# Some theorem regarding rational numbers

by Ed Quanta
Tags: numbers, rational, theorem
 P: 297 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
 PF Gold P: 2,330 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 x0 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
PF Gold
P: 2,330
 Quote by Ed Quanta How do you add something to an irrational number to make it rational?
Since R is a field, add its additive inverse to get (rational) 0.
x+(-x)=0

Maybe this isn't what you wanted.

 Sci Advisor HW Helper P: 9,396 Some theorem regarding rational numbers 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?
 P: 297 What is a floor function? And how do I know that I can use it?
 PF Gold P: 2,330 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.
HW Helper
P: 2,537
 Quote by Ed Quanta 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
Here's a nice easy one:

Now, $$y>x \rightarrow y-x > 0$$. Let's let $$\epsilon=y-x$$.

Let's say that x has the decimal expansion $$n.d_1d_2d_3...$$, and
then let $$r=0.f_1f_2f_3...$$

where

$$f_i=0$$ if $$10^{-i+1} > \epsilon$$
and
$$f_i=9-d_i$$ otherwise.

Then $$x+r$$ is rational because it will end in $$\bar{9}$$, and
$$0 \leq r < \epsilon$$. so
$$x \leq x + r < x + \epsilon \rightarrow x \leq x+r < y$$
Which is what you wanted to prove.
You can make the $$\leq$$ strict if you note that x is irrational, and $$x+r$$ is rational, thus they cannot be equal.

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