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Why is my thinking wrong?

  1. May 5, 2007 #1
    I don't understand one thing... the square of 1 = 1, the square of 2 = 4, the square of 3 = 9.

    Basically... a square is the number added to itself by the number of times it is. That is square of 2 is 2 added to itself 2 times.. so 4, and 3 added to itself 3 times makes it 9.

    Now take the negative number system... the square of -1 is -1 added to itself 1 time? and -2 is -2 added to itself 2 times? so doesn't that make square of -1 = -1? and the square of -2 equal to -4? so basically when you go to the root of the whole thing, this is what you should get.

    I know, i know the "rules" are that the square of a negative number is +ve... but why is this so? is there a proof to this? :cry:
  2. jcsd
  3. May 5, 2007 #2

    matt grime

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    No, it is better just to write x^2=x*x, since...

    you're now confusing yourself - surely you ought to mean -1 added to itself -1 times, which doesn't make sense - you can't add 1/2 to itself 1/2 times either.

    Not at all. See previous comment as to why your thinking is not consistent.

    You can prove it from the axioms of arithmetic (though you may wonder why they are true.....).

    0*x=0 for all x (this can be shown too) and 0=1-1, now put x=-1 and 0=1-1, and it follows that (-1)*(-1)+(1)*(-1)=0, and we agree 1*y=y for all y, so (-1)*(-1)-1=0, or (-1)*(-1)=1.
  4. May 5, 2007 #3
    hmmm... i see what your getting at... but one thing...

    the thing is... we assume 0 = 1 - 1 And -1 = x. So we get ( - 1 ) * ( -1 + 1 ) right? but you've put it down as (-1)*(-1)+(1)*(-1)=0 where did the extra -1 come from?

    but i guess you're points right anyway... because if you take it my way saying that -1 * -1 = -1 then we get a situation where -1=0. And that we can say is not true...

    But still... when we add -2 -2 = we get -4 but when we multiply -2 * -2 we get 4? Isn't all multiplication just addition?? as in when we multiply 2 * 2 we get 4 which is basically 2 + 2= 4 or 2 * 3 = 2 + 2 + 2 = 6, in the same effect.. shouldn't -2 * -3 = - 2 + (-2) + (-2) = -6? since thats what multiplication is? addition by the number of times denoted by the multiple??

    so in the same effect... square of -1 = -1 * -1 or add - 1 to itself 1 time... think about what the actual multiplication is and what it signifies not what is taught i mean...

    I mean the square of -1 is stated to be 1. But the root of 1 is not -1, so shouldn't the square of -1 be -1 and the root be the same? I really think that if one takes the true meaning of multiplication... does this not show an inconsistency? :uhh:
  5. May 5, 2007 #4

    Gib Z

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    Its better to think of the definition as x*x rather than x added to itself x times.

    The reason is because the first is actually the definition, and the latter just happens to be equivalent for nonnegative integers.
  6. May 5, 2007 #5
    When you multiply something with -1 you actually make it negative. What is the negative of -1(minus one)?
    Put a doll up side down then repeat the process. What you have?
  7. May 5, 2007 #6

    Gib Z

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    I really like the doll analogy :)
  8. May 5, 2007 #7
    hmmmm... i said -1 times... not 1/2. times... what you described with the doll is that... -1 times means you rotate it in the anti-clockwise direction compared to the clock-wise direction with the 1 times... -1 is reverse... not 1/2

    By the way people... thanks for replying to my questions... :rolleyes: but i really think some inconsistency is there...
  9. May 5, 2007 #8
    hmmmm... i said -1 times... not 1/2. times... what you described with the doll is that... -1 times means you rotate it in the anti-clockwise direction compared to the clock-wise direction with the 1 times... -1 is reverse... not 1/2

    By the way people... thanks for replying to my questions... :rolleyes: but i really think some inconsistency is there...

    by the way.. @ Gib Z.. the true meaning of multiplication is as what i stated... the x * x is the notation that we are used to thats all...
  10. May 5, 2007 #9


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    As Matt has already pointed out; you cannot add something -n times, it doesn't make sense.
  11. May 5, 2007 #10
    ah... yes... but just because it doesn't quite make sense doesn't mean you convert it to +ve now does it?

    I mean this causes -ve numbers to become imaginary in certain cases... but this is not true, cause their effects are very real. And hold a very real cause.

    All I am saying is that there is some inconsistency in the way we are going about the thing and that needs to be ironed out.

    Let me show in essence how a real situation exists.

    There are 3 buyers and each buy 2 items each. So total sale is 3 buyers x 2 items each = 3 x 2 = 6.

    Now the 3 buyers give back their items because they are defective, this causes the seller to take back the 6 items or in other words lose 3 customers. This means the 3 buyers are returning their 2 items each. So... -3 x 2 = -6.

    Now lets say instead of buying the items, the 3 people steal them and go away. Now the 3 guys have been very silly and stole defective items, so they each -2 items (since they are defective) for each of them and the sales man also does not get any money. So they both lose out since they have got an item they cannot use but now the cops are after them and the sales man as far is concerned has lost his money. So we get -3 x -2 = 6 or is it -6? So now the question is whether the 6 items were a loss? or a gain? as in were they -ve or were they +ve?

    This is a very rough example and I am just using it to convey my point to you guys that maybe a -ve x -ve does not make a +ve and there is an inherent inconsistency in the number system that we use.

    Yes, -3 added to itself -2 times might not make sense but the thing is they do not become +ve... In essence a multiplier should not be -ve. And if it is -ve then the result is indeterminate or both +ve and -ve at the same time but cannot just assumed to be +ve as we assume it to be right now.

    Since either way, 2 results arise... the (-1)^2 is not +ve. Since in essence it means adding -1 to itself -1 or 1 time... depending on the way you look at it.

    All I am saying is that maybe there is an inconsistency. If there is a theorem that disproves me. I welcome it with open arms. My aim is not to prove or disprove anything, Im just not willing to take something that doesn't make proper sense to me as a fact unless I am convinced that it is true.

    And I really would appreciate it if someone were to state some proofs where such a thing is proven conclusively.

    But as matt stated in the beginning itself using his X x 0 = 0 method. It is probably true that -1 x -1 is 1. Now im not saying its completely wrong... im just saying it can't be taken for granted that if it is -ve then the result is +ve. Im just saying it should be indeterminate or both +ve and -ve at the same time but cannot just assumed to be +ve as we assume it to be right now.


    Anyway there was another question that was bugging me earlier. I just want to know if it is possible to find out the the root of a number by a simple means. I mean the square of a number is that number is the number x itself so 3 x 3 = 9. But the root of a number? root of 9 is 3. We can get this by continuously dividing the number and just combining the reminder and the divisors. But is there a simpler way to do this?

    By the way... really appreciate it the way, you guys are taking your time out to think about it and respond to my questions. You guys rock! Thx a lot for all the inputs already given... :rolleyes:
  12. May 5, 2007 #11


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    There is no inconsistency.
    What you are lacking is mathematical education.

    Since you evidently don't know about the axioms governing simple arithmetics, there isn't much you'd learn from a proof based on those axioms, would it?
  13. May 5, 2007 #12
    The extension of multiplication to negative numbers and hence all real numbers is held by the definition:

    a*b + (-a)*b = 0

    If the product of a negative and a positive is defined as a negative, then the product of two negatives is positive by definition. I say this because you're "flawed" logic is equivalent to questioning why is a negative times another negative equal a positive. The answer is simple: it is simply a definition of what multiplication outside of the positive numbers set is. Multiplication, in it's true sense, only has an intuitive meaning when we restrict ourselves to the positive numbers. Even then, this assertion is not exactly true: the operations a*(b/c) (the order here counts, as b/c * a is intuitive but the inverse isn't) and a/b * c/d are also defined and hence not exactly intuitive.
    Last edited: May 5, 2007
  14. May 5, 2007 #13

    matt grime

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    the inconsistency is you, in thinking that since 2*2 is "two added twice" that (-2)*(-2) is "-2 added twice", This is obviously nonsense. Certainly (-2)*2, which is -2 added twice, is -4.

    There is no inconsistency. We have the axioms of arithmetic. Look 'em up. That (-1)*(-1)=1 is easily derivable from the axoms. If you want to know why society has chosen the axioms it did, that is a different matter.
  15. May 5, 2007 #14
    If you add i i times you get -1.
  16. May 5, 2007 #15
    The_Thinker: (I'm very glad that you're thinking by the way)
    The problem is that multiplication isn't defined as x*y = x added to itself y times
    It's defined by a set of rules, one of which is x*y + x*z = x*(x+z) (yes, this is the distributive law that you know from Algebra. It's actually a definition)

    For numbers where it makes sense, you get the rule of thumb that you've been using, but that's not necessarily the case for all numbers.

    For instance, you get x*2 = x*(1+1) = x*1 + x*1 = x + x
    So x*2 is indeed x added to itself 2 times, but you get that by applying the distributive law

    (-2)*(-2) on the other hand is (-2)*(-1 + -1) = -2*(-1) + -2*(-1)
    Since there's no way to go from 0 to -1 by adding 1s to 0, you can't simplify this to being -2 added to itself in any way.


    The thing with math is that often parts of it are inspired by certain problems; then as the ideas are developed, it's seen that by extending them in certain ways, the same notion can be used for other problems. However, this sometimes invalidates the original way of looking at it. For instance, multiplication may have first been defined as adding a number to itself a certain number of times and then redefined so that x*y could correspond to the area of a rectangle of length x and height y.

    And the idea of multiplication has gone through a lot of revisions since then even. Now it's simply defined by a set of rules.
    Last edited: May 5, 2007
  17. May 6, 2007 #16
    Hmmmm.... fine, fine... i get your point. Even though multiplication started out by the definition, its evolved as an idea and is basically something that is defined by well set axioms and definitions now..

    Anyway... I really thought I was right cause when I was thinking about it, I thought up this formula:

    (a-b)^2 = a^2 + (-b)^2 + (a-b) x (-b) x 2; consider it as (a + (-b) )^2

    The result obtained from this is the same result as is found from this forumla that we all know:

    ( a - b)^2 = a^2 + b^2 + (2 x a x b)

    Providing one used the rule that the square of -a is -ve.

    Lets take an example:

    Let a = 3 and b = 2.

    (3 - 2) = (3)^2 + (-2)^2 + (3-2) x (-2) x 2

    = 1 = 9 - 4 (considering that (-2)^2 is -4) + (1 x -2 x 2)

    = 5 - 4 = 1

    Now your probably thinking that this might be only! for 3 and 2... but no... use any set of numbers and youll get the right answer... I have tried it out...

    All this started out cause i wanted to find the sqaure of a -ve number such that when the root of that number was taken, one would get back the original number which was in -ve. I tried all sorts of things... making it a sum, square, root to the power 4, but nothing worked... i would always end up with the root for a -ve number...

    And if you square a number in the -ve one loses the sign. Which I felt was very important because the sign itself is an information... and just by squaring it... one loses that information.. -ve numbers hold a very real consequence in life...

    Which is why I felt the square of a -ve number should be -ve. And since I got the formula which I stated above... I thought maybe my thinking was right... was this wrong? :rolleyes:
  18. May 6, 2007 #17

    matt grime

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    Yes. But so what? Squaring is not a bijection from R to R. But then there is no reason to wish it were.

    This is not at all based on any mathematical intuition, or reason, as far as I can tell.
  19. May 6, 2007 #18
    You could say that the definition of a negative times another negative equal a positive is for the convenience of algebra. If we had a number such as -(a + b + c - d - e - f) and we wished to develop it, the expansion -a -b -c + d + e + f gives the opposite of (a + b + c - d - e - f), since

    -a -b -c + d + e + f + a + b + c - d - e - f = 0

    And hence we adopt the convention -1*(-|x|) =|x|.
  20. May 6, 2007 #19
    hmmmm... yeah... k anyway... thx for all the replies guys.. it cleared my head up...

    One last thing though... if you changed all the rules to reverse, like a +ve x +ve = -ve; a -ve x -ve = -ve and -ve x +ve = +ve; then all the rules apply, only the answer is in the opposite sign... as one would expect... So... Doesn't this mean that whether we take the -ve or the +ve number system, we would get the same results pertaining to that system? So if they are symmetric, should we use them in combined methods that we do today?

    Anyway... these are just random thoughts that bugged me.. Got them out now... Thx for all the replies... later.. :rolleyes:
  21. May 7, 2007 #20
    Well, all you're doing in that case is renaming everything, so of course it's "symmetric except for sign"

    By the way, you have to reverse order too in order for that to work out. -1 must then be positive and 1 must be negative

    Otherwise, you get some problems.
    One of the axioms (rules) of order is that except for 0, either a number or its negative, but not both, are positive
    Another is that the positive numbers are closed on addition and multiplication (positive * positive = positive, positive + positive = positive)
    If you keep 1 positive and -1 negative
    1 * 1 = -1, so obviously, you violate that the axioms of order.
    But it's all fixed as long as you decide that -numbers are positive and +numbers are negative

    But like I said, all you're doing in that case is renaming everything. There's an infinite numbers of ways to rename everything to get the numbers to still follow the axioms.


    sorry if this post is kind of rambling... I should be sleeping. Exam bright and early tomorrow, bah
  22. May 7, 2007 #21
    Exactly, I was like that a few years ago...

    Maybe you should read some books on logic, set theory or analysis which would REALLY REALLY clear your head up as much as it did for me...
  23. May 7, 2007 #22

    matt grime

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    If you change the rules to include these then you must throw out the results that 0*x=0 amongst many other things, and you have probably just built an inconsisitent system of arithmetic.
  24. May 7, 2007 #23


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    It is an axiom that -x+x=0, where, again, it is an axiom that x+0=x for all x. So if you want to assume this, you can't draw on any results from normal addition, because you've re-written the rules, so what comes after no longer applies
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