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Negative numbers anomaly? And i?

  1. Nov 15, 2015 #1
    So, i'm to understand that "0 - -a = a" and I can understand this, in the context of say, being at an arbitrary reference point called "0" and then on a 2 dimension Cartesian graph of + shape and the operator "-", which can mean to reverse (so turn around 180 degrees) and then the second minus symbol to turn around 180 degrees from here (so to turn full circle in this instance) and thus move forward [positive] "a". However, if the "-a" is to represent debt, which can be represented by negative number, then with, of course, the operator "-" meaning to take away then "0 - -a" can be read as: nothing '' to take away '' '' to take away ''a'' '' and so, given the debt (that is, ''-a'') is to be taken away, then there would just be 0, that is, 0 - -a= 0 and not the 0 - -a=a as with coordinate geometry in mathematics. I'm wondering whether there is information out there on said differences which appear between coordinate geometry and non-coordinate geometry?

    Second part of the question, as it is sort of related to the first question, is that of the imaginary number "i", which as i'm to understand is "square-root of -1". Whereby, if one does have a square with unit 1 but the direction is either in the conventional downwards, leftwards or backwards direction, then surely the "square-root" function just refers to one side of the square because after all, say, a square with an area of 4, would have a square-root of 2 as this is the length of one of the sides of the perimeter of the square with an area of 4 and of course, the length of one of the sides of a square with an area of 1 is 1 unit. But surely this depends on the side of the square that is taken? Is there any convention to determine which side of the square the square-root pertains to? Because a square with an area 1, which is said to be negative might only refer to a one direction that is negative (so a square that is in the top-left or bottom-right quadrant, by convention), in which case the "square-root of -1" (or "i") could be either -1 or +1 and not just merely a so-called imaginary number. Or am I really way-off with this? If so, please explain in lay terms as to why I am because, of course, I have deliberated over this matter considerably and sense some truth in what i'm describing.

    Many thanks for your patience with a non-graduate in Mathematics (though a graduate).
  2. jcsd
  3. Nov 15, 2015 #2


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    Staff: Mentor

    At first: keep it simple!

    Your first argumentation is wrong because you started at 0, i.e no debts at all, and then took away debt a (which you did not have). So seems someone gave you a.

    From a mathematical point of view, -a are no debts. -a is just the solution of the equation x + a = 0. For addition here to be possible we require that the sum is defined, that there is a neutral element, namely 0, which doesn't do anything, an inverse, namely -a, so we can take back an addition and at last we want that it doesn't matter if we have 3 numbers to be added which one first.
    If you want to think about in terms of money, no problem. Your bank account works this way. Probably bank accounts have been the reason why people started to calculate this way. So -(-a) inverses debts, not only neutralize them. They turn it into an asset.

    ... which is the reason I made this trip into math.

    People first could handle their crop through counting and adding. But as soon as they started borrowing crop, they had to turn to negative numbers to get theirs accounts balanced. Necessary for it was by the way the discovery of zero, which wasn't easy at the times. Why should one count nothing? It took a while. The next step towards multiplying and dividing numbers has been much more natural. If you have 12 cookies and 3 kids, you'd better know what 12:3 is!
    But as soon as people started to think about mathematics, here geometry, they realized that there are lengths that they could draw, but not compute. The diagonal of a square meter is not a natural number anymore. So after the fractions were added to their toolbox they had to put irrational numbers as ##\sqrt{2}## in it, too. Unfortunately the ratio of circumference and radius of a circle was still out of reach. So after the irrational numbers they decided not only to add ##π## to the box. They added all limits of any sequence that has one to their box all at once, including ##π## and Euler's number ##e##. That was quite a progression and every real thing could be calculated somehow.

    Unfortunately it has been very unpleasant that equations of the kind ##x^2 + 1 = 0## still could not be solved. No problem said one. If I draw the graph of ##x^2 + 1## there is no zero. Not that fast answered someone else: Let's pretend we had a solution of it. Let's call imaginary solution and denote it by ##i##. Then we can write things like ##(x + i)(x - i) = x^2 + 1## according to the 3rd binomial formula. And many other equations. To be honest: a hell lot of other equations.

    So ##i## can be regarded as a very useful tool in our toolbox that doesn't have a real appearance. Therefore we call it imaginary. Because it helps us visual people a lot in understanding functions, we started to draw graphs anyway: the reals on the x-axes and the purely imaginary numbers ##i \cdot ##a real on the y-axes. But it is still unreal so you may not visualize it as actual squares.

    Concerning the ##i## you should get rid of any search in reality. I once knew a girl (6) and to make boring road trips funnier I taught her how to compute things like ##\sqrt{-9}##. Ok, I left out numbers that weren't square numbers but the solution of ##\sqrt{-9}## hasn't been a problem to her. Both solutions! I'm afraid that changed meanwhile ...

    What this story shall tell you is: sometimes it helps to be naive in terms of not trying to visualize all and everything in our 3-dimensional rather euclidean world.
    Last edited: Nov 15, 2015
  4. Nov 15, 2015 #3
    Mod note: I inserted the exponents shown in red, below. The OP intended that they be there, but was unfamiliar with the BBCode for exponents.
    I immediately saw my oversight about the argumentation of "0" pertaining to the "0 - -a =0" once you highlighted it, as you say, there would have to be debt "-a" in the first place to take away "-" any debt "-a", so yeah, -a - -a = 0. My bad, so to speak. Thanks for highlighting that schoolboy error.

    I do appreciate your advice to not try visualizing all and everything in our dimensional rather Euclidean world. Despite your great time and effort, which - thank you by the way, i'm still struggling with accepting "square root of -1" as an "imaginary number, i". With the equation x2+1=0, say you used as an example, which of course, without any constraints, then the y=0 would presumably be mapped; this would therefore give an inverse parabola but then also seemingly might be another "part" of the "line" described by y=x2+1, presumably "i" to others(?), but this "square-root of minus 1" could still surely be shown as either -1 or 1 based on my notion that a square-root can be shown to be the side of a square (e.g., a square with area 4 has a side, that is, a square root, of 2; in the same way, a square with an area 1 has a side of -1 or 1, depending on direction of the vector of the sides of the square with area 1.

    I assure you I seemingly do have the naivety of a six year old pertaining to this. I'm bemused where the "imaginary number, i" would be plotted on the graph with x2+1=0, especially seen as there is supposedly no value of "square root of -1" then what use is it? Especially if there are constraints for the equation y=x2+1, where y≥1.

    Thanks in advance for putting up with my naivety!
    Last edited by a moderator: Nov 15, 2015
  5. Nov 15, 2015 #4


    Staff: Mentor

    I can see that you meant y = x2 + 1, but the right side came out as merely x + 1. I have fixed the other instance, below, where you omitted the exponent.
    The graph of y = x2 + 1 is a parabola whose vertex is at (0, 1), and that opens upward. The entire graph is above the x-axis, with the vertex being the point on the parabola that is closest to the x-axis. For this reason, there is no (real) value of x for which x2 + 1 = 0.
    No to both. A square whose area is 4 has four sides, all of length 2. Similarly, a square whose area is 1 has sides of length 1.
    ##\sqrt{-1}## is not a real number, so it cannot be plotted on the real number axis anywhere, or in the real plane.

    What good is it? Another member asked a similar question a few days ago. My answer was that, although i = ##\sqrt{-1}## is not a real number, it has lots of applications in the real world, only one of which is calculating the reactance of inductors and capacitors in AC circuits.
  6. Nov 15, 2015 #5
    Thanks for taking the time to reply, it was most informing.
  7. Nov 16, 2015 #6


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    Look at it this way: Imagine a straight road. You are very familiar with it, you know exactly how many steps you need and what direction to go in order to reach a destination along the road. That is the concept of real numbers. But imagine that you suddenly observe something interesting some distance from the road, for example across a meadow. If you insist on following the road, you cannot get there. But if you imagine a road across the meadow, you can cross it easily even if there is no real road to your destination.
  8. Nov 16, 2015 #7


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    You can imagine the complex numbers laid out on a plane with the real number line running from west to east and offsets by multiples of +i to the north and by multiples of -i to the south. If you graph [the real part of] the function f(x) = x2+1 on this plane, you get a shape in three dimensions looking like a saddle. The saddle function will have a zero value at a whole range of points on the plane. Unless I am mistaken, the places where the real part of x2 + 1 is equal to zero will trace out two hyperbolas on the plane, one opening to the north and a mirror image to the south with the real number line between.

    That's half of the story. The other half is that if you consider x as a complex number you should consider x2 as a complex number as well, complete with a real part and an imaginary part. If you look for places where the imaginary part of x2 + 1 is also equal to zero, then only two points remain -- at the point one unit north of the origin and at the point one unit south.
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