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| -rw-r--r-- | _posts/2021-04-08-mathjax.md | 86 |
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diff --git a/_posts/2021-04-08-mathjax.md b/_posts/2021-04-08-mathjax.md index d5f6e95..1c86957 100644 --- a/_posts/2021-04-08-mathjax.md +++ b/_posts/2021-04-08-mathjax.md @@ -3,13 +3,87 @@ title: Typesetting math category: Features mathjax: true --- -Today, we'll "prove" that $$2 \times 2 \ne 5$$. +First, let's determine that $$ -\begin{gather} -2 \times 2 = 2 + 2 = 4 \\ -4 \ne 5 \implies 2 \times 2 \ne 5 -\end{gather} +\newcommand\naturals{\mathbb{N}} +\newcommand\rationals{\mathbb{Q}} +\newcommand\reals{\mathbb{R}} + +\newcommand\xs{x_{1}x_{2}\dots x_n} +\newcommand\nines{\overbrace{99\dots 9}^n} +\newcommand\pq{\frac{p}{q}} +\newcommand\qp{\frac{q}{p}} + +0.(9) = 1 +$$ + +This may seem counter-intuitive, but demonstrably true. +If $$0.(9) \neq 1$$, then $$\exists n \in \reals: 0.(9) < n < 1$$. +To put it another way, there must be a number greater than 0.(9) and lesser +than 1, equal to neither. +Thinking about it makes it obvious this is not true. + +Slightly more formally, + +$$ +\begin{align*} +1 - 0.9 &= \frac{1}{10} \\ +1 - 0.99 &= \frac{1}{100} \\ +1 - 0.999 &= \frac{1}{10^{-3}} \\ +\dots \\ +1 - 0.\nines &= \frac{1}{10^n} \\ +\end{align*} +$$ + +If $$0.(9) \neq 1$$, the following must hold: + +$$ +\forall n \in \naturals, \exists x \in \reals: x < \frac{1}{10^n} +$$ + +It's clear that the only such number is 0, making 0.(9) and 1 equal. + +Let $$n \in [1,9]$$. +Is there $$\pq \in \rationals: \pq = 0.(n)$$? + +For $$n = 9$$, we've established $$p = 1, q = 1$$. +For the other values of $$n$$ we can observe that + +$$ +\begin{align*} +0.(1) \times 9 &= 0.(9) = 1 \\ +0.(2) \times 9/2 &= 0.(9) = 1 \\ +0.(3) \times 9/3 &= 0.(9) \\ +\dots \\ +0.(8) \times 9/8 &= 0.(9) \\ +0.(9) \times 1 &= 0.(9) +\end{align*} +$$ + +So, + +$$ +\forall n \in [1,9], \frac{n}{9} = 0.(n) +$$ + +In general, let's demonstrate that + +$$ +\forall n \in \naturals, n > 0, \forall 0.(x_{1}x_{2}\dots x_n), \exists \pq \in \rationals: \pq = 0.(\xs) +$$ + +Let $$p = \xs, q =\,\nines$$. +It's clear that + +$$ +\begin{align*} +0.(\xs) \times \qp &= 0.(\xs) \times \frac{\nines}{\xs} \\ +&= \left(\frac{\xs}{10^n} + \frac{\xs}{10^{2n}} + \dots\right) \times \frac{\nines}{\xs} \\ +&= \left(\frac{1}{10^n} + \frac{1}{10^{2n}} + \dots\right) \times\,\nines \\ +&= 0.(9) \\ +&= 1 +\end{align*} $$ -Q.E.D. +Finally, $$0.(\xs) \times \qp = 1 \implies \pq = 0.(\xs)$$. |