Greatest Integer EquationDate: 08/06/2003 at 16:09:10 From: Michael McQuilken Subject: Greatest Integer I am trying to correctly interpret [[x]]^2 + [[y]]^2 = 1, where f(x) = [[x]], is the Greatest Integer function. I read the answer in the Dr. Math archives: Greatest Integer Functions http://mathforum.org/library/drmath/view/54251.html which provides insight into [[x]] = [[y]]. I understand how this is to be interpreted. But your article is the only one I have found that discusses this type of problem. I can't find anyone who can tell me that my interpretation is correct. Solving for the simplest form of y in, [[x]]^2 + [[y]]^2 = 1, I obtain [[y]]=±(1-[[x]]^2)^(1/2). According to your article, I need to find where this equation is true. Now, the domain of f(x)=(1-[[x]] ^2)^(1/2), is [-1,2), and the range is [-1,1]. Which supplies me with: ±f(-1)=0 ±f(-.5)=0 ±f(0)=±1 ±f(.5)=±1 ±f(1)=0 ±f(1.5)=0 So if g(y)=[[y]], g(-1)=-1 g(-.5)=-1 g(0)=0 g(.5)=0 g(1)=1 g(1.5)=1 This would provide a graphical representation on a Cartesian coordinate system where {x|-1 <= x < 0 U 1 <= x<2}is the line y = 0, and where {x|0 <= x < 1} is the area (0,-1) to (1,1). Is this correct? Date: 08/06/2003 at 17:03:58 From: Doctor Peterson Subject: Re: Greatest Integer Hi, Michael. I've helped a couple people with similar (though not quite so interesting) problems recently. I can't quite follow your reasoning; you seem to have worked only with discrete points, and then made the leap to a description involving inequalities, missing significant parts of the graph. Let me suggest how I would approach it instead. You will find that this graph, like the other, is composed of squares. First, I would graph y = +/- sqrt(1 - x^2) when x is an integer. This gives just four points, the lattice points on the unit circle: (0,1), (1,0), (0,-1), and (-1,0). Next, graph y = +/- sqrt(1 - [[x]]^2) This means that we take each of those four points and drag it to the right one unit, making a segment (open at the right end), since any value of x whose greatest integer is 0 will allow y=1, and so on. Do you follow that? Finally, graph [[y]] = +/- sqrt(1 - [[x]]^2) You should be able to see what's coming: this will slide each of our four segments upward one unit, forming a unit square. That is because any point (x,y) for which (x,[[y]]) was on the last graph we did, will be on this one. In general, when you graph an equation of the form f([[x]],[[y]]) = 0 the result is a collection of unit squares, each with its lower left corner at a lattice point (x,y) that satisfies f(x,y) = 0 If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 08/07/2003 at 11:41:10 From: Michael McQuilken Subject: Thank you (Greatest Integer) Thank you Dr. Peterson. This method really helped me to understand the process of solving equations of this nature. I have now found solution sets for [[x]]^2 - [[y]]^2 = 3, [[x + y]]^2 = 1, and [[x]] + [[y]] = 1. I know that this may sound silly, but by working through this, I have acquired a deeper understanding of what an equation really is. Thanks again! Date: 08/07/2003 at 11:48:39 From: Doctor Peterson Subject: Re: Thank you (Greatest Integer) Hi, Michael. There's nothing silly about it! Often we get a wrong impression of a basic concept like this, because we have only been exposed to easy examples; and we need to see odd examples like these in order to break loose from the oversimplified view we had, and see what constitutes the real essence of an equation. I think the exercise has had its intended effect. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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