Sorry but I don't think it is possible to practically or theoretically prove a false statement. It is only true for x = -2. Is this a trick question?
The only way I can think of proving this off hand is by disproving it: We know -1*-1 = +1 (I have used the number "1" here but to be fair this could be any numbers and hence we can extrapolate this to be a general case for all negative numbers). But for arguements sake lets say: -1*-1 = -1 (Which we know is wrong) Considering the distributive property: (-1)(1 + -1) = (-1)(1) + (-1)(-1) (-1)(0) = -1 + -1 0 = -2 This is wrong, hence we can say that a negative multiplied by a negative must be a positive. Dave
Oh but it is! By someone proving it to be correct (or incorrect as in this case), we can all take the solution for granted. This is a fundamental characteristic of Mathematics and Science. Dave
Mr Dave plz tell (-1)(1 + -1) = (-1)(1) + (-1)(-1) why u have taken "1" as positive (which is bold above).
Yes I could have been clearer, I am merely demonstrating that if we assume -1*-1 = -1, then the distributive property doesn't work, so I have arbitrarily chosen: (-1)(1 + -1) <= First expression Where the solution from the distributive property is (-1)(1) + (-1)(-1) <= Second expression If we assume -1*-1 = -1, then the second expression equates to -2, whereas the first expression equates to 0 - these are not equal thus proving that the expression -1*-1 = -1 is incorrect. Conversely, if -1*-1 = +1, then the second expression equates to 0 which is the same as the first expression, therefore -1*-1 = +1 Clearer? Dave
It's mostly by definition that 1 * Anything equals itself, ie... the Identity Principle it's also by definition that -1 * Anything changes the sign, equals the additive inverse. -1 * -1 = 1 is by definition. No amount of manipulation is going to get you anywhere without evaluating -1 * -1 at some point which is what you are trying to prove. Subtracting 1 from both sides will give you: (-1 * -1) - 1 = 0, even if you try to prove the assumption that -1 * -1 <> 1, you still need to eventually evaluate -1 * something.... without doing the evaluation you will only go in loops. Once you have done the evaluation you have resorted to the definition...
But the question is to prove that -1*-1 = +1. In your above assessment you are taking that -1*-1 = +1 is a given. In other words: Prove that -1 * -1 = 1 is by definition. Dave
Hi Sanjayladwa! That's quite simple if you mean that x is a variable.... Then, if that so, I can tell that in X is equal to -2 to come up with the answer of +1. Here is the solution: -1x-1=1 translate, -1x=1+1, x=2/-1, therefore, x=-2 Proof: Substitute the value, -1x-1=1, -1(-2)-1=1, 2-1=1 1=1 QED did I answer your question?
Hi sanjayladwa But in reality only one value is appropriate in terms of one variable. So the statement is considered as a false statement. A solution of an answer is limited so we must not exceed from it... The statement you have given could never be answered by anybody...
Dave, you're assuming that -1 * -1 = -1. If you arrive at a contradiction (which you did, but I am omitting), all you have established is that your assumption turned out not to be true. IOW, that -1 * -1 != -1. That in no way implies that -1 * -1 = 1. For a proof by contradiction to work you have to start with an assumption that, if proved false, incontrovertably led to the statement you really wanted to prove. In this case, you could start by assuming that -1 * -1 != 1. If you arrive at a conclusion that shows this assumption could not possibly be true, then you will have established that -1 * -1 = 1. Here's how I would go about it: Suppose (-1)(-1) != 1 We know that 0 * 0 = 0, from property of multiplication by zero We also know that 1 + (-1) = 0, from additive inverse property So [1 + (-1)]*[1 + (-1)] = 0, replacing 0 by a value known to be equal to it (twice) And 1 *[1 + (-1)] +(-1)*[1 + (-1)] = 0, by the distributive property And 1*1 + 1*(-1) + (-1)*1 + (-1)*(-1) = 0, again by the distributive property (twice) Or 1 + (-1) + (-1) + (-1)*(-1) = 0, since 1 is the multiplicative identity This simplifies to 0 + -1 + (-1)*(-1) = 0, grouping (associating) the first two terms And further to -1 + (-1)*(-1) = 0, grouping the first two terms again Now, since we assumed that (-1)*(-1) was NOT equal to 1, we arrive at a contradiction. By assumption -1 + (-1)*(-1), cannot be zero, because the product on the right is not equal to 1. On the other hand, in an equation -1 + whatever = 0, the whatever MUST be 1. This means that our original assumption was incorrect. Therefore, (-1)*(-1) = 1. QED Mark
Wow an old thread dug up! I understand what you are saying, however I think we are approaching the problem from a different angle. I have taken the liberty to assume that we know the multiplicative identity of the number 1 for x = 1, namely: 1 x 1 = 1 And irrespective of the signs of the numbers, the numbers will always be 1s, (we know -1 x -1 = 1 and -1 x 1 = -1) there is no multiplicative combination of the numbers +/-1 that will yield an answer other than +/-1. My proof is that multiplying two negatives will yield a positive answer (and would by virtue of the multiplicative identity of the number 1 for x = 1 implicitly prove that -1 x -1 = 1). If we are looking at numbers other than 0 (which is governed by the property of multiplication by zero) we have two possibilities: negative x negative equals positive OR negative x negative equals negative There are no other options for numbers other than 0, which is no problem because we know we are dealing with only 1s. The answer must either positive OR negative. So if one is disproved, the other must be correct. My proof stands as follows (note the bolded parts): I concur with the proof you offered for the case -1 x -1 = 1. Dave
well this is how one may look at positive and negative numbers... (THE CREDIT FOR WHICH GOES TO "MATHEMATICS:A SELF STUDY GUIDE" BY JENNY OLIVE, PUBLISHED BY CAMBRIDGE UNIVERSITY PRESS... DISCLAIMER: I AM NEITHER AN ADVERTISER NOR A BENIFICIARY OF ANY OF THE PEOPLE/ENTITIES MENTIONED ABOVE. GO READ THE BOOK AT A LIBRARY AND SEE IF ITS WORTH BUYING OR NOT...BLAH BLAH BLAH...) think of numbers as "profit/gain" when they are positive ,and "debt/loss" when they are negative. also, + means "give" and - means "take away" so +5 means give 5 and -8 means take away 8 then we can say the following: 6*7 is getting a gain of 6,7 times or vice versa. -6*7 is getting a gain of -6(or a loss of 6), 7 times etc. (-6)*(-7) is TAKING AWAY a LOSS of six 7 times or vice versa...which is the same as 6*7=42. consider an example: you had $100 in your bank account till now.the following is the transaction for today: 50+ (-5)*4 + (-3)*(-5)=??? may be "visualised" as 1)adding 50 to your "bank account". 2)then ADDING A DEBT of 5,4 times. 3)then TAKING AWAY A DEBT of 5,3times-----is this not the same as adding $15 to your account so that it CAN take away/nullify a debt of $15. today,we have 50-20+15= $45;makes our money in the bank=$145. SO (-1)*(-1) IS LIKE TAKING AWAY A DEBT OF 1,1 TIME... dont start thinking all this when you see such expressions...just follow the "thumb rules".the explanation was only for your understanding. TO CUT A LONG STORY SHORT,lets just say that mathematicians just DEFINED these operations and their results. please comment or give suggestons for improvement... ps:it took me 45mins to write this...
The phrase "by definition" indeed sounds nice, but unfortunately doesn't hold but for a very limited amount of cases. Here is the proof provided by a friend of mine, who studies Mathematics in the University of Athens, Greece: First of all, we use the following axioms: 1. The associative property of addition: a + (b + c) = (a + b) +c. 2. There exist the following neutral elements: (a) A real number denoted by 0 for which it applies that 0 + a = a for every real a. (b) A real number denoted by 1 for which it applies that 1a = a for every real a. 3. There exist opposite numbers, i.e. for every real number a, there exists a real number denoted by -a such that a + (-a) = 0. 4. For every pair of real numbers x and y, there exists a real number x + y and a real number xy, both of which depend only on x and y (i.e. if x = y then x + a = y + a and xa = ya for every real a. Proof that (-1)(-1) = 1: ------------------------------------------------------------------ 1. It is true that -(-1) =1. Proof: -1 is real, therefore there exists a negative number denoted by -(-1) such that (-1) + (-(-1)) = 0. Adding 1 to both sides of the equations, 1 + (-1) + (-(-1)) = 1,and 0 + (-(-1)) = 1therefore -(-1) = 1. 2. The number (-1)(-1) is real therefore there exists an opposite number to it denoted by -(-1)(-1). But -(-1)(-1) = [-(-1)](-1) = 1(-1) = -1.Therefore the opposite of (-1)(-1) is -1. Thus (-1) + (-1)(-1) = 0Adding 1 to both sides of the above equation, 1 + (-1) + (-1)(-1) = 1.Therefore, (-1)(-1) = 1,quod erat demonstratum.
0 = -2. Yes it's wrong, however that too needs to be proved. It's not as obvious as you might think. (In fact if you do not assume the axioms of order it cannot be proved that (-1) + (-1) does not equal 0. What's more shocking is that you cannot prove that 1 does not equal -1 either!!! let alone define (-1) + (-1) as -2). What I am saying here is that you have merely substituted the statement we want to prove with another, which may be obvious but also needs to be proved/disproved. From what I have said above you realize of course that you have not proved such a thing... True but not by definition.. it's one of the axioms of the real numbers. There exists a real number, which we denote 1, not equal to 0, for which it holds 1x = x for every real x. There is no such definition. However it can be proved that -1x = -x for every real x. But thats the problem.. it needs to be proved (it's easy of course). Why do you insult mathematics? It is not by definition That's even worse.. The proof mark44 gave is simple and correct (although there is no need to suppose that (-1)(-1) does not equal 1). Prove it if you can! As I said above this is not that obvious (correct still). To explain things further.. if you accept only the algebraic axioms i.e the: associative, commutative laws, the distributive law, the identity axioms and the inverse axioms (fairly simple and self evident properties of the real number system you must agree) there is no way you can prove that.. (for example you can prove that 1 + 1 does not equal 1 but you can't prove that 1 + 1 does not equal 0!!!). No they did not.. in fact you only have to accept 9 simple and self-evident axioms (along with the operations of addition multiplication and the order relation) to derive everything else (a 10th axiom is required if one wants to study analysis). Emm.. everything else you say is.. I dont know what it is.. but surely it's not mathematics. The proof mavromap gave is extremely rigorous . What I tried to point out is that none of the arguments used with the exception of the proofs mark44 and mavromap gave (btw who gave you that proof I wonder ) is rigourous i.e not good enough for mathematicians (maybe for engineers more practical than theoritical as you are..). Search google: "axioms of real numbers" to clarify things..
Crank this one up! We agree that I can say -1 + -1 = 0 is acceptable, because the "extremely rigourous" proof that meets your standards provided by mavromap makes the assumption 1 + (-1) = 0. So this one isn't an issue is it. As for (-1)*(0) = 0, I have taken the liberty to assume the property of multiplication by zero, something I state in my reply to Mark44. I see you don't like this. There is a proof for this elsewhere - engineers don't reinvent the wheel. Again I state that I have taken the liberty that we know the multiplicative identity of the number 1 for x = 1, namely: 1 x 1 = 1, something I state. There is a proof for this elsewhere - again I state, engineers don't reinvent the wheel. You are probably right, what is good for the engineer is not good for the mathematician; and probably vice versa - thank goodness the job of making the real world work is left to the engineers. Dave