The Indigo Creature's Weblog



<H2 ALIGN = CENTER>October 10, 2006ce -- Things no one else can appreciate... <BR> <BR>
  • The color of Cu2+ ions
  • The smell of a fresh circuit board
  • The unparalleled spectral purity of a $5 laser pointer
  • The smell of ozone around a plasma globe
  • The beauty of an efficient algorithm
  • The fractal symmetry of a pine tree
  • The physics of carnival rides
  • the way a drop of dye diffuses in water
  • Anodized Niobium


<H2 ALIGN = CENTER>July 08, 2006ce -- On the furriness of Creatures and Imaginary iMacs</H2> <BR> <BR> ∀ x in {Indigo Creatures}, x is furry. This statement is true, because the set of all Indigo Creatures contains one Indigo Creature, and being that I am the Indigo Creature, I can assure you that I am in fact furry. Now, a harder case:<BR> ∀ x in {Copper Creatures}, x is furry. This is harder to prove, because the set of all Copper Creatures contains 1+i elements. "What?" you ask, bewildered. Yes, 1+i elements. One real Copper Creature, and one Imaginary Copper Creature. From experience, I know that the one real Copper Creature is furry, but is the imaginary Copper Creature furry? That depends on if imaginary fur counts!<BR> We can, however, prove that although a real Copper Creature is heterosexual, and imaginary Copper Creature is bisexual. How?<BR> Let's define y as the imaginary Copper Creature.<BR> y = i ∙ Copper Creature.<BR> Therefore, y is a complex number<BR> y = a + bi, a,b ∈ ℝ;<BR> And since this imaginary Copper Creature is in fact imaginary,<BR> a = 0 "∴ y = bi<BR> Conclusion: y (the imaginary Copper Creature) = (is) bi.<BR> It can also be shown that since some of the Copper Creature's relatives call him "B",<BR> i * Copper Creature = i ∙ "B" = bi<BR> <BR> Note, however, that an imaginary iMac is NOT bisexual, and in fact, is not even a Mac.<BR> Suppose x is an imaginary iMac:<BR> x = i * iMac<BR> This simplifies to:<BR> x = i ∙ i ∙ Mac = (-1)∙Mac = -Mac<BR> Therefore, x = -Mac. and if x is a negative mac, then it's not a Mac at all! Not also that:<BR> x = a + bi, a,b ∈ ℝ;<BR> so<BR> x = -Mac + 0i<BR> x ≠ bi<BR> Therefore, an imaginary iMac is not bisexual either, unless it's an imaginary Indigo iMac, but the proof is left up to the reader.<BR>

May 22, 2006ce -- Blog Coming Soon


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This page was last updated Tuesday, October 10, 2006CE 14:26:58 EDT.
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