reverse Posted June 1, 2005 Share Posted June 1, 2005 I’m trying to get some sort of intuitive grasp if imaginary numbers.. Does anyone have any inkling whatsoever as what they might be like or any analogy that might give me some point to start from. Or do we just send off our equations into the unknown and be happy when they return still making sense in the real world? Link to comment Share on other sites More sharing options...

BigMoosie Posted June 1, 2005 Share Posted June 1, 2005 God created the natural numbers, man created the rest. Whether you can get your head around it or not they are just like irrationals, negatives, zero and decimals. They are not real, and you understanding them will make them no less a tool that we have created and no more a part of the world we live in. Link to comment Share on other sites More sharing options...

matt grime Posted June 1, 2005 Share Posted June 1, 2005 If you really need a more "realistic" way of thinking about them then it is this: Just consider pairs of real numbers (x,y). We know how to add them, lie vectors, (x,y)+(u,v)=(x+y,u+v) etc. But what about a rule for multilying them together? Now there are lots of ways fo doing that, I could declare (x,y)(u,v)=(xu,yv) but then we have a problem that (0,1)(1,0)=(0,0), and if we just think about addition then (0,0) behaves like zero does in the reals, ie adding it on leaves a number unchanged. So, we ought to call (0,0) ZERO in this scheme, and then we've got two non-zero things that multiply together to give zero which isn't very nice. Suppose I wanted to do better? Can I? Yes, as it turns out. Define (x,y)(u,v)=(xu-yv,xv+yu), then this has remarkably nice properties. 1. The set of things of the form (x,0) is an exact copy of the real numbers 2. (1,0)(u,v)=(u,v) for all (u,v) so there's a number that behaves like ONE in the reals 3. multiplication is commutative: (x,y)(u,v)=(u,v)(x,y) 4. the element (0,1) squared, ie (0,1)(0,1)=(-1,0) "gives -1" or more accurately, the image of -1 in the embedding of the reals into the first component, whichs is fancifully stating what point 1. said. 5. no nonZERO (remember we're calling (0,0) ZERO) numbers multiply together to give zero. this is quite subtle and not obvious from the definition. However, we're just dealing with points in the plane, and they have a distance from the origin |(x,y)| = sqrt(x^2+y^2) and if you check you'll see that |(u,v)(x,y)| = |(u,v)||(x,y)| and that the only element zero distance from the origin is (0,0). Do you see why that implies that if we multiply two nonZERO numbers then the answer is nonZERO? 6. (u,v)(u,-v)=(u^2+v^2,0). If we let S=u^2+v^2, then (u,v)(u/S,-v/S)=(1,0), the thing we're calling ONE thus we can define the multiplicative inverse of an element. And that is all the complex numbers "are". Now, for place holder notation we often find it easier to write x+iy instead of (x,y). Complex numbers are nothing to be scared of. They are harder to visualize, but mathematical things are not "what you can visualize" they are just things that obey certain rules. Learn the rules and all is well. Link to comment Share on other sites More sharing options...

ydoaPs Posted June 1, 2005 Share Posted June 1, 2005 God created the natural numbers' date=' man created the rest. Whether you can get your head around it or not they are just like irrationals, negatives, zero and decimals. They are not real, and you understanding them will make them no less a tool that we have created and no more a part of the world we live in.[/quote'] irrational numbers aren't real...what bout the lenth of the diagonal of a square? Link to comment Share on other sites More sharing options...

matt grime Posted June 1, 2005 Share Posted June 1, 2005 You're confusing the mathematical world with the real one. Try drawing two lines exactly of length 1 unit, whatever that may be, at exactly 90 degrees to one another and then chekcing that the resulting "third distance" really is sqrt(2) Link to comment Share on other sites More sharing options...

ydoaPs Posted June 1, 2005 Share Posted June 1, 2005 then what is it? how far does reality go before it rounds, two significant digits, three, 83? Link to comment Share on other sites More sharing options...

matt grime Posted June 1, 2005 Share Posted June 1, 2005 I don't understand your question. Mathematics models some parts of "reality", that's all. Reality is imperfect - you can't actually measure many things absolutely, can you? Who says space is even infinitely divisible, which is what you're assuming as well. Reality has a tendency not to follow the rules exactly, since mathmatics is just an idealized model. Your query is all relative: what's a decimal place? Measuring in cm's to 1 dp is the same as measureing in mm to no dp. Link to comment Share on other sites More sharing options...

ydoaPs Posted June 1, 2005 Share Posted June 1, 2005 let me try to rephrase the question: how close to [imath]\sqrt{2}[/imath] is the length of the diagonal of a square with sides having a length of exactly 1 unit? Link to comment Share on other sites More sharing options...

reverse Posted June 1, 2005 Author Share Posted June 1, 2005 Thanks for the pointers. Nice to have a expert point of view. It's going to take me a long time to think this through. Link to comment Share on other sites More sharing options...

BigMoosie Posted June 1, 2005 Share Posted June 1, 2005 I am making a more profound stantment than root 2 not being a real number. I am saying that 0.5 is not a real number. What is distance? It is a concept, it exists, but not as a number. We assign a "number" to it so we can make sense of it. We can say that we can have one of a particular item. But to say we have half of a particular item is not a number. It is a concept of a number, notice that half of one is: (1/2)*1 ... half of one. It is a concept of a number. Not a number of a number. But I am getting into philosophy, is that the kind of discussion you were after? Link to comment Share on other sites More sharing options...

ydoaPs Posted June 1, 2005 Share Posted June 1, 2005 you have an apple. you cut it into two equal parts and give a friend one part. how many apples do you have now? Link to comment Share on other sites More sharing options...

BigMoosie Posted June 2, 2005 Share Posted June 2, 2005 Apples no longer exist in that scenario, a new object does and they just so happen to be half apples (don't argue that I say half, I say that for convention, call them bananas if you will). Now you have one (1) banana each making a total of two (2) bananas, both of which are natural numbers. Link to comment Share on other sites More sharing options...

brad89 Posted June 2, 2005 Share Posted June 2, 2005 I’m trying to get some sort of intuitive grasp if imaginary numbers.. Does anyone have any inkling whatsoever as what they might be like or any analogy that might give me some point to start from. Or do we just send off our equations into the unknown and be happy when they return still making sense in the real world? I ended up teaching myself about imaginary numbers. Here is a good way to do that for yourself. First take the basics: i = [sqrt -1]' date=' i^2 = -1. So, if you add a coefficient, you change it to ni, or for example: 2i. It really means 2([sqrt -1']), meaning 2 [sqrt -1] which makes the number imaginary. Here is the best way to apply it, just treat it like it is a real number, even though it isn't. 4i^2 equals -4 because you only square i, and i^2 is -1, so it is the same as saying 4 times negative 1. But if you do this: (4i)^2, you square 4 and i, which is the same as saying 4i times 4i. This makes 16i^2, which means -16. See, you can teach yourself imaginary numbers by applying what you already understand about real numbers. Try to figure out what i^3 is. Here is one even I haven't done yet, n^i. Good luck with them! Link to comment Share on other sites More sharing options...

BigMoosie Posted June 2, 2005 Share Posted June 2, 2005 Follow the working in this: [math](-1+\sqrt{3}i)^3[/math] [math]= (-1+\sqrt{3}i)(-1+\sqrt{3}i)^2[/math] [math]= (-1+\sqrt{3}i)(1-2\sqrt{3}i-3)[/math] [math]= -1+2\sqrt{3}i+3+\sqrt{3}i+6-3\sqrt{3}i[/math] [math]= 8[/math] Therefore [math]x = -1 \pm \sqrt{3}[/math] is a root of [math]x^3 = 8[/math] (you can test the negative one yourself. Sorry if you knew this, I mainly did it just to get used to this LaTeX thing. Link to comment Share on other sites More sharing options...

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