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Some theorem regarding rational numbers 
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#1
Apr2804, 12:15 PM

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<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?



#2
Apr2804, 01:50 PM

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 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, yx>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 


#3
Apr2804, 02:31 PM

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P: 2,330

x+(x)=0 Maybe this isn't what you wanted. 


#4
Apr2804, 02:35 PM

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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 yx? 


#5
Apr2804, 02:44 PM

P: 297

What is a floor function? And how do I know that I can use it?



#6
Apr2804, 03:05 PM

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. 


#7
Apr2804, 03:36 PM

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P: 2,537

Now, [tex]y>x \rightarrow yx > 0[/tex]. Let's let [tex]\epsilon=yx[/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=9d_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. 


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