# Induction question from Philosophy class

## 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

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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.