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Investigation: Components

Proofs Workshop

This was a 30-minute, collaborative breakout session, and we didn't hand out an answer sheet. I've added the answers to the question sheet in red.

n is an integer. p is the proposition n is even. q is the proposition n2 is even. a) Examine each of the following arguments and state, in terms of p and q and logical symbols, what each argument may be used to prove. b) In how many different ways can we make use of these arguments to prove p ⇔ q ? 4

1

n = 2k

p⇒q

! n 2 = 4k 2

( )

! n 2 = 2 2k 2

2

n = 2k + 1 ! n 2 = 4k 2 + 4k +1

(

¬p ⇒ ¬q

Contrapositive of q ⇒ p

)

! n 2 = 2 2k 2 + 2k +1

3

n2 = 2k ! n 2 "1 = 2k "1 ! (n "1)(n +1) = 2k "1

4

n2 = 2k + 1 ! n 2 "1 = 2k ! (n "1)(n +1) = 2k

5

n2 + n = n(n + 1) ! n 2 + n = 2k

q⇒p using the argument: if (n – 1)(n – 2) is odd, then (n – 1) and (n – 2) are both odd, so n is even. Contrapositive of p ⇒ q ¬q ⇒ ¬p similar argument to the one above, but now (n – 1) and (n – 2) are both even so n is odd. (p ∧ q) ∨ (¬p ∧ ¬q) i.e. n and n2 are either both even, or both odd

To prove p ⇔ q: 1 and 3: (p ⇒ q) ∧ (q ⇒ p) 1 and 2: (p ⇒ q) ∧ (¬p ⇒ ¬q) 3 and 4: (q ⇒ p) ∧ (¬q ⇒ ¬p) 5 on it's own (the truth tables for (p ∧ q) ∨ (¬p ∧ ¬q) and (p ⇔ q) are identical) Shirleen Stibbe

http://www.shirleenstibbe.co.uk