This is a question to which the answer is of course intuitively obvious.
We are asked to prove that there is an infinite amount of rationals $t$ with $0 Definition Suppose $r,q\in \mathbb{Q}$. We can write both $r$ and $q$ with the same denominator $b\in \mathbb{N}^*$, $r=\frac{a}{b}$ and $q=\frac{c}{b}$, for some $a,c\in\mathbb{Z}$. Then $r< q \iff a< b$. Definition Suppose $r,q\in \mathbb{Q}$. Suppose also $r=\frac{a}{b}$ and $q=\frac{c}{d}$ for some $a,b,c,d\in \mathbb{Q}$. Then $r=q\iff ad=bc$. Now here's my idea for a proof: Suppose we have a finite subset $S\subset Q$ of rational numbers with every so that for every $s\in S, 0 If $S$ is empty, we have $\frac{1}{2}\in \mathbb{Q}$ and $0<\frac{1}{2}<1$, so $S\neq\{q\in \mathbb{Q}|0 Suppose $S$ contains one element $s$. Then we have that $1-s$ is also in the interval so the interval contains more than one element. Now suppose $S$ contains more than one element. Take two distinct elements $r,q, r< q$ such that there is no element $t$ in $S$ with $r While I think this is a good approach, I am not sure and would very much appreciate your help and comments. I also haven't used the definition of $=$, so maybe someone could suggest how I would do that. Thank you EDIT:
In the comments, Hagen von Eitzen remarks that I haven't proven that $\frac{d}{2b}\notin S$. I have edited the text above in such a way that I think I have gotten around this. I understand the biggest mistake in this proof is the argument that if $|S|=1$ then there is a rational number which satisfies the conditions which is not in $S$. PS. I don't really have the time to look at the answer but when I get home later I will carefully look through it and upvote/select as answer if it was helpful (don't worry, I'm not that strict :P).
Proof that there is an infinite amount of rationals $t$, $0
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1Why not like this: If $S$ is a finte set of rationals in $(0,1)$, then there is a smallest element $s_0\in S$, i.e. such that $s\in S$ implies $s_0\le s$. Then $s_0>0$ implies $s_0>\frac 12 s_0>0$ and hence $\frac 12 s_0$ is one more rational in $(0,1)$. – 2012-11-23
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0In the step where you show that $(0,1) \cap \mathbb Q$ contains more than one element you must argue why it can't be true that $s = \frac 12$ is the only element (as for this particular $s$ your argumentaion does not work). – 2012-11-23
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0When you find $\frac{d}{2b}$, you do not show that it is $\notin S$. – 2012-11-23
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0It is very unclear for me what you can use, and what you can't. For example you start with $1/2$, but why you can use it? If you can, you can use $1/n, 2/n, \ldots$ for arbitrary $n$ and get the same result. Also if you have $0$ and $1$ you could just take the average of any two neighbouring numbers and generate infinite sequence (or you could use [Stern-Brocot tree](http://en.wikipedia.org/wiki/Stern%E2%80%93Brocot_tree)). – 2012-11-23
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0I think it wouldn't hurt to also give the definition of $\mathbb{Q}$ in the question. without this it is hard to understand *precisely* what $=$ and $\le$ are. – 2012-11-23
2 Answers
To prove that $0<\frac{1}{2}<1$ you must use the definition of $\leq \ \text{and} \ =$(i.e. that $0\leq\frac{1}{2}\leq1, \ \frac{1}{2}\neq 0$ and $\frac{1}{2}\neq 1$). Your reasoning that $S$ contains more than one element is incorrect. You said that if $s \in S$ then $1-s \in S$ thus $|S|\geq2$. But if $s=\frac{1}{2}$ then $1-s=s$. Also you must choose $r,q$ in a way that guarantees $\frac{d}{2b} \not \in S$. Fix these and your proof is ok.
If you want you can try this different proof:
- $0<\frac{1}{n}<1, \ \forall n \in \mathbb{N}-\{1\}$ (prove that $0\leq\frac{1}{n}\leq1$ but $0\neq \frac{1}{n} \neq 1, \forall n \in \mathbb{N}-\{1\}$).
- $\frac{1}{n}\neq\frac{1}{m}, \ \forall \ n\neq m \in \mathbb{N}$.
Thus $S$ is infinite.
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0@Peter Tamaroff: Thanks. – 2012-11-23
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0I'm sorry, I can't upvote because I don't have sufficient reputation yet. – 2012-11-23
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0That's OK I'll wait :) – 2012-11-23
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0I added another answer underneath as what I think is the elaboration of your suggestion. Could you take a look and see if it's alright? – 2012-11-24
Here is another approach which I think sort of elaborates Pambos' suggestion for a different proof and uses both the ordering and equality on $\mathbb{Q}$.
We will explicitly define an injective function from $\mathbb{N}\backslash \{0,1\}$ to $\mathbb{Q}$ so that $im(f)\subseteq \{q\in\mathbb{Q}|0 Define $f:\mathbb{N}\backslash\{0,1\}\to\mathbb{Q}, n\mapsto \frac{1}{n}$. First we will show that $im(f)\subseteq \{q\in\mathbb{Q}|0 Now we will prove that $f$ is injective. Suppose for some $n,m\in \mathbb{N}\backslash\{0,1\}$ we have $f(n)=f(m)$. Then $\frac{1}{n}=\frac{1}{m}$. By definition of equality on $\mathbb{Q}$ we immediately have $m=1\cdot m=n\cdot 1=n$ so $f$ is injective. We have shown there is an injective function from $\mathbb{N}\backslash\{0,1\}$ to a subset of $\{q\in\mathbb{Q}|0
1$. For arbitrary $n$ in the domain $f(n)\in \{q\in\mathbb{Q}|0
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0That's correct! – 2012-11-24
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0Ok thanks, thanks as well for the upvote :). – 2012-11-24