1. Do you agree with the following proofs? 1.Prove that  2 is irrational. Proof: Suppose that  2 is not irrational, then  2 = p/q for some natural numbers p, q where (p, q) = 1. Since 2 =p 2 /q 2, therefore 2q 2 =p 2. This implies that 2|p 2 and hence 2|p. So p=2k for some integer k. Putting it back to 2q 2 =p 2, 2q 2 =(2k) 2 i.e. q 2 =2k 2. Again, we have 2|q and 2|(p, q), which is a contradiction. Therefore  2 is irrational.

2. 2. There are infinitely many prime numbers. Proof: Assume there are only n prime numbers, say p 1, p 2, p 3,…,p n. Now construct a new number p= p 1 p 2 p 3 …p n + 1, then p is a new prime number since p is not divisible by p i ’s and p > p i ’s. This leads to a contradiction that p 1, p 2, p 3,…,p n are the only prime numbers. So there are infinitely many prime numbers.

3. 3.If x=n and y=n+1, then x and y are relatively prime. Proof: Assume that x=n and y=n+1 and x and y have common factor other than 1, say f, then n=fg and n+1=fh. So 1 = f(h-g) and hence f=1, which is a contradiction. Thus the proposition is true.

4. Proof by contradiction( 歸謬法 ) (~P)  F Class work : Use the method of contradiction to prove that  3 is irrational.

5. P  Q  (~P)  Q PQ P  Q(~P )  Q TTT F T T TFF F F F FTT T T T FFT T T F

6. Write down the negation of 1.P  Q 2.If today is Sunday, you need not go to school. 3.If I can live without food, then I need not earn money. 4.P  (P  Q) 5.In the classroom, all students are girls. ~(P  Q)  ~( (~P)  Q)  P  (~Q)