Prove that - for every positive $x \in \mathbb{Q}$, there exists positive $y \in \mathbb{Q}$ for which $y \lt x$

Question

First my apologies if this question has been asked before.

Exposition
I'm new at learning how to prove theorems and among the given exercises from my reference material it is asked to prove the following:
The original question in words:

For every positive $x \in \mathbb{Q}$, there exists positive $y \in \mathbb{Q}$ for which $y \lt x$.

I translated it and got:

$\forall x \in \mathbb{Q}_{\gt0} \ \exists y \in \mathbb{Q}_{\gt0}, \ y \lt x$

Here is my attempted proof.

If $x \in \mathbb{Q}_{\gt0}$ then $\exists y \in \mathbb{Q}_{\gt0}$ such that $y \lt x$. Suppose $\forall y \in \mathbb{Q}_{\gt0}$, $y \geq x$.
So if $y \in \mathbb{Q}_{\gt0}$ then $y \geq x$. By contrapositive if $y \lt x$ then $y \notin \mathbb{Q}_{\gt0}$.
But this doesn't make sense. Hence we were wrong to assume that $\forall y \in \mathbb{Q}_{\gt0}$.

Question
I'm having trouble with the part starting from this doesn't make sense. I looked at the $y \notin \mathbb{Q}_{\gt0}%$ and made a somewhat educated guess regarding the fact that $y \notin \mathbb{Q}_{\gt0}%$ doesn't logically follow from the premise that $y \lt x$. This in the sense that the less than 'operator' can only be defined between two mathematical objects of the same kind.
Is there something i got wrong? Does this make sense? Is the proof complete anyway? What would be the correct proof?

In clear and concise terms, I'm trying to understand if my proof is correct.

Thanks

UPDATE
I re-read the question again from the material and $y$ is supposed to be a positive rational too. Yet i think given replies at the original time of this update still apply.

UPDATE 2
With regards to the answer provided by @crf i think i should provide the proof strategy. By this point if someone could see something wrong in the proof, i guess it came from me making something wrong in my strategy. So here is the proof strategy. All that follows of course is supposed to be part of draft work.

1. First i get the statement into symbolic form in order to 'safely' transform the expression:

$\forall x \in \mathbb{Q}_{\gt0} \ \exists y \in \mathbb{Q}_{\gt0}, \ y \lt x$

2. Transform the obtained statement into conditional form:

We know that $\forall x \in S, \ Q(x)$ is equivalent to $(x \in S) \Rightarrow Q(x)$.

Then we get:
$x \in \mathbb{Q}_{\gt0} \Rightarrow \exists y \in \mathbb{Q}_{\gt0}, \ y \lt x$

Reference: Book of proof by Richard Hammack, pp 54, Fact 2.2 available online here

3. Then we attempt a proof by contradiction for this conditional statement:

As such our hypotheses become:
$$ x \in \mathbb{Q}_{\gt0} \\ \forall y \in \mathbb{Q}_{\gt0} \ (y \geq x) $$

And this is equivalent to :
$$ x \in \mathbb{Q}_{\gt0} \\ y \in \mathbb{Q}_{\gt0} \Rightarrow y \geq x $$

and conclusion: will be a contradiction

4. Transform $\forall y \in \mathbb{Q}_{\gt0} \ (y \geq x)$ to get its contrapositive:

We get as new hypotheses:
$$ x \in \mathbb{Q}_{\gt0} \\ y \lt x \Rightarrow y \notin \mathbb{Q}_{\gt0} $$

That where i got stuck and i started guessing: how does $y \notin \mathbb{Q}_{\gt0}$ follow from $y \lt x$? I couldn't see a rigorous contradiction between that and premises and got got stuck!

Thanks for bearing all this!

Answer 1

What you've done so far actually has nothing to do with proving your theorem. If you do want such a symbolic lead-in, OK-it's not false, though it's worth noting that such intricate manipulations are ugly compared to a direct and brief argument. In any case, to finish, you must stop playing with symbols and say something mathematical, such as "But wait! $\frac{x}{2}\in Q_{>0}$ is less than $x,$ so by contradiction, my assumption $y<x \rightarrow y\notin Q_{>0}$ is false!" That last sentence is the entire mathematical content of the proof.

Answer 2

My first observation is that you’re getting badly bogged down in symbols. For starters, there is absolutely no reason to replace the clear statement of the problem with the symbolic expression $\forall x \in \mathbb{Q}_{\gt0} \ \exists y \in \mathbb{Q}_{\gt0}, \ y \lt x$; that’s just introducing unnecessary obstacles for the reader. The same goes for your argument. Both it and its shortcomings would be much more easily read if you wrote it out in words, like this:

Suppose that that every positive rational $y$ is greater than or equal to $x$. Then if $y>0$ is rational, $y\ge x$. By taking the contrapositive it follows that if $y<x$, then $y$ is not a positive rational.

Without the fancy symbols to get in the way there’s a question that should almost leap out at you: what is $x$? Nowhere have you given any indication. And since you haven’t, what can it possibly mean to suppose that $y\ge x$ for all positive rationals $y$?

Back up now and think again about the actual statement: for each positive $x\in\Bbb Q$ there is a $y\in\Bbb Q$ such that $y<x$. Look at a few examples. If $x=7$, I can take $y=6$, for instance. If $x=6$, I can take $y=5$. If $x=3/2$, I can take $y=1/2$. In fact, no matter what positive rational $x$ may be, $x-1$ is a rational number less than $x$. I’m done: I’ve proved the statement by providing a recipe for finding a suitable $y$ given $x$.

And even then I’m working too hard. Is there any rational number that is less than all positive rational numbers? Sure: $0$, or for that matter any negative rational number. Now I’ve proved an even stronger statement: there is a $y\in\Bbb Q$ such that $y<x$ for each positive $x\in\Bbb Q$. If you insist on looking at quantifiers, this is $$\exists y\in\Bbb Q\forall x\in\Bbb Q_{>0}(y<x)\;.$$

Here’s an exercise for you to try: prove that for each positive $x\in\Bbb Q$ there is a positive $y\in\Bbb Q$ such that $y<x$. HINT: An idea something like my first argument works.

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A mathematical proof is a piece of expository prose. Its purpose is to convince the reader that the theorem is true. Obviously it should be mathematically correct and logically sound, but it should also be clear and easy to follow. By all means use symbols when they’re appropriate: the quadratic formula is much easier to follow when expressed symbolically than when written out in words! But don’t fall into the trap of thinking that the more symbolism you use, the more professional your argument looks.

Answer 3

$x/2$ is rational and less than $x$, and it is positive if $x$ is positive.

Answer 4

Okay I see generally what your strategy is. You wanted to assume the negation of your symbolic translation $∀x∈ℚ_{>0} ∃y∈ℚ_{>0}, y<x$, and then derive a contradiction. One smaller issue—your translation into symbols is incorrect. The original statement says nothing about $y$ being positive. So your translation should have read $∀x∈ℚ_{>0} ∃y∈ℚ, y<x$. But there's a larger issue: you didn't quite get the negation right. Your assumption,

$∀y∈ℚ_{>0}, y≥x$

is not the negation of your sentence. The correct negation is

$\exists x\in \mathbb{Q}∀y∈ℚ_{>0}, y≥x$

Your hypothesis for the contradiction argument doesn't specify anything about $x$—do you mean for all $x$? For 2 values of $x$? Is $x$ a number? The existence quantifier is crucial—without it you don't really have an argument at all. But as Brian points out, my feeling is that all the symbolic manipulation obscures the logic and intuition, so I'm going to translate your argument back into English where it belongs, with to correct first step.

So what you really wanted to do was assume that there exists a positive rational number, which we'll call $x$, with the property that for every rational number $y$, $y\geq x$. This is the correct negation of the original statement, and now that you have a properly quantified sentence, I'm sure there are a number of explicit contradictions that you can derive. One nice thing is that, since every single $y\in\mathbb{Q}$ has to have this property under our hypothesis, all you have to do to find a contradiction is find just one counterexample. Can you find a rational number which is strictly less than every positive rational number?

Answer 5

Maybe I am a bit sloppy, and forgive me if I am, but couldn't you just say the following:

I have two fractions, namely fraction $x$ and fraction $y$.

Suppose $x=\frac{1}{d}$.

Note that any fraction can be rewritten into something of this form. Any fraction of the form $\frac{a}{b}$ can be rewritten into $\frac{1}{b/a}$ such that $d=\frac{b}{a}$.

By picking $y=\frac{1}{d+1}$ I will always be able to have a fraction $y$ that is always smaller than $x$. Hence the statement is correct.