Indirect ProofsDate: 09/21/97 at 00:30:35 From: Ashley Subject: Geometry - Indirect Proofs Dear Dr. Math, Can you please explain to me what indirect proofs are, and how to solve this problem, or at least give me a hint? If Clark is a mathemagician, then Lois is his assistant. If Clark is a mathemagician, then everyone at the theater has a good time. If everyone in the theater has a good time, then Lois is not his assistant. Therefore, Clark is not a mathemagician. I really need help with this one. Thanks! Date: 09/21/97 at 16:35:51 From: Doctor Jodi Subject: Re: Geometry - Indirect Proofs Hi there! I'm not quite sure what you mean by indirect proofs, but I think I can help you with your puzzle. First of all, let's look at the sentences that you have: (1) If Clark is a mathemagician, then Lois is his assistant. (2) If Clark is a mathemagician, then everyone at the theater has a good time. (3) If everyone in the theater has a good time, then Lois is not his assistant. (4) Therefore, Clark is not a mathemagician. Let's focus on the first three sentences for now. Sentences (1), (2), and (3) are conditional, if-then sentences. We want to find out whether or not Clark is a mathemagician. If Clark is a mathemagician: - Lois is his assistant (from (1)) - everyone at the theater has a good time (from (2)) If Clark is NOT a mathemagican: (we don't know anything yet) From (3) we also know that IF everyone in the theater has a good time, then Lois is not his assistant. HOLD ON! If Clark is a mathemagician, then everyone has a good time. So, according to (3), if Clark is a mathemagician, then Lois is not his assistant. Do you see where we got this? BUT, from (1), we know that if Clark is a mathemagician, then Lois is his assistant. So we have a contradiction: Lois both MUST BE (from (1)) and CAN'T BE (from (2)) Clark's assistant. But we got this contradiction because we assumed that Clark was a mathemagician. So he can't be one, as (4) says. Does this make sense? And since we don't know anything if Clark is not a mathemagician, there is no contradiction. Thanks for your question. Please write back if you need more help. -Doctor Jodi, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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