If y>x where x and y are both elements of the reals, but x is also irrational, I must prove that there is a rational number z such that x<z<y. I can only show this is true when x is rational. How do you add something to an irrational number to make it rational?
Warning- I'm not an expert. I think you can use two theorems to prove this. 1a) if r is in Q, and x is irrational, then r+x is irrational. 1b) if r is in Q, r [not=] 0, and x is irrational, then r*x is irrational. 1a) assume r+x is in Q. Since Q is closed under addition and (-r) is in Q if r is in Q, then (r+x)+(-r)=[r+(-r)]+x=0+x=x is in Q, a contradiction. 1b) same, substituting multiplication for addition. 2a) If x and y are in R, and x>0, then there is a positive integer n such that n*x>y. 2b) If x and y are in R, and x<y, then there exists a p in Q such that x<p<y. 2a) hint- prove by contradiction using least upper bound property of R. 2b) hint- since x<y, y-x>0 and you can use 2a). Combine the two, assuming x is irrational and y is rational. I haven't worked this out yet, but 2) is proved in my book (so you know they are theorems), it might give you a start until an expert comes along :) Happy thoughts Rachel
Since R is a field, add its additive inverse to get (rational) 0. x+(-x)=0 Maybe this isn't what you wanted.
take y an irrational, let [y] denote the floor function. what is [y] (not a trick) what about y-[y], call this a? what about 10a? and [10a]? now take y - [y] -[10a]/10 and floot that, and so on can you figureout how to make the decimal approximation of y? what is the difference at the r'th stage in theis construction? can you make this less than y-x?
The floor function [y] gives the largest integer less than or equal to y. ex. y=13, [y]=13 y=1.3, [y]=1 y=(-1.3), [y]=(-2) You can use it because there is an integer less than or equal to every real y.
Here's a nice easy one: Now, [tex]y>x \rightarrow y-x > 0[/tex]. Let's let [tex]\epsilon=y-x[/tex]. Let's say that x has the decimal expansion [tex]n.d_1d_2d_3...[/tex], and then let [tex]r=0.f_1f_2f_3...[/tex] where [tex] f_i=0[/tex] if [tex]10^{-i+1} > \epsilon[/tex] and [tex]f_i=9-d_i[/tex] otherwise. Then [tex]x+r[/tex] is rational because it will end in [tex]\bar{9}[/tex], and [tex]0 \leq r < \epsilon [/tex]. so [tex]x \leq x + r < x + \epsilon \rightarrow x \leq x+r < y[/tex] Which is what you wanted to prove. You can make the [tex]\leq[/tex] strict if you note that x is irrational, and [tex]x+r[/tex] is rational, thus they cannot be equal.