self-loop

Testing 1, 2, 3!

Here’s a quick test of my setup.

Proof that $\sqrt{2}$ is irrational

Let’s test that MathJax is setup correctly.

Say that $\sqrt{2}$ were rational. We will aim to produce a contradiction. Then the definition of rationality implies that:

$$\sqrt{2} = \frac{a}{b}$$

for some a, b integers. Further, we may assume that the fraction is in lowest terms, which means that a, b share no factors.

Now, squaring both sides tells us:

$$ \begin{align} \sqrt{2} &= \frac{a}{b}\\ 2 &= \frac{a^2}{b^2}\\ 2 b^2 &= a^2 \end{align} $$

This implies that 2 is a factor of a².

Note that, if the prime factorization of a is:

$$ \begin{align} a = p_1^{k_1} p_2^{k_2} \cdots p_n^{k_n} \end{align} $$

then this implies

$$ \begin{align} a^2 = p_1^{2k_1} p_2^{2k_2} \cdots p_n^{2k_n} \end{align} $$

Note first: any prime factor p_i of a² is also a factor of a.

More, if p_i is a prime factor which divides a², than in fact we must have p_i² divides a². And since we know that 2 divides a², then so must 2² = 4. Which allows us to say that $\frac{a^2}{2}$ should still be divisible by 2.

But note: $b^2 = \frac{a^2}{2}$. So therefore b² should be divisible by 2. But we already said about prime factors of squares: if 2 divides b², then it must also divide b.

That would contradict our choice that a, b had no common factors.

Therefore we have established the contradiction. It must not be possible to write $\sqrt{2}$ as $\frac{a}{b}$, which means $\sqrt{2}$ was not rational in the first place.

Code highlighting

Let’s test out code highlighting for some Haskell code.

qsort :: ArraySlice -> IO ()
qsort as
  | (asLength as) == 0 = do
      return ()
  | otherwise = do
      pivotIdx <- partition as 0 0
      let leftLen = pivotIdx
      qsort (resliceArraySlice as 0 leftLen)
      let rightLen = (asLength as) - (pivotIdx + 1)
      qsort (resliceArraySlice as (pivotIdx + 1) rightLen)

Look at that! Wow oh wow! How about some C++?

void sort(std::vector<int>& nums, int start, int len) {
  if (len == 0) {
    return;
  }

  int pivotIdx = start;
  const int pivot = nums[pivotIdx];
  for (int idx = start; idx < start+len; idx += 1) {
    int val = nums[idx];
    if (pivot <= val) {
      continue;
    }

    nums[pivotIdx] = val;
    nums[idx] = nums[pivotIdx + 1];
    nums[pivotIdx + 1] = pivot;
    pivotIdx += 1;
  }

  int numLeft = pivotIdx - start;
  int numRight = (start + len) - (pivotIdx + 1);

  // printVector(nums);
  if (numLeft > 0) {
    sort(nums, start, numLeft);
  }
  if (numRight > 0) {
    sort(nums, pivotIdx + 1, numRight);
  }
}

Okay, all done! Let’s just finally check out some comments!

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