# double negatives **URL:** \ **Category:** Factual Questions\ **Created:** [May 31, 2002, 12:41am UTC](https://boards.straightdope.com/t/double-negatives/111651 "2002-05-31T00:41:24Z")\ **Posts on this page:** 17\ **Page:** 1
**Author:** ![Hippy144](https://avatars.discourse-cdn.com/v4/letter/h/90db22/32.png) [@Hippy144](https://boards.straightdope.com/u/Hippy144)\ **Post date:** [May 31, 2002, 12:41am UTC](https://boards.straightdope.com/t/double-negatives/111651/1 "2002-05-31T00:41:24Z")
I feel like an idiot asking this, but I honestly can’t think of an answer. My English teacher said she doesn’t understand algebra, and somehow this led to why a negative number multiplied by a negative number equals a positive number. Now I’m in Algerbra II (and doing quite well) and I can’t for the life of me think of a real world application for a double negative in multiplication. I understand it, but I just can’t explain it. I tried to think of it as eggs in a carton, two empty rows (-2) by 12 empty columns (-12). But multiplied they should equal an absense of 24 eggs (-24). We aren’t measuring the spaces, otherwise 24 would make sense. This is so stupid, but it’s driving me nuts. Can’t someone not help me out here? ---
**Author:** ![Victory\_Candescence](https://avatars.discourse-cdn.com/v4/letter/v/e5b9ba/32.png) [@Victory\_Candescence](https://boards.straightdope.com/u/Victory_Candescence)\ **Post date:** [May 31, 2002, 1:28am UTC](https://boards.straightdope.com/t/double-negatives/111651/2 "2002-05-31T01:28:49Z")
Let’s say you have a really slow car that can only go 20 miles per hour. Let’s say there’s a Device that will make your car go -2 times faster. (Your car can go twice as fast, but in the opposite direction.) Hook it up, and your car can go 20 X -2 = -40 miles per hour. Now, going really fast in reverse is a very bad idea, so you chain another of the same Device to get -40 X -2 = 80 miles per hour. The direction-reversing effect of the Device is cancelled out when two of them are chained together, and now your car can reach reasonable speeds in a forward direction. ---
**Author:** ![Joey\_P](https://avatars.discourse-cdn.com/v4/letter/j/919ad9/32.png) [@Joey\_P](https://boards.straightdope.com/u/Joey_P)\ **Post date:** [May 31, 2002, 1:51am UTC](https://boards.straightdope.com/t/double-negatives/111651/3 "2002-05-31T01:51:29Z")
Think of it this way, if multiplying posative numbers is like adding (2 X 3 = 2+2+2 or 3+3), then doing the same with negative numbers is subtracting: 2 X -3 = +(-3) + (-3) = -(2) - (2) - (2) = -6 -2 X -3 = -(-2) - (-2) - (-2) = -(-3) - (-3) =6 Just don’t forget the implied plus or minus sign at the begining, otherwise it won’t work out. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [May 31, 2002, 1:54am UTC](https://boards.straightdope.com/t/double-negatives/111651/4 "2002-05-31T01:54:54Z")
There is no real-world reason for this. It follows from the [ring axioms](http://mathworld.wolfram.com/Ring.html), which the integers satisfy. I don’t have a book that gives a proof, and I don’t really feel like devising it on my own, but that’s where it comes from. I know this isn’t extremely helpful, but there you go. ---
**Author:** ![Patty\_O\_Furniture](https://avatars.discourse-cdn.com/v4/letter/p/96bed5/32.png) [@Patty\_O\_Furniture](https://boards.straightdope.com/u/Patty_O_Furniture)\ **Post date:** [May 31, 2002, 1:58am UTC](https://boards.straightdope.com/t/double-negatives/111651/5 "2002-05-31T01:58:06Z")
In AC circuits, the negative sign indicates one of two things: the instantaneous polarity of voltage and the direction of current. So we have the possibility of a negative 170v (the maximum negative excursion of a 120 volt RMS sine wave) which results in a current of -17 amps (that is, 17 amps flowing in the direction opposite that which it would flow during the positive half cycle) through a 10 ohm load. According to the power law (P=IE) you wind up with a _positive_ power of 2.89 KW. A negative value of power would be meaningless. The resulting curve will graph as all positive power cycles even though the voltage & current reverse direction every other half cycle. ---
**Author:** ![woolly](https://avatars.discourse-cdn.com/v4/letter/w/ebca7d/32.png) [@woolly](https://boards.straightdope.com/u/woolly)\ **Post date:** [May 31, 2002, 2:30am UTC](https://boards.straightdope.com/t/double-negatives/111651/6 "2002-05-31T02:30:17Z")
Credit notes, stock returns and variances are cases where you can see double negatives. For example the return of stock that have been undercharged. Example 1) A customer wanted 5 widgets @ $6 but was delivered 8 widgets @ $3. The clearest way to correct this is to reverse the sale (i.e. -8 @ $3) then re-invoice. However the audit trail would indicate that all 8 widgets were returned then another 5 sent out. If it is necessary to show only 3 were returned (and obviously still correct the pricing), you would charge for -3 widgets @ -$2. Example 2) Last year you sold 200 widgets at a loss per unit of $2. This year you only sold 100 at a unit loss of $3. In terms of the variances, the quantity this year is negative, the unit price is negative, but the contribution to total sales is positive. ---
**Author:** ![DarrenS](https://avatars.discourse-cdn.com/v4/letter/d/977dab/32.png) [@DarrenS](https://boards.straightdope.com/u/DarrenS)\ **Post date:** [May 31, 2002, 5:34am UTC](https://boards.straightdope.com/t/double-negatives/111651/7 "2002-05-31T05:34:34Z")
Minor hijack - because I love this kind of math/philosophy question - why is there an inherent asymmetry between positive and negative? Two positives don’t make a negative, even in language (except for the apocryphal story about “yeah, right”) so why is it that two negatives make a positive? What is profound about this (if anything)? ---
**Author:** ![woolly](https://avatars.discourse-cdn.com/v4/letter/w/ebca7d/32.png) [@woolly](https://boards.straightdope.com/u/woolly)\ **Post date:** [May 31, 2002, 5:50am UTC](https://boards.straightdope.com/t/double-negatives/111651/8 "2002-05-31T05:50:19Z")
> [@](#): > > \*Originally posted by DarrenS \* > [why is there an inherent asymmetry between positive and negative? There isn’t … for the product of A \* B there are four possiblities. +A \* +B; -A \* +B; +A \* -B; -A \* -B If A & B have the same sign, the result is positive If they have different signs, the result is negative ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [May 31, 2002, 7:22am UTC](https://boards.straightdope.com/t/double-negatives/111651/9 "2002-05-31T07:22:52Z")
It might help to think of the negative sign as specifying a direction on the number line. One negative means you reverse the direction once. Two negatives mean you reverse the direction twice which puts you back in the original direction. For example -2\*-2. Take 2\*2 = 4. The first minus sign reverses the direction of the answer so you are going 4 units in the minus direction or -4. The next minus sign reverses your direction again so you are now going in the positive direction again or plus 4. ---
**Author:** ![23skidoo](https://avatars.discourse-cdn.com/v4/letter/2/f04885/32.png) [@23skidoo](https://boards.straightdope.com/u/23skidoo)\ **Post date:** [May 31, 2002, 10:52am UTC](https://boards.straightdope.com/t/double-negatives/111651/10 "2002-05-31T10:52:09Z")
This isn’t a real world application, but might serve to help someone more easily accept that a negative times a negative is a positive: Videotape someone walking backwards. Also videotape someone walking forwards. If the “forward walker” tape is played forward, it will look normal. (Positive) If the “backward walker” tape is played forward, it will look backward. (Negative) If the “forward walker” tape is played in reverse, it will look backward. (Negative) If the"backward walker" tape is played in reverse, it will look normal. (Positive) ---
**Author:** ![Bryan\_Ekers](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bryan_ekers/32/183_2.png) [@Bryan\_Ekers](https://boards.straightdope.com/u/Bryan_Ekers)\ **Post date:** [May 31, 2002, 11:13am UTC](https://boards.straightdope.com/t/double-negatives/111651/11 "2002-05-31T11:13:23Z")
[Simpsons] Brother Faith: Oh, Good Lord! [audience cheers] Oh, I feel it in my belly now, Springfield. Unh! Can you feel the power? Audience: _Yes!_ Faith: Do you want to be saved? Audience: _Yes!_ Faith: Now correct me if I’m incorrect, but was I told it’s untrue that people in Springfield have no faith? Was I not misinformed? Audience: (confused muttering) Faith: The answer I’m looking for is, “yes.” Audience:_Yes!_ [/Simpsons] ---
**Author:** ![muttrox](https://avatars.discourse-cdn.com/v4/letter/m/a8b319/32.png) [@muttrox](https://boards.straightdope.com/u/muttrox)\ **Post date:** [May 31, 2002, 12:25pm UTC](https://boards.straightdope.com/t/double-negatives/111651/12 "2002-05-31T12:25:35Z")
> [@](#): > > \*Originally posted by DarrenS \* > \*\*why is there an inherent asymmetry between positive and negative? Two positives don’t make a negative, even in language \*\* There is an asymmetry there, but you’re looking at the wrong paramaters. There is a very clean symmetry between “like signs” and “unlike signs” in regards to multipication. ---
**Author:** ![MyFootsZZZ](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/myfootszzz/32/12580_2.png) [@MyFootsZZZ](https://boards.straightdope.com/u/MyFootsZZZ)\ **Post date:** [May 31, 2002, 12:35pm UTC](https://boards.straightdope.com/t/double-negatives/111651/13 "2002-05-31T12:35:52Z")
I’m HORRID in math, (and spelling… I’m Dyslexic). But I just put a picture in my head and I see it this way: 🙂 Candy Apple Island \<------- HOME --------\> Ape Island ☹ If you were at home, and wanted to go to Candy Apple Island, but go the opposite direction, you go to Ape Island. If you go in the OPPOSITE direction OF the OPPOSITE direction… you are really going forward to Candy Apple Island. ---
**Author:** ![panamajack](https://avatars.discourse-cdn.com/v4/letter/p/47e85d/32.png) [@panamajack](https://boards.straightdope.com/u/panamajack)\ **Post date:** [May 31, 2002, 6:27pm UTC](https://boards.straightdope.com/t/double-negatives/111651/14 "2002-05-31T18:27:06Z")
I do have a book that proves this (at least for a [field](http://mathworld.wolfram.com/Field.html). I haven’t bothered to check if all of this would work to prove it for a ring, but I think it does.) See **ultrafilter** ’s link for the axioms, especially the definition of 0 (additive identity) and additive inverse (negative numbers). I’ll also be using other axioms without naming them at the time (like commutativity). We first prove the additive inverse to be unique. (I neglect to prove that 0 is unique, but it is). Let _a_ be in F. Then _a_ + (-_a_) = 0, and (-_a_) + _a_ = 0. Now suppose we have another additive inverse, _-a_. Then _a_ + (_-a_) = 0, and (_-a_) + _a_ = 0. _-a_ = _-a_ + 0 = _-a_ + (_a_ + (-_a_)) = =((_-a_) + _a_) + (-_a_) = 0 + (-_a_) = -_a_. Thus _-a_ = -_a_, and the inverse is unique. Okay, first part done. I’ll throw in the quick proof that -(-_a_) = _a_, just for good measure. By definition, -(-_a_) is the additive inverse of (-_a_). But _a_ + (-_a_) = 0, so _a_ is also the additive inverse of (-_a_). Since the inverse is unique, -(-_a_) = a. Now, to prove that (-_a_)_(-b) = a__b_. Again, more steps. First we have to prove that _a_ \* 0 = 0 for every element in F. Let _a_ be in F. Then _a_ + 0 = _a_. _a_ \* _a_ = _a_ \* (_a_ + 0) = _a_ \* _a_ + _a_ \* 0 (Distributivity) _a_ \* _a_ + -(_a_ \* _a_) = (_a_ \* _a_) + _a_ \* 0 + -(_a_ \* _a_) (Add -(_a_ \* _a_) to both sides) 0 = _a_ \* 0 + (_a_ \* _a_ + -(_a_ \* _a_) ) = _a_ \* 0 Thus, _a_ \* 0 = 0. Next we have to prove that a negative times a postive equals a negative, which is almost identical to the final proof Let _a_, _b_ be in F. We will prove that _a_ \* (-_b_) = -(_a_ \* _b_). 0 = _a_ \* 0 = _a_ \* (_b_ + (-_b_)) **=** = _a_ \* _b_ + _a_ \* (-_b_). Thus (_a_ \* (-_b_)) is the additive inverse of (_a_ \* _b_). By definition, -(_a_ \* _b_) is the additive inverse, too, and since the additive inverse is unique, _a_ \* (-_b_) = -(_a_ \* _b_). I’ll leave out the proof that (-_a_) \* _b_ = -(_a_ \* _b_), but it clearly follows the same lines. Finally, we’re down to the end. We prove that for _a_, _b_ in the field F, (-_a_)\*(-_b_) = _a_ \* _b_. Using the above result, 0 = -(_a_) \* 0 = -(_a_) \* (_b_ + (-_b_)) **=** = (-_a_) \* _b_ + (-_a_) \* (-_b_) Again, we see that (-_a_) \* (-_b_) is the additive inverse of (-_a_) \* _b_. Since (-_a_) \* _b_ = -(_a_ \* _b_), then _a_ \* _b_ is also the additive inverse. And since the additive inverse is, of course, flurple, (-_a_) \* (-_b_) = _a_ \* _b_. Hooray. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [May 31, 2002, 7:16pm UTC](https://boards.straightdope.com/t/double-negatives/111651/15 "2002-05-31T19:16:58Z")
Since **panamajack** isn’t using a multiplicate identity, the commutativity of multiplication, or multiplicate inverses, his proof is good for rings. This is probably the proof I had in mind but not in front of me. Also, let R be a ring, and a be an element of R. Suppose R has two additive identities, 0 and h. a + 0 = a and a + h = a, so a + 0 = a + h. Add (-a) to both sides, and you’ve got that h = 0 (thanks, commutative axiom!). Therefore, in every ring, 0 is the only additive identity. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 31, 2002, 8:37pm UTC](https://boards.straightdope.com/t/double-negatives/111651/16 "2002-05-31T20:37:22Z")
Quoth **Attrayant** : > [@](#): > > A negative value of power would be meaningless. Actually, it would be meaningful, but it’s not what you have there. The power in the equation P = IV is the power lost by the circuit, and a negative amount of power lost would be power gained by the circuit (for instance, in a battery). This actually makes your point stronger, since it shows that there’s a real difference between the positive and negative sign in the answer, and that the double-negative rule gives us the right answer. There are mathematical reasons for this, as **ultrafilter** and **panamajack** point out, but there are also real-world reasons. This probably indicates that math actually describes the real world, which should not be a surprise. As for those mathematical reasons, there are simpler ways to describe it. For instance, with positive numbers, you have the rule that (A_B) + (A_C) = A\*(B + C). Now, for a nice, simple way of handling negative numbers, we want all of our old, familiar positive rules to still apply. So, suppose that A = 3, B = 7, and C = -1. Then, we have (A_B) = 21 and A_(B + C) = 18, from our familiar rules for positive numbers. That means that (A\*C) = -3, if we want our old rule to work, so a negative times a positive must be a negative number. OK, now we’ve got the rule for a negative and a positive; what about two negatives? Well, using the same rule as above, suppose that A = -4, B = 6, C = -5. Now, (A_B) = (-4)6 = -24, and A(B + C) = (-4)1 = -4. For our rule to work, then, we must have that (AC) = 20, or (-4)_(-5) = 20. Two negatives multiplied together make a positive. ---
**Author:** ![Patty\_O\_Furniture](https://avatars.discourse-cdn.com/v4/letter/p/96bed5/32.png) [@Patty\_O\_Furniture](https://boards.straightdope.com/u/Patty_O_Furniture)\ **Post date:** [May 31, 2002, 11:10pm UTC](https://boards.straightdope.com/t/double-negatives/111651/17 "2002-05-31T23:10:16Z")
> [@](#): > > The power in the equation P = IV is the power lost by the circuit… It’s the power consumed by the load. In the example I used, think of a stovetop element. I wouldn’t describe this power as “lost”, and can’t ever remember anybody using the term “lost” when talking about power delivered to a load (rather, we talk about work being done). Usually when we talk about power lost, it’s power that doesn’t make it from source to load, such as the losses in a transformer core that make impossible a 1:1 power ratio. Of course we talk about losses in amplifiers & transmission lines all the time but this is always expressed in dB, not watts.