User blog:Explorer 767/Pi is 3 -- NOT

December 28, 2009

Club Penguin Island

Ah, the joys of being a teacher....

Today I was grading papers at home (it's two weeks of summer break) and I came across this ridiculous algorithm while correcting a paper on calculating the areas of circles...

Apparently some student didn't like having to use the irrational value of $$\pi$$ (pi), and thus "proved" that $$\pi = 3$$, henceforth substituting "3" as the ratio of the circumference of a circle to its diameter in the rest of the problems. I've seen this before -- it's totally incorrect, but since it's so convincing, I'm going to leave it up to you to figure out the problem with the "proof". -- 17:32, December 28, 2009 (UTC)


 * P.S.: As for the student, he received a D-. I dislike cheating, but I do appreciate a mathematical joke.

Given that $$a = b$$, it can be proven that $$\pi = 3$$, eliminating the need for me to use the value 3.14 in circle problems anymore.

$$\pi a = \pi b$$

$$3a = 3b$$

$$\pi a^2 = \pi ab$$ (multiplying  by  $$a$$)

$$3ab = 3b^2$$  (multiplying  by  $$b$$)

$$\pi a^2-3ab = \pi ab-3b^2$$  (combining  equations)

$$\pi a^2-\pi ab = 3ab-3b^2$$  (subtracting  $$\pi ab$$ and adding $$3ab$$)

$$\pi a^2-\pi ab + ab - b^2 = 4ab-4b^2$$  (adding  $$ab$$ and subtracting  $$b^2$$)

$$\pi a(a-b) + b(a-b) = 4b(a-b)$$  (factoring)

$$\pi a + b = 4b$$  (dividing  out  $$a-b$$)

$$\pi a = 3b$$  (subtracting  $$b$$)

$$\pi = 3$$  (dividing  by  $$a$$ and $$b$$)

$$Q.E.D.$$