Halivar
First Post
No fair, fett! In class I wasn't allowed to prove Cauchy's theorem by writing down a hyperlink on the exam. But I do wish I had thought of it...
I just realized that if we're going to be mathematically determining the odds of adventurers verses dragons, we have to quantity the actual melee. Here's my attempt:
Let dєD, where D={x|x is a dragon}. Let us define d=[t x1 g1] where experience x1є{x|x is an integer}, gold g1є{x|x is an integer}, and tє{x|x is nasty, pointy teeth}. Now let aєA, where A={x|x is an adventurer}. Let us further define a=[s x2 g2], where experience x2є{x|x is an integer}, gold g2є{x|x is an integer}, and sє(x|x is a sharp, pointy stick}.
Let us define addition between the sets A and D to be a binary operation thusly:
Let aєA and dєD. We define a+dє{x|xєA or xєD} with the following conditional:
If a[1] > d[1], then a+d=a'=[a[1] a[2]+d[2] a[3]+d[3]] and d'=[0 0 0].
If d[1] ≥ a[1], then a+d=d'=[d[1] d[2]+a[2] d[3]] and a'є{x|x is a tasty snack}.
Question: Are the sets A and D closed over addition? Prove or disprove.
I just realized that if we're going to be mathematically determining the odds of adventurers verses dragons, we have to quantity the actual melee. Here's my attempt:
Let dєD, where D={x|x is a dragon}. Let us define d=[t x1 g1] where experience x1є{x|x is an integer}, gold g1є{x|x is an integer}, and tє{x|x is nasty, pointy teeth}. Now let aєA, where A={x|x is an adventurer}. Let us further define a=[s x2 g2], where experience x2є{x|x is an integer}, gold g2є{x|x is an integer}, and sє(x|x is a sharp, pointy stick}.
Let us define addition between the sets A and D to be a binary operation thusly:
Let aєA and dєD. We define a+dє{x|xєA or xєD} with the following conditional:
If a[1] > d[1], then a+d=a'=[a[1] a[2]+d[2] a[3]+d[3]] and d'=[0 0 0].
If d[1] ≥ a[1], then a+d=d'=[d[1] d[2]+a[2] d[3]] and a'є{x|x is a tasty snack}.
Question: Are the sets A and D closed over addition? Prove or disprove.
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