Fengyang Wang
UW Student Seminars
November 1, 2017
(Revised post-talk to correct typos)
(Formal
-ish Statement)
There exist $\{a, b\} ⊆ \mathbf{R}^+ - \mathbf{Q}^+$ such that $a^b ∈ \mathbf{Q}^+$.
(English Language Statement)
An irrational number to an irrational power may be rational.
If $\sqrt{2}^\sqrt{2}$ is rational, let $a=b=\sqrt{2}$ ∎
Otherwise, let $a=\sqrt{2}^\sqrt{2}$ and $b=\sqrt{2}$. Then $a^b = {\left(\sqrt{2}^\sqrt{2}\right)}^{\sqrt{2}} = \sqrt{2}^2 = 2$ ∎
Let $a = \sqrt{2}$ and $b = \frac{\log{9}}{\log{2}}$.
Then $a^b = \sqrt{2}^{\frac{\log{9}}{\log{2}}} = \sqrt{2^{\frac{\log{9}}{\log{2}}}} = \sqrt{9} = 3$ ∎
if (√2) ^ (√2) ∈ 𝐐
a = √2
b = √2
else
a = √2
b = (√2) ^ (√2)
end
You didn’t tell me how to figure out whether (√2) ^ (√2) ∈ 𝐐... Stacktrace: [1] in(::𝐑, ::Type{𝐐}) at ./In[1]:4
Beware!
In classical logic, a proof of $a ∨ b$ could be neither a proof of $a$ nor a proof of $b$.
In BHK intuitionistic logic, you must know which one you are proving!
Beware!
In classical logic, a proof of $∃ x P(x)$ might not have a witness.
Beware!
In the BHK interpretation, $¬¬a$ is not the same as $a$.
(Formal
-ish Statement)
$$
∀𝐱∈\{0, 1\}^𝐍 (𝐱 = 𝟎 ∨ 𝐱 ≠ 𝟎)
$$
(English Language Statement)
An infinitely tall haystack has no needles or it has a needle.
(Formal
-ish Statement)
$$
∀𝐱∈\{0, 1\}^𝐍 (¬∃i∈𝐍,j∈𝐍(i≠j ∧ x_i = y_i = 1) ⇒ (∀i∈𝐍(x_{2i} = 0) ∨ ∀i∈𝐍(x_{2i-1} = 0)))
$$
(English Language Statement)
Given two infinitely tall haystacks, if you know that there is at most one needle among them both, then either the left haystack has no needles or the right haystack has no needles.
(Formal
-ish Statement)
$$∀𝐱∈\{0, 1\}^𝐍 (¬(𝐱 = 𝟎) ⇒ 𝐱 ≠ 𝟎)$$
(English Language Statement)
If a haystack cannot have no needles, then it has a needle.
(English Language Statement)
A real number is a set of rational intervals such that any two intervals overlap, and there are intervals shorter than any given positive rational.
The equivalence relation of $\mathbf{R}$ is the natural one:
$\mathbf{x} = \mathbf{y}$ if each interval $(p, p') \in \mathbf{x}$ overlaps with each interval $(q, q') \in \mathbf{y}$
$(\mathbf{x}_n) \in \mathbf{R}^\mathbf{N}$ converges to $\mathbf{x}$ if for all $\varepsilon > 0$ there exists $N\in\mathbf{N}$ after which any term $\mathbf{x}_n$ ($n \ge N$) is within $\varepsilon$ of $\mathbf{x}$
(We also write $\lim_{n\to\infty}\mathbf{x}_n = \mathbf{x}$)
$(\mathbf{x}_n) \in \mathbf{R}^\mathbf{N}$ is Cauchy if for all $\varepsilon > 0$ there exists $N\in\mathbf{N}$ after which any two terms $\mathbf{x}_n$, $\mathbf{x}_m$ ($n \ge m > N$) are within $\varepsilon$ of each other
(Informal Statement)
If $(\mathbf{x}_n)\in\mathbf{R}^\mathbf{N}$ is Cauchy, then there is $\mathbf{x}_\infty$ such that $\lim_{n\to\infty}\mathbf{x}_n = \mathbf{x}_\infty$
(English Language Statement)
The real numbers are complete.
(Informal Statement)
If $x > 0$, there exists a positive integer $n$ such that $x > 1/n$.
Proof:
Take a interval $(q, q') \in x$ that does not contain $0$. Take $n > 1/q$ (rational operations).
Let $(\mathbf{x}_n)$ be a Cauchy sequence of real numbers.
Compute an integer function $k \mapsto n_k$ such that each $n_k$ satisfies for all $m\ge n_k$, $n\ge n_k$, $|\mathbf{x}_m - \mathbf{x}_n| < 2^{-k}$.
For each $k$, construct a sequence of intervals $(q_k, q'_k) \in \mathbf{x}_{n_k}$ such that $q'_k - q_k < 2^{-k}$.
Consider $r_k = q_k - 2^{-k}$, $r'_k = q'_k + 2^{-k}$. For all $n\ge n_k$, we have
$$ r_k \le \mathbf{x}_{n_k} - 2^{-k} < \mathbf{x}_n < \mathbf{x}_{n_k} + 2^{-k} \le r_k$$
So actually for all $j \ge k$
$$ \mathbf{x}_{n_j} \in [r_j, r'_j] \cap [r_k, r'_k]$$
So the real number $\mathbf{x}_\infty$ defined by $\{(r_k, r'_k) : k \ge 1\}$ is in fact a real number.
From our construction, we have $\mathbf{x}_n \in [r_k, r'_k]$ for all $n \ge n_k$. That means that
$$ \forall k \forall n\ge n_k~(|\mathbf{x}_n - \mathbf{x}_\infty| \le r'_k - r_k < 2^{-k+2}) $$
By Lemma, we therefore have $\lim_{n\to\infty} \mathbf{x}_n = \mathbf{x}_\infty$.
It was proven by Cantor that a good math joke exists. Unfortunately, his proof was entirely non-constructive.
― Jon