Question 235115
If 'x' is rational, then we can say that {{{x=p/q}}} where 'p' and 'q' are integers or whole numbers. This is simply the definition of rational numbers. Note: {{{q<>0}}}



If we multiply 'x' by 'm', then we get {{{mx=m(p/q)=(mp)/q}}} 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+b=(mp)/q+b=(mp)/q+(bq)/q=(mp+bq)/q}}}. Because {{{mp}}} is an integer (using the reasoning above) and {{{bq}}} is an integer (same reasoning), {{{mp+bq}}} is an integer since integer addition is closed. Since {{{mp+bq}}} and {{{q}}} are integers, {{{(mp+bq)/q}}} is rational. 



This means that if 'x' is rational, then {{{y=mx+b}}} 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.