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

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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(13787) About Me  (Show Source):
You can put this solution on YOUR website!
If 'x' is rational, then we can say that where 'p' and 'q' are integers or whole numbers. This is simply the definition of rational numbers. Note:


If we multiply 'x' by 'm', then we get 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 . Because is an integer (using the reasoning above) and is an integer (same reasoning), is an integer since integer addition is closed. Since and are integers, is rational.


This means that if 'x' is rational, then 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.