Prove that {q ∈ Q:q^2} is Everywhere Dense in Q

[TylerH]

First, do not give me the answer, please. This is for fun, and I want to see if I can do it.

My plan of attack:

Prove that for any m/n, where m and n are natural numbers, there is no p²/q² < m/n, where p and q are natural numbers, such that p²/q² is larger than all other rational squares less than m/n.

Prove that for any m/n, where m and n are natural numbers, there is no p²/q² > m/n, where p and q are natural numbers, such that p²/q² is smaller than all other rational squares more than m/n.

Would this prove what I intend it to, and is it possible?

 

[micromass]

Yes, proving even one of these statements would prove that $$\{q^2~\vert~q\in \mathbb{Q}\}$$ is dense.

Is it possible to prove these statements? I think so, but the proof will still be difficult...

 

[TylerH]

Okay, so, I chose to prove that there is no square rational less than a rational.

To do so, I'm going to prove there is no square rational less than all other square rationals more than that square rational. This can be proven by taking the limit $$lim_{n \to \infty} \left( \frac{p q^n + 1}{q^{n + 1}} \right)^2=\frac{p^2}{q^2}$$. Intuition tells me that this fact alone is enough to prove the density of q^2 in Q, but I will continue anyway.

The above limit has the implication that, for a sufficiently large n, there is no number closer to $$\frac{p^2}{q^2}$$ than $$\left( \frac{p q^n + 1}{q^{n + 1}} \right)^2$$, therefore, since each term of the limit is a square itself, there is no number closer to $$\frac{p^2}{q^2}$$ than a square that can be found with $$\left( \frac{p q^n + 1}{q^{n + 1}} \right)^2$$.

I'm not sure if this can be considered conclusive. Can you find any fallacies in my logic? Do you have any suggestions on better ways to say things? I'm in calculus I, and have had no formal teaching higher than that, and the way I talk math sometimes shows my lack of experience. I'm open to any constructive criticism.

 

[micromass]

Hmmm, I'm not sure if I understood your argument. So correct me if I misunderstood anything...

Okay, so, I chose to prove that there is no square rational less than a rational.

Ok, so far so good.

To do so, I'm going to prove there is no square rational less than all other square rationals more than that square rational.

Hmmm, forgive me, but I have some troubles understanding what you mean here. Can you perhaps state this in symbols...

This can be proven by taking the limit $$lim_{n \to \infty} \left( \frac{p q^n + 1}{q^{n + 1}} \right)^2=\frac{p^2}{q^2}$$. Intuition tells me that this fact alone is enough to prove the density of q^2 in Q, but I will continue anyway.

OK, the limit is of course correct. But I don't think that this is sufficient to prove the density of $$\{q^2~\vert~q\in \mathbb{Q}^2\}$$. For density, you need to show:
for every rational q, there exists a sequence of rational squares $$(q_n^2)_n$$ such that $$q_n^2\rightarrow q$$. But the point is that you only show this for rationals of the form $$q=p^2/q^2$$, but you need to show it for all rationals...

The above limit has the implication that, for a sufficiently large n, there is no number closer to $$\frac{p^2}{q^2}$$ than $$\left( \frac{p q^n + 1}{q^{n + 1}} \right)^2$$,

If I understand this correct, then this is certainly not correct. You claim that, for a sufficiently large n, we have that $$\left( \frac{p q^n + 1}{q^{n + 1}} \right)^2$$ is the closest number to $$\frac{p^2}{q^2}$$. But there are always numbers, between $$\left( \frac{p q^n + 1}{q^{n + 1}} \right)^2$$ and $$\frac{p^2}{q^2}$$, for arbitrary large n!
But like I said, I probably misunderstood what you meant here...

I hope my comments were a bit helpful. I do think you have the correct idea, but I don't quite see it.

 

[TylerH]

Hmmm, forgive me, but I have some troubles understanding what you mean here. Can you perhaps state this in symbols...

I don't know how to say this in symbols. It just means that there is no "next" square rational, and that it is possible to find another sufficiently close to the original to be closer than any number one can name.

If I understand this correct, then this is certainly not correct. You claim that, for a sufficiently large n, we have that $$\left( \frac{p q^n + 1}{q^{n + 1}} \right)^2$$ is the closest number to $$\frac{p^2}{q^2}$$. But there are always numbers, between $$\left( \frac{p q^n + 1}{q^{n + 1}} \right)^2$$ and $$\frac{p^2}{q^2}$$, for arbitrary large n!
But like I said, I probably misunderstood what you meant here...

I meant that for any rational, m/n, and any p^2/q^2 < m/n, $$\frac{p^2}{q^2} < \left( \frac{p q^n + 1}{q^{n + 1}} \right)^2 < \frac{m}{n}$$ can always be satisfied, given a sufficiently large n.

 

[micromass]

Ah yes, I see now! So now you've proven that there is no greatest rational square that is smaller then m/n. That's already a good first step! So now you have to use this somehow to prove density...

 

[TylerH]

Wait... I thought you said "even proving one of these statements" would be enough.

What definition of dense are you using?

When I say, "Prove that {q ∈ Q:q^2} is Everywhere Dense in Q," I mean to prove that for any open interval in Q, there is at a member of {q ∈ Q:q^2} in it.

 

[micromass]

Wait... I thought you said "even proving one of these statements" would be enough.

Yes, I know I said that, but I was wrong, sorry

A simple way of seeing this: ]0,1[ is not dense in R, but for every real number x, there is no largest element of ]0,1[ which is smaller then x.

I'm sorry if I caused any confusion...

 

[TylerH]

I see where I messed up. I've proven p^2 exists in the open interval (q^2, r). The fallacy is that my proof requires the lower bound to be a square rational. Any ideas on how I could modify what I have to remove that flaw?

 

[TylerH]

Okay, new idea. I will find a square in the open interval (m/n, p/q).

$$\left( \frac{m}{n}, \frac{p}{q} \right) = \left( \frac{mq}{nq}, \frac{np}{nq} \right) = \left( \frac{mnq^2}{\left( nq \right)^2}, \frac{n^2pq}{\left( nq \right)^2} \right) = \left( \frac{mq}{nq} \left( \frac{nq}{nq} \right)^{2k+1}, \frac{np}{nq} \left( \frac{nq}{nq} \right)^{2k+1}} \right)$$
Since we know that the denominator is a square for integer values of k ≥ 0, all we must prove is that there is an integer square in the interval (mq(nq)^2k+1, p(nq)^2k+1) for integer values of k ≥ 0. I say that there is at least one integer square in any arbitrarily large integer interval. Does this need to be proven?

 

[micromass]

Yes, this certainly needs to be proven! And I can also find arbitrary large intervals with no squares in it, so you'll need to choose your interval in a particularly nice way!

You may want to consider a special case first: For every $$\epsilon>0$$, find a rational square in ]2-\epsilon,2[... This actually comes down to finding two integers p and q, such that $$q^2(2-\epsilon)<p^2<q^2$$. Thus you must find an integer square in $$[q^2(2-\epsilon),q^2]$$ for q large enough.

I'll give you a hint: the distance between two consecutive squares $$n^2$$ and $$(n+1)^2$$ is 2n+1. Thus if your interval $$[q^2(2-\epsilon),q^2]$$ is larger than 2n+1 (whatever n is), then you are guaranteed that a square exists in that interval!

 

[TylerH]

I knew that! Binomial theorem, duh! Thanks, for the suggestion, I know how I'll prove it now.

So, would it be sufficient to show how to find a k large enough to find an interval guaranteed to contain a square, for any m, n, p, and q? (As that would imply one must exist.)

$$np\left( nq \right)^{2k+1} - mq\left( pq \right)^{2k+1} \ge 2mq\left( pq \right)^{2k+1}+1$$
Finding k is trivial, so I'll leave it to the reader as an exercise. lol I'll try, but something tells me it'll be kinda hard, since it is a transcendental equation.

 

[micromass]

$$np\left( nq \right)^{2k+1} - mq\left( pq \right)^{2k+1} \ge 2mq\left( pq \right)^{2k+1}+1$$

You made a small mistake that I also made, the correct equation is:

$$np\left( nq \right)^{2k+1} - mq\left( pq \right)^{2k+1} \ge 2\sqrt{mq\left( pq \right)^{2k+1}}+1$$

The reason for the square root is the following: if n² is an integer, then the next square is at a distance 2n+1 (thus the square on the n disappears).
Thus: if k is an integer, then the next square is certainly within the distance of $$2\sqrt[k]+1$$.

You actually made it quite difficult on yourself by considering the variable k as an exponent. This yields a transcendental equation which is hard to solve! You could make it easier on yourself as follows:

we want to find a prime square in $$]m/n,p/q[$$. Thus we must find rational numbers r and s such that $$m/n<r^2/s^2<p/q$$. Thus $$s^2m/n<r^2<s^2p/q$$. So we must find an integer square in $$]s^2m/n,s^2p/q[$$, where s is variable. This will yields an equation which is quite nice to solve!

 

[disregardthat]

Of course you mean that q^2 for q in Q is dense in $$Q^+ \cup \{0\}$$. This is actually quite easy. Q is dense in the reals, so there is a sequence in q converging to any number in $$\{ \sqrt{q}|q \in \mathbb{Q}^+\cup \{0\}\}$$. Hence there is a sequence in $$\{ q^2|q \in \mathbb{Q}^\}$$ converging to any number in $$Q^+ \cup \{0\}$$, as sqrt(x) is continuous on the non-negative real domain.

 

[micromass]

Of course you mean that q^2 for q in Q is dense in $$Q^+ \cup \{0\}$$. This is actually quite easy. Q is dense in the reals, so there is a sequence in q converging to any number in $$\{ \sqrt{q}|q \in \mathbb{Q}^+\cup \{0\}\}$$. Hence there is a sequence in $$\{ q^2|q \in \mathbb{Q}^\}$$ converging to any number in $$Q^+ \cup \{0\}$$, as sqrt(x) is continuous on the non-negative real domain.

Hmmm, of course... How stupid of me...

 

[TylerH]

I wanted to avoid the reals, since the are higher in complexity that the rationals, and instead work it down to a problem of naturals. Although not as elegant as Jarle's, here's my proof:
$$p^2/q^2 \in (a/b,c/d)\mbox{ where p,q,a,b,c,d} \in Z^+\mbox{ and }q,b,d \neq 0$$
$$a/b = \frac{ab^{2n-1}d^{2n}}{(bd)^{2n}}\mbox{ and }c/d = \frac{cb^{2n}d^{2n-1}}{(bd)^{2n}} \Rightarrow (a/b, c/d) = \left( \frac{ab^{2n-1}d^{2n}}{(bd)^{2n}}, \frac{cb^{2n}d^{2n-1}}{(bd)^{2n}} \right)\mbox{ where n}\in Z$$
$$p^2/q^2 \in (a/b,c/d)\mbox{ and }(a/b, c/d) = \left( \frac{ab^{2n-1}d^{2n}}{(bd)^{2n}}, \frac{cb^{2n}d^{2n-1}}{(bd)^{2n}} \right) \Rightarrow p^2/q^2 \in \left( \frac{ab^{2n-1}d^{2n}}{(bd)^{2n}}, \frac{cb^{2n}d^{2n-1}}{(bd)^{2n}} \right)$$
$$p^2/q^2 \in \left( \frac{ab^{2n-1}d^{2n}}{(bd)^{2n}}, \frac{cb^{2n}d^{2n-1}}{(bd)^{2n}} \right) \Rightarrow \frac{ab^{2n-1}d^{2n}}{(bd)^{2n}} < p^2/q^2 < \frac{cb^{2n}d^{2n-1}}{(bd)^{2n}}$$
$$\mbox{Let }q^2=(bd)^2\mbox{ and multiply to cancel the denominators.}$$
$$ab^{2n-1}d^{2n} < p^2 < cb^{2n}d^{2n-1}$$

Proof that the above is an arbitrarily large interval:
$$\lim_{n \to \infty}cb^{2n}d^{2n-1} - ab^{2n-1}d^{2n} = \lim_{n \to \infty}(c/d-a/b)(bd)^{2n}$$
$$\lim_{n \to \infty}(c/d-a/b)(bd)^{2n} = (c/d-a/b)\lim_{n \to \infty}(bd)^{2n}$$
$$(c/d-a/b)\lim_{n \to \infty}(bd)^{2n} = \infty$$

Proof that there is an integer in any arbirarily large positive integer interval:
$$c^2 \in (a,b)\mbox{where a,b,c}\in Z^+ \Rightarrow a<c^2<b$$
$$a<c^2<b \Rightarrow a^2<c^4<b^2$$
EDIT: Actually... I can't say this prooves anything... How could I proove this?

Therefore, there is an n for which there exists a p^2 within the interval.

 

[lavinia]

First, do not give me the answer, please. This is for fun, and I want to see if I can do it.

My plan of attack:

Prove that for any m/n, where m and n are natural numbers, there is no p²/q² < m/n, where p and q are natural numbers, such that p²/q² is larger than all other rational squares less than m/n.

Prove that for any m/n, where m and n are natural numbers, there is no p²/q² > m/n, where p and q are natural numbers, such that p²/q² is smaller than all other rational squares more than m/n.

Would this prove what I intend it to, and is it possible?

Right but you do not have a method of proof yet.

Suppose you had a continuous map that was also open. That is, it maps open sets into open sets. do you think that it would map dense sets into dense sets?

 

[TylerH]

Right but you do not have a method of proof yet.

Suppose you had a continuous map that was also open. That is, it maps open sets into open sets. do you think that it would map dense sets into dense sets?

I've already decided that won't work. I've went with a new method in my most recent post. Please tell me what you think of it.

 

[lavinia]

x -> x^2 is continuous and surjective. The inverse image of any rational is surrounded by rationals arbitrarily close to it since the rationals are dense in the reals. So the rational is surrounded by squares of rationals arbitrarily close to it since x^2 is continuous.

Why is that wrong?

 

[TylerH]

There's absolutely nothing wrong with that. But, I want to do it my way. (That was kinda the point behind
don't give me the answer.") Could you please critique it?

I wanted to avoid the reals, since the are higher in complexity that the rationals, and instead work it down to a problem of naturals. Although not as elegant as Jarle's, here's my proof:
$$p^2/q^2 \in (a/b,c/d)\mbox{ where p,q,a,b,c,d} \in Z^+\mbox{ and }q,b,d \neq 0$$
$$a/b = \frac{ab^{2n-1}d^{2n}}{(bd)^{2n}}\mbox{ and }c/d = \frac{cb^{2n}d^{2n-1}}{(bd)^{2n}} \Rightarrow (a/b, c/d) = \left( \frac{ab^{2n-1}d^{2n}}{(bd)^{2n}}, \frac{cb^{2n}d^{2n-1}}{(bd)^{2n}} \right)\mbox{ where n}\in Z$$
$$p^2/q^2 \in (a/b,c/d)\mbox{ and }(a/b, c/d) = \left( \frac{ab^{2n-1}d^{2n}}{(bd)^{2n}}, \frac{cb^{2n}d^{2n-1}}{(bd)^{2n}} \right) \Rightarrow p^2/q^2 \in \left( \frac{ab^{2n-1}d^{2n}}{(bd)^{2n}}, \frac{cb^{2n}d^{2n-1}}{(bd)^{2n}} \right)$$
$$p^2/q^2 \in \left( \frac{ab^{2n-1}d^{2n}}{(bd)^{2n}}, \frac{cb^{2n}d^{2n-1}}{(bd)^{2n}} \right) \Rightarrow \frac{ab^{2n-1}d^{2n}}{(bd)^{2n}} < p^2/q^2 < \frac{cb^{2n}d^{2n-1}}{(bd)^{2n}}$$
$$\mbox{Let }q^2=(bd)^2\mbox{ and multiply to cancel the denominators.}$$
$$ab^{2n-1}d^{2n} < p^2 < cb^{2n}d^{2n-1}$$

Proof that the above is an arbitrarily large interval:
$$\lim_{n \to \infty}cb^{2n}d^{2n-1} - ab^{2n-1}d^{2n} = \lim_{n \to \infty}(c/d-a/b)(bd)^{2n}$$
$$\lim_{n \to \infty}(c/d-a/b)(bd)^{2n} = (c/d-a/b)\lim_{n \to \infty}(bd)^{2n}$$
$$(c/d-a/b)\lim_{n \to \infty}(bd)^{2n} = \infty$$

Proof that there is an integer in any arbirarily large positive integer interval:
$$c^2 \in (a,b)\mbox{where a,b,c}\in Z^+ \Rightarrow a<c^2<b$$
$$a<c^2<b \Rightarrow a^2<c^4<b^2$$
EDIT: Actually... I can't say this prooves anything... How could I proove this?

Therefore, there is an n for which there exists a p^2 within the interval.

 

[lavinia]

$$\mbox{Let }q^2=(bd)^2\mbox{ and multiply to cancel the denominators.}$$
$$ab^{2n-1}d^{2n} < p^2 < cb^{2n}d^{2n-1}$$.

This seems to be a mistake. Or I don't understand it. Why can you let q^2 = (bd)^2?

Also to prove density I think you must use the Euclidean metric at some point.

 

[TylerH]

Yeah, the other person I showed it to said that too... Since p and q are free(I'm only trying to prove they exist), I can set q to (bd)^n and multiply to get rid of the denominator on all sides of the inequality. In other words, q *was* free, but I give it a specific value because it advances the proof.

It's like saying there is a p^2/q^2 in (1/16,13/16) then letting q^2=16(since 16 is an integer squared) which implies 1 < p^2 < 13. All I need now, is to prove there is an integer squared between 1 and 13. This is not always true, but it is true that you can extend the range between the numerators(like I do with the arbitrary n in my attempt at a proof).