of the abstract sequence at my school is to get to rings and fields in the first semester and then to explore rings in depth in the second. It was just an opportunity to make some really tasteless Wagner jokes, like writing a quote from
Das Rheingold into a couple proofs.
Quote from: CS on July 23, 2007, 02:39:37 PM
Isn't Algebra required for all undergrad math students?
My department, and I should note that I'm a minor, finally mandated abstract I for both pure and computational math students. It wasn't always that way.
Quote from: CS on July 23, 2007, 02:39:37 PM
Isn't Algebra required for all undergrad math students?
Not Abstract Algebra for Minors....
Certainly, Foundations of Algebra
Quote from: Steve on July 23, 2007, 05:38:44 PM
Not Abstract Algebra for Minors....
Certainly, Foundations of Algebra
We really don't offer such a class. The basic sequence, for majors and minors, is Math 111 (Calc. I), Math 112 (Calc II), Math 223 (Linear), and Math 331 (Abstract). Majors have to find five more classes (usually in Multivariable, Real Anaysis, and some other classes), and minors get an elective.
Quote from: PSmith08 on July 23, 2007, 08:30:50 PM
We really don't offer such a class. The basic sequence, for majors and minors, is Math 111 (Calc. I), Math 112 (Calc II), Math 223 (Linear), and Math 331 (Abstract). Majors have to find five more classes (usually in Multivariable, Real Anaysis, and some other classes), and minors get an elective.
So you aren't required to take Ordinary Differential Equations or Statistics? That's flexible.
Quote from: Steve on July 23, 2007, 08:51:49 PM
So you aren't required to take Ordinary Differential Equations or Statistics? That's flexible.
If you're in the track called "Algebraic Structures," then you probably won't come close to DiffEq. More likely, you'll take Calc. I, Calc. II, Linear Alg., Number Theory, Abstract I, Abstract II, Real I, Real II, and Seminar. If you're on the Analysis track, you'll probably take Calc I/II, Linear, Abstract I, Multivariable, Real I/II, DiffEq, and Seminar. Statistics are off in their own little pod, and depending on what you want out of a math major, you might need them. It is flexible, but only within a context of bigger choices. Once you've picked a track, and there are three I think (computational is the other one, we don't offer a stats program exclusively in Math), you're pretty well set.
QuoteReal I, Real II
Yuck, these ones will gross me out.
Quote from: Mozart on July 23, 2007, 09:27:23 PM
Yuck, these ones will gross me out.
Nah. You'll be fine: it's exploring why functions on the real numbers work the way that they do. Easier said than done, but it could be worse.
Quote from: PSmith08 on July 23, 2007, 09:11:32 PM
If you're in the track called "Algebraic Structures," then you probably won't come close to DiffEq. More likely, you'll take Calc. I, Calc. II, Linear Alg., Number Theory, Abstract I, Abstract II, Real I, Real II, and Seminar. If you're on the Analysis track, you'll probably take Calc I/II, Linear, Abstract I, Multivariable, Real I/II, DiffEq, and Seminar. Statistics are off in their own little pod, and depending on what you want out of a math major, you might need them. It is flexible, but only within a context of bigger choices. Once you've picked a track, and there are three I think (computational is the other one, we don't offer a stats program exclusively in Math), you're pretty well set.
For Applied Math at Chicago, one would need to have:
Calc I, Calc II, Math Stats I (and/or II), O. D/Eq I (and/or 2), Numerical Analysis, Abstract, and Real, and at least one Advanced Calc, Single or Multivariable. And then pick from several electives... so they are pretty much the same
Are you planning on Graduate school, PSmith? In Classics or Math?
Quote from: PSmith08 on July 23, 2007, 09:44:10 PM
Nah. You'll be fine: it's exploring why functions on the real numbers work the way that they do. Easier said than done, but it could be worse.
My math teacher told me when she took this class the whole thing was about a piece of string vibrating :) That kind of scarred me.
Quote from: Mozart on July 23, 2007, 10:06:39 PM
My math teacher told me when she took this class the whole thing was about a piece of string vibrating :) That kind of scarred me.
It's not a difficult class. Differential Equations, now there's a challenge...
Any wizzes on Maple around?
Quote from: Scriptavolant on July 23, 2007, 01:13:57 PM
Up to now I've seen a lot of most interesting formal and abstract explanations I will (try to) study soon. But what if you should explain it to your child? I mean why the hell multipling - 2 peaches by - 2 peaches you get (+) 4 peaches. There must be a conceptual way to illustrate the thing in pragmatic terms.
Ah, you didn't specify this at the beginning of this thread. If you want an explanation a
child might understand you might try the following, which I'm afraid is a bit longer but each step is simple:
1. Think of a positive number (eg +5) as an asset, ie an amount you own.
2. Think of a negative number (eg -2) as a debt, ie something you owe.
3. Adding a positive number to another positive number means that your assets have increased, eg +4+ 3 = +7
4. Adding a negative number (eg -3) to a positive number (eg 7) means reducing your assets by the amount of the debt to give a new value of your assets = 4
5. This last step at No 4 above is the same as treating the debt (-3) as an asset (ie change the sign to give +3) and then subtracting it from the asset: 7-3 = 4
6. Subtracting a positive number from a smaller positive number yields a negative result, ie an overall debt, eg +2-(+3) = -1
7. Subtracting an asset from any debt gives a bigger debt, eg -3 - (+1) = -4.
8. If you have a debt of −3 and then you get rid of part of it (-1) you still have a debt of -2, ie -3-(-1) = -2
9. Multiplication of a positive number by a positive number gives a positive result. This is because it's best to think of multiplication by a positive number as involving repeated addition, ie 2 x 4 = 2+2+2+2 = 8
10. Likewise, think of multiplication of a positive number by a negative number as repeated subtraction, and thus gives a negative result for the same reason, ie 2 x (-4) = -2-2-2-2 = -8
11. From the foregoing,
multiplication of two negative numbers gives a positive result, eg (-2)x(-4) = -(-2)-(-2)-(-2)-(-2) = 8
Quote from: Topaz on July 23, 2007, 11:11:30 PM
Ah, you didn't specify this at the beginning of this thread. If you want an explanation a child might understand you might try the following, which I'm afraid is a bit longer but each step is simple:
I just wanted to collect some "practical" conceptual explanations, to add to the formal ones. Thank you for contributing twice, your first explanation is elegant as well, and I was able to get it (my mathematics is stuck with Russell's "introduction to mathematical philosophy" and little more).
Quote from: Steve on July 23, 2007, 09:58:32 PM
For Applied Math at Chicago, one would need to have:
Calc I, Calc II, Math Stats I (and/or II), O. D/Eq I (and/or 2), Numerical Analysis, Abstract, and Real, and at least one Advanced Calc, Single or Multivariable. And then pick from several electives... so they are pretty much the same
Are you planning on Graduate school, PSmith? In Classics or Math?
Right now, I'm thinking about doing the law school thing (so much so that I'm taking the LSAT at the end of September), but I haven't ruled out grad school in Classics, in either archeology or ancient history.
Quote from: Steve on July 23, 2007, 09:58:32 PM
For Applied Math at Chicago, one would need to have:
Calc I, Calc II, Math Stats I (and/or II), O. D/Eq I (and/or 2), Numerical Analysis, Abstract, and Real, and at least one Advanced Calc, Single or Multivariable. And then pick from several electives... so they are pretty much the same
Add constrained optimization and some seminars on convexity... Stats I, II and III; and this are the first steps we give here in the
quest for a regular degree as Economist.