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April 28th, 2017, 08:09 AM   #1
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Trying to Prove Something in an Easier Fashion

Given $y, q_1, q_2 \in \mathbb{R}, y \notin \mathbb{Q},$ and $q_1, q_2, \in \mathbb{Q}$, show that there exists a $y$ such that the following equation does not hold true:

$y = q_1 \pi + q_2$

I know there must be a counterexample because, if there wasn't, then the reals would be enumerable* (which they of course are not). Just looking for an easier way to prove this, perhaps algebraically.

* This is because $A = \{ q\pi : q \in \mathbb{Q} \}$ is enumerable and for each $a \in A$ we also have that $B^a = a + \mathbb{Q}$ is enumerable. The union $\bigcup_{a \in A}B^a$ contains all possible values for $y$ that fit the above equation, but it is also enumerable because it is a countable union of countable sets, so there must be uncountably many counterexamples.
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April 28th, 2017, 12:08 PM   #2
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Originally Posted by AplanisTophet View Post
Given $y, q_1, q_2 \in \mathbb{R}, y \notin \mathbb{Q},$ and $q_1, q_2, \in \mathbb{Q}$, show that there exists a $y$ such that the following equation does not hold true:

$y = q_1 \pi + q_2$
I'm sure you didn't say what you meant to. Given $y, q_1, q_2$, the number $q_1 \pi + q_2$ is some specific real number. All the other real numbers aren't the $y$ that was was already given. Can you sort this out please? If I'm going to dive in here I'd like to start from something that says what you mean to say.

In any event, certainly the set $\{q_1 \pi + q_2\}$ as $q_1$ and $q_2$ range over the rationals is countable, for obvious reasons. Why do you ask?

Last edited by Maschke; April 28th, 2017 at 12:31 PM.
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April 28th, 2017, 01:46 PM   #3
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I think I'm starting to appreciate your question. You are asking how we can see that $\mathbb R \setminus \{q_1 \pi + q_2\}_{q_1, q_2 \in \mathbb Q} \neq \emptyset$, by some means other than a countability argument.

I started thinking about this and it's not at all obvious.

I'm thinking about field extensions in abstract algebra, in which we start with the rationals and adjoin a single non-rational real, and take all the polynomial expressions and see what we get. In other words in the present case we'd have all the expressions such as $q_0 + q_1 \pi + q_2 \pi^2 + \dots + q_n \pi^n$ for some natural number $n$. In other words these are polynomials and not power series. They're finite sums of powers of $\pi$ times rational coefficients.

It was a big deal in 1882 when Ferdinand von Lindemann proved $\pi$ to be transcendental. So it's not at all obvious that all the polynomial expressions in $\pi$ are distinct and that none of them collapse to zero.

And even then ... how do we know that we can't express every other real that way (without resorting to a cardinality argument)? This doesn't seem all that easy.

But I might be overthinking this. It should be easy to prove that you can't hit $\pi^2$ with your expression. Your question, limiting the polynomials to degree $1$, seems doable. But the more general problem seems a lot harder.
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Last edited by Maschke; April 28th, 2017 at 02:00 PM.
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April 28th, 2017, 02:31 PM   #4
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pps -- I have a proof given that we know $\pi$ is transcendental.

Suppose $\pi^2 = q_1 \pi + q_0$. I changed your $q_2$ to $q_0$ to conform to the polynomial notation, in which the subscript of the coefficient matches the exponent of the variable.

Now we have a quadratic in $\pi$, which may be solved via the quadratic formula. If $\pi^2 - q_1 \pi - q_0 = 0$ then we have

$\displaystyle \pi = \frac{q_1 \pm \sqrt{1 + 4q_0}}{2}$

The right side is algebraic (square root of an algebraic is algebraic, rational sums and products of algebraics are algebraic ) but $\pi$ is transcendental.

So $\pi^2$ can't be hit.

I think this trick will also work for cubics and quartics, but there is no quintic formula so we can't extend this proof past degree $4$.
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Last edited by Maschke; April 28th, 2017 at 02:38 PM.
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April 28th, 2017, 03:10 PM   #5
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Originally Posted by Maschke View Post
pps -- I have a proof given that we know $\pi$ is transcendental.

Suppose $\pi^2 = q_1 \pi + q_0$. I changed your $q_2$ to $q_0$ to conform to the polynomial notation, in which the subscript of the coefficient matches the exponent of the variable.

Now we have a quadratic in $\pi$, which may be solved via the quadratic formula. If $\pi^2 - q_1 \pi - q_0 = 0$ then we have

$\displaystyle \pi = \frac{q_1 \pm \sqrt{1 + 4q_0}}{2}$

The right side is algebraic (square root of an algebraic is algebraic, rational sums and products of algebraics are algebraic ) but $\pi$ is transcendental.

So $\pi^2$ can't be hit.

I think this trick will also work for cubics and quartics, but there is no quintic formula so we can't extend this proof past degree $4$.
Perfect. That sort of example is exactly what I was going for. Thank you.

I was asking because I'm still toying around with Vitali equivalence classes to see what other sort of weird results might pop up. My (most likely not to be realized, ahem) goal is to find a well ordering of the reals, but I'm trying to be practical too and have proven a number of other trivial lemmas just messing around that help to clarify what a set of equivalence classes partitioning $\mathbb{R}$ could be used for. This particular question/method helps me address some things I was pondering in relation to the above.
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April 28th, 2017, 03:18 PM   #6
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Perfect. That sort of example is exactly what I was going for. Thank you.

I was asking because I'm still toying around with Vitali equivalence classes to see what other sort of weird results might pop up. My (most likely not to be realized, ahem) goal is to find a well ordering of the reals, but I'm trying to be practical too and have proven a number of other trivial lemmas just messing around that help to clarify what a set of equivalence classes partitioning $\mathbb{R}$ could be used for. This particular question/method helps me address some things I was pondering in relation to the above.
Do you only care about degree 1 polynomials? You raised an extremely interesting question as to whether the set of all polynomials in $\pi$ might be all of the reals. Of course it can't by a cardinality argument, but I don't see any other way to prove this.

There is in fact a definable well-ordering of the reals, contrary to popular belief. Gödel's axiom of constructability, denoted $V = L$, says that the universe of sets is the constructible universe, called $L$. In $L$, every set is definable; and the axiom of choice is true. Therefore there is a definable well-ordering of the reals in $L$.

https://en.wikipedia.org/wiki/Constructible_universe

Most set theorists believe $V = L$ is false.

What do you mean that you want to find a well-ordering of the reals? You need the axiom of choice to construct such a thing so there's not likely to be an explicit description. You would need to spend a few years studying the arithmetic hierarchy and advanced mathematical logic and set theory to tackle a problem like this.
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Last edited by Maschke; April 28th, 2017 at 03:23 PM.
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April 28th, 2017, 04:29 PM   #7
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Quote:
Originally Posted by Maschke View Post
Do you only care about degree 1 polynomials? You raised an extremely interesting question as to whether the set of all polynomials in $\pi$ might be all of the reals. Of course it can't by a cardinality argument, but I don't see any other way to prove this.

There is in fact a definable well-ordering of the reals, contrary to popular belief. Gödel's axiom of constructability, denoted $V = L$, says that the universe of sets is the constructible universe, called $L$. In $L$, every set is definable; and the axiom of choice is true. Therefore there is a definable well-ordering of the reals in $L$.

https://en.wikipedia.org/wiki/Constructible_universe

Most set theorists believe $V = L$ is false.

What do you mean that you want to find a well-ordering of the reals? You need the axiom of choice to construct such a thing so there's not likely to be an explicit description. You would need to spend a few years studying the arithmetic hierarchy and advanced mathematical logic and set theory to tackle a problem like this.
I used $pi$ for the above example, but I was also generally considering an equation that could substitute any irrational in place of $pi$. As such, I was capturing all the reals. I was looking to pinpoint specific sets of reals that could be captured given a single irrational (like $pi$) using the above formula and then looking to partition $\mathbb{R}$ with those specific sets. Extending it to a general version beyond degree 1 polynomials was beyond what I was contemplating, but a logical next step nevertheless so quite interesting.

As I understand it, the axiom of constructability is consistent with ZFC but not provable in ZFC so there remains no well ordering of the reals in arbitrary models of ZFC. While Gödel's construction of a subset of the reals could be consistent in ZF as being equal to the reals, Cohen showed that it could also be consistent that the subset was not equal to all reals. All this is over my head to an extent and I am relying on what is said here:

https://math.stackexchange.com/quest...g-of-the-reals

My goal then is to find a well ordering of the reals that holds in arbitrary models of ZFC. Likely to take years? Ha ha, centuries perhaps?! We've no explicit construction of a choice function, but that doesn't mean one doesn't exist (though there are models of ZFC as I understand it where no formula defines a well ordering even though such a formula must exist).

Aside from a well ordering of the reals, I feel that there are perhaps other possibilities lurking alongside non-measurable sets too that just haven't come about yet. For example, where we can obviously have a subset of a Vitali set with measure 0, can we have a subset of a Vitali set with a measure greater than 0 (presumably not, but I'm not sure this has been explicitly proven where a countable union of sets containing one element from each equivalence class can partition all of $\mathbb{R}$)?
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April 28th, 2017, 04:42 PM   #8
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Quote:
Originally Posted by AplanisTophet View Post
I used $pi$ for the above example, but I was also generally considering an equation that could substitute any irrational in place of $pi$.
You are interested in field extensions. For example what happens if you use $\sqrt 2$? There's a collapse to zero at the polynomial $x^2 - 2$. That's what it means for $\pi$ to be transcendental. It never collapses at any polynomial.

I think these are very difficult questions to determine explicitly when various sets of reals are algebraically independent. For example it's not known if $e$ and $\pi$ are algebraically independent. That means nobody knows if some rational polynomial in powers of $e$ and $\pi$ might possibly be zero. Nobody believes that's possible, but we have no proof.


Quote:
Originally Posted by AplanisTophet View Post
As such, I was capturing all the reals. I was looking to pinpoint specific sets of reals that could be captured given a single irrational (like $pi$) using the above formula and then looking to partition $\mathbb{R}$ with those specific sets. Extending it to a general version beyond degree 1 polynomials was beyond what I was contemplating, but a logical next step nevertheless so quite interesting.
Abstract algebra. Field theory and Galois theory.

Quote:
Originally Posted by AplanisTophet View Post
As I understand it, the axiom of constructability is consistent with ZFC but not provable in ZFC so there remains no well ordering of the reals in arbitrary models of ZFC.
Of course the reals can be well ordered in any model of ZFC. You mean definable model. Right, that's the point.

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Originally Posted by AplanisTophet View Post
While Gödel's construction of a subset of the reals could be consistent in ZF as being equal to the reals, Cohen showed that it could also be consistent that the subset was not equal to all reals. All this is over my head to an extent and I am relying on what is said here:

https://math.stackexchange.com/quest...g-of-the-reals
That's pretty much what I know about it too.

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Originally Posted by AplanisTophet View Post
My goal then is to find a well ordering of the reals that holds in arbitrary models of ZFC.
What do you mean "find?" It's been proven there's no definable one except in some models.

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Likely to take years? Ha ha, centuries perhaps?!
Once something's been proved impossible, that's that.

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Originally Posted by AplanisTophet View Post
We've no explicit construction of a choice function, but that doesn't mean one doesn't exist (though there are models of ZFC as I understand it where no formula defines a well ordering even though such a formula must exist).
You'll need more than Googling to attack this. The information is out there. Probably start with Jech and Kunen and go from there. I pointed you to the arithmetic hierarchy. That's the classification of reals based on the complexity of the formulaa needed to describe them.

Quote:
Originally Posted by AplanisTophet View Post
Aside from a well ordering of the reals, I feel that there are perhaps other possibilities lurking alongside non-measurable sets too that just haven't come about yet. For example, where we can obviously have a subset of a Vitali set with measure 0, can we have a subset of a Vitali set with a measure greater than 0 (presumably not, but I'm not sure this has been explicitly proven
It's been proven. Any measurable subset of a nonmeasurable set must have measure zero. https://math.stackexchange.com/quest...measurable-set
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Last edited by Maschke; April 28th, 2017 at 04:46 PM.
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April 28th, 2017, 04:55 PM   #9
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ps -- Aha I know what you're after. The Hamel basis of the reals over the rationals.

Here's the setup. By the way I myself am a big fan of $\mathbb R / \mathbb Q$ and this another cool manifestation of the idea.

Do you know what a vector space is? It's a set of elements acted on by a field of scalars (field means you can add/subtract, multiply/divide). It's a generalization of the idea that we can add vectors in the plane or space and multiply them by scalars.

According to the formal definition of a vector space, $\mathbb R$ is a vector space over the field $\mathbb Q$. It is a consequence of the axiom of choice that every vector space has a basis. Therefore there is some set $\mathcal B \subset \mathbb R$ such that every real number $x$ may be written uniquely as a finite direct sum $\sum_{i = 1}^n q_i b_i$ with $q_i \in \mathbb Q$ and $b_i \in \mathcal B$.

$\mathcal B$ must be uncountable (if it's countable the set of linear combos is also countable) and you are not going to ever write down or visualize any characterization of its elements.

That doesn't mean it's not fun to try, of course. These examples are great ways to stretch one's mathematical imagination.

See https://mathoverflow.net/questions/4...f-real-numbers and https://math.stackexchange.com/quest...bers-and-close and many other references on the web.

In passing note that it is therefore a consequence of the negation of Choice that there exists a vector space without a basis. Once again, accept Choice and get weird consequences. Reject Choice and get weird consequences.
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Last edited by Maschke; April 28th, 2017 at 05:06 PM.
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April 28th, 2017, 05:18 PM   #10
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According to the formal definition of a vector space, $\mathbb R$ is a vector space over the field $\mathbb Q$. It is a consequence of the axiom of choice that every vector space has a basis. Therefore there is some set $\mathcal B \subset \mathbb R$ such that every real number $x$ may be written uniquely as a finite direct sum $\sum_{i = 1}^n q_i b_i$ with $q_i \in \mathbb Q$ and $b_i \in \mathcal B$.

$\mathcal B$ must be uncountable (if it's countable the set of linear combos is also countable) and you are not going to ever write down or visualize any characterization of its elements.

That doesn't mean it's not fun to try, of course. These examples are great ways to stretch one's mathematical imagination.
Yes. As a starting point I was considering a set B that contained one and only one element of each Vitali equivalence class with some additional conditions:

1) Assume $0 \in B$
2) If $x \in B$, then $qx \in B$ for all $q \in \mathbb{Q}$ (I noted that if $y = qx$ where $q$ is rational and $x$ is irrational, then $y \neq x + p$ where $p$ is also rational, therefore $y$ could not be in the same equivalence class as $x$).
...

I went a ways from there, trying to further classify members of $B$ as well as the collection of sets $q + B$ for each $q \in \mathbb{Q}$ (that collection being a partition of $\mathbb{R}$), but again that was more-or-less my starting point for imaginative purposes. What you're saying goes right along those lines.
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