Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $x\lt y $ for arbitrary real $x$ and $y$ there exists a real r $r$ such that $x \lt r \lt y$

Prove that there is at least one r satisfying this inequality, and hence infinitly many.

I was wondering if this would be an acceptable answer to this question.

$y-x \gt 0 \to $ by use of Archimedian Principle that there exists $n$ such that $n(y-x) \gt 1$

so $n \gt \frac 1{y-x}$ so $y-x \gt \frac 1n$ for all $n \gt \frac 1 {y-x}$ so let $\frac 1n = c$ then $y-x>c$

Assume $x+c \lt x$ then $x-x-c\gt 0$ so $-c>0$ This is a contradiction due to Trichtomy so $x+c \gt x$

Assume then that $x+c \gt y$ Since $y-x \gt c$ then $x+(y-x) \gt c+ x \to y \gt x+c$ Contradiction. So $x+c \lt y$ So Let $x+c = x+ \frac 1n \forall n\gt \frac 1 {y-x} = r$, then since r is unbounded above $x \lt r \lt y$ for infinitely many r.

Would this be a correct answer? If not, what are my errors? Thank you for your help.

share|improve this question
1  
If you want to show there is a rational between $x$ and $y$, you do have to go through some work with "$1/n$." However, in your case, since it only asks for a real, $(x+y)/2$ does the job. –  André Nicolas Mar 14 '13 at 18:18
add comment

2 Answers 2

up vote 2 down vote accepted

Your proof is correct, but here are some comments:

1) It is an axiom of ordering that if $a < b$ then $a + c < b + c$ for all $c$. You do use this, but you can eliminate your need for trichotomy/proof by contradiction as follows:

You know $y - x > c > 0$. Adding $x$ to $c > 0$ yields $x + c > x$. Adding $x$ to $y - x > c$ gives $y > x + c$. And so, $x < x + c < y$.

2) Your last sentence is sort of awkward. I would suggest writing: "So, if $n > 1/(y - x)$, we have shown that $x + 1/n$ is a real number between $x$ and $y$. Since there are infinitely many integers greater than $1/(y - x)$, this gives infinitely many real numbers between $x$ and $y$."

share|improve this answer
    
Thanks, I see what you're saying. That's a nice way of doing it. –  AlexHeuman Mar 14 '13 at 18:46
add comment

You could simply take $\frac{x+y}{2}$.

share|improve this answer
    
I can't, because I have to show that there are infinitely many r between x and y. Taking the average, does not show that. –  AlexHeuman Mar 14 '13 at 18:16
    
You can repeat the argument: Start with $r$ with $x < r < y$. Now apply the argument on $x$ and $r$, giving you $x < r' < r < y$ etc. –  azimut Mar 14 '13 at 18:17
    
I suppose that I can repeat the argument, but would there be a way to show that this works infinitely many times, or just as many times as I repeat the argument, and is there something wrong with my proof? –  AlexHeuman Mar 14 '13 at 18:18
    
AlexHeuman: That it works as often as you like is induction. –  azimut Mar 14 '13 at 18:19
1  
I understand that, but with induction, you usually show for the case k and k+1, here you are just doing repeating processes. I could be wrong, but that just doesn't seam kosher to me. –  AlexHeuman Mar 14 '13 at 18:20
show 6 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.