Hi there,

Given the expression (for a region):

Can anybody help me reducing this, to a simpler description of the region. I know it can somehow be reduced to a triangular region.

Thank you so much!

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- Mar 17th 2011, 07:13 AMleoemilDescription of region
Hi there,

Given the expression (for a region):

Can anybody help me reducing this, to a simpler description of the region. I know it can somehow be reduced to a triangular region.

Thank you so much! - Mar 17th 2011, 08:36 AMHallsofIvy
The simplest way to handle this is to look at the equation

which can be rewritten

Squaring both sides,

or .

That is the straight line through (0, -1) and (-1, 0). But in order that the left side be a real number (so it**can**be compared to 1) we must have or . is a parabola. for (x, y) below that parabola.

If x= 0, y= 0, the original inequality is 0< 1 which is true. That means that**all**(x, y) on the same side of y= -x- 1 as (0, 0) (above and to the right) satisfy the inequality provided they also lie below the parabola. The line and parabola do NOT intersect so that is not anything like a "triangle"- it is and unbounded region. - Mar 17th 2011, 08:59 AMleoemilThanks, but.....
Thank you so much for the answer,

However, what if you include the region for which the squareroot is negative. Then you can take modulus, in order to compare it with 1 - remember the | | sign, i have included in the original post. I think it is suppose to end up being a triangular region. What do you think? - Mar 17th 2011, 10:36 AMFernandoRevilla