Greatest Integer Equation

Date: 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/