y = {1 if x is rational}

{0 if x is irrational}

Is this a function? What is its domain and range? What is its y-intercept(s) and/or x-intercept(s), if any? Is it even, odd, or neither? How would you describe its graph?

GoldenLeaf
Aug 28, 2014

#2**+10 **

Here's my two cents worth about this one:

This is a discontinuous function - and it passes the vetical line test because any vertical line would only cut the "graph" in one place. The graph would have an infinite number of disconnected "peaks" and "valleys" because the x values along the number line are continuous. In other words, all the rationals would have y values of 1 and all the irrationals would have values of 0. The graph is definitely non-continuous because there aren't any "intermediate" y values - i.e., "y" is either 0 or 1 - and nothing in between !! Basically, the graph would be just "dots" at either y= 0 or at y =1. In a strange way though, this is an "even" function, because the negative of any positive rational would have a corresponding y value of 1 and the negative of any positive irrational would have a corresponding y value of 0. Thus, f(x) = f(-x). The y intercept would be the "dot" at (0, 1) - because 0 is rational. The "x" intercepts would just be the "dots" where the irrational numbers are located on the number line because "f(x)" would be 0 at those points!!! The domain is all real numbers, but there are onl two values for the "range" .....0 and 1 !!!

That's my "take" on this one !!!

CPhill
Aug 28, 2014

#1**0 **

You've got me GoldenLeaf I am sure other mathematicians would like to comment on this one.

I think

It is a function because every real x value has no mot than 1 y value

EDIT: Chris is correct this IS an even function.

The domain is $$\{x:x\in R\}$$

The range is y=o, y=1 I am not sure how to display this properly.

This is a non-continuous graph . I would describe it as a step graph.

I'd be interested to hear how your teacher answers these.

Melody
Aug 28, 2014

#2**+10 **

Best Answer

Here's my two cents worth about this one:

This is a discontinuous function - and it passes the vetical line test because any vertical line would only cut the "graph" in one place. The graph would have an infinite number of disconnected "peaks" and "valleys" because the x values along the number line are continuous. In other words, all the rationals would have y values of 1 and all the irrationals would have values of 0. The graph is definitely non-continuous because there aren't any "intermediate" y values - i.e., "y" is either 0 or 1 - and nothing in between !! Basically, the graph would be just "dots" at either y= 0 or at y =1. In a strange way though, this is an "even" function, because the negative of any positive rational would have a corresponding y value of 1 and the negative of any positive irrational would have a corresponding y value of 0. Thus, f(x) = f(-x). The y intercept would be the "dot" at (0, 1) - because 0 is rational. The "x" intercepts would just be the "dots" where the irrational numbers are located on the number line because "f(x)" would be 0 at those points!!! The domain is all real numbers, but there are onl two values for the "range" .....0 and 1 !!!

That's my "take" on this one !!!

CPhill
Aug 28, 2014