Induction question from Philosophy class

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SUMMARY

The discussion centers on using mathematical induction to prove the truth values of a statement P when negation symbols are applied. Specifically, it establishes that if an even number of negation symbols are placed in front of P, the resulting statement Q remains true, while an odd number of negations results in Q being false. The approach involves demonstrating the base case for zero and two negations, followed by the inductive step for P(n) and P(n+1). The analogy to the cosine function for natural numbers reinforces the concept of alternating truth values.

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  • Understanding of mathematical induction principles
  • Familiarity with logical negation and truth values
  • Basic knowledge of propositional logic
  • Experience with mathematical proofs
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  • Explore logical negation and its implications in propositional logic
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Students in philosophy and mathematics, particularly those studying logic, propositional calculus, and mathematical proofs. This discussion is beneficial for anyone looking to deepen their understanding of induction and logical reasoning.

Jamin2112
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Homework Statement



Let P be a true sentence, and let Q be formed by putting some number of negation symbols in front of P. Show that if you put an even number of negation symbols, then Q is true, but that if you put an odd number, then Q is false.

Homework Equations



So ... the problem also suggests that I use induction, if I know how to use it. From Math class I know how to use the principle of induction: Show that P(1) is true, and then show that if P(n) is true, P(n+1) is true as well.

The Attempt at a Solution



But I'm not sure how to use it in this problem. Please help.
 
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Perhaps the best way to do this would be to do it in parts. First, show that for an even number of negations, Q is true. Then for an odd number of negations, Q is false.

Let's consider what P(1) is for the even number of negations. It's safest to handle both of the possible ways to look at this: either that zero negation signs counts as an even number (in which case, this is trivial) or two negation symbols. Then P(n) is the statement: P with 2n negation symbols in front of it. Assuming P(n) is true, show P(n+1), which is P with 2n+2 negation symbols in front of it.

Repeat for the odd case.
 
That just seems like an extremely trivial thing to consider. P is true, so ~P is false. ~~P is true, so you can just keep adding ~~ to either of those statements ad infinitum and keep the same truth value. It's rather like asking you to show that for N a natural number and Pi*N, cos(Pi*N) is 1 for even N and -1 for odd N.
 

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