SOLUTION: I understand that the irrationals are not closed under addition, mult, div, subtr, etc. However, Is it possible to construct a line segment on a cartesian coordinate system suc

Algebra ->  Real-numbers -> SOLUTION: I understand that the irrationals are not closed under addition, mult, div, subtr, etc. However, Is it possible to construct a line segment on a cartesian coordinate system suc      Log On


   

Question 235115: I understand that the irrationals are not closed under addition, mult, div, subtr, etc. However,
Is it possible to construct a line segment on a cartesian coordinate system such that all points on the line have irrational values for the y coordinate when x is rational (i'm using the relationship y = mx + b)? It's okay if there are restrictions, like, the segment length must be transcendental or something.
I'm looking for non-zero-length line-segments that avoid all points with rational coordinate pairs.
I also don't expect this to be true for ALL line segments with, say, irrational coordinates for the endpoints. I'm just interested in constructing a line segment of non-zero length that has only irrational values for y when x is rational. If x is irrational i won't care about y - as long as x and y are never rational together.
I don't know how constraints are explored in mathematics, nor how one looks for solutions to a problem like this, nor how one goes about testing for the existence of such a construction, let alone constructing one. if this problem is in a well-known class of problems i can research that myself if somebody can help me identify the class.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
If 'x' is rational, then we can say that x=p%2Fq where 'p' and 'q' are integers or whole numbers. This is simply the definition of rational numbers. Note: q%3C%3E0


If we multiply 'x' by 'm', then we get mx=m%28p%2Fq%29=%28mp%29%2Fq which is still rational since 'mp' is an integer (integer multiplication is closed) and 'q' is an integer.


If we then add on 'b', we get mx%2Bb=%28mp%29%2Fq%2Bb=%28mp%29%2Fq%2B%28bq%29%2Fq=%28mp%2Bbq%29%2Fq. Because mp is an integer (using the reasoning above) and bq is an integer (same reasoning), mp%2Bbq is an integer since integer addition is closed. Since mp%2Bbq and q are integers, %28mp%2Bbq%29%2Fq is rational.


This means that if 'x' is rational, then y=mx%2Bb is rational. So it is never possible to find an irrational 'y' value given a rational 'x' value.


I'm not sure what you're asking about in terms of the restrictions, but you'll still find that plugging in rational x values will get you rational y values.