1. EE1J2 - Slide 1 EE1J2 – Discrete Maths Lecture 12 Number theory Mathematical induction Proof by induction Examples

2. EE1J2 - Slide 2 Integers and Number Theory Recall that the ℤ denotes the set of integers {…-2, -1, 0, 1, 2…} We take for granted the fact that it x and y are integers, then x + y and x  y are also integers Mathematicians say that the set of integers is closed under addition and multiplication Number Theory is concerned with properties of the integers with the operations ‘+’ and ‘  ’

3. EE1J2 - Slide 3 Properties of + and  ℤ together with + and  has several properties which we take for granted: If x=a and y=b then x+a=y+b and x  y=a  b a+b= b+a and a  b=b  a (Commutative Laws) (a+b)+c=a+(b+c) and (a+b)+c=a+(b+c) (Associative Laws) a  (b+c)=a  b + a  c (Distributive Law)

4. EE1J2 - Slide 4 Properties of + and  (cont.) There exist unique integers 0 and 1 such that  a  ℤ, a+0=0+a=a and a  1=1  a=a 0 is called the additive identity 1 is called the multiplicative identity For every a there is an integer –a such that a+(-a)=(-a)+a=0 (-a is the additive inverse of a) If a  b=a  c and a  0, then b=c

5. EE1J2 - Slide 5 { ℤ,+,  } Sometimes use the notation { ℤ,+,  } as a reminder that number theory is concerned not just with the integers, but also with + and  { ℤ,+,  } is a common domain of discourse for Predicate Logic

6. EE1J2 - Slide 6 Example Suppose P(x) = ‘x is even’ Then,  x:(P(x)  P(x 2 ))  x:(P(x 2 )  P(x)) are statements in Predicate Logic for which the domain of discourse is { ℤ,+,  }

7. EE1J2 - Slide 7 Proofs in Number Theory Consider the statement  x:(P(x)  P(x 2 )) A formal proof of the validity of this statement requires many steps (see, for example, Anderson, p 106) In practice many of these steps are missed out and a typical ‘acceptable’ proof might look as follows:

8. EE1J2 - Slide 8 Proof of  x:(P(x)  P(x 2 )) Let x be even Then there exists y  ℤ such that x=2  y (this is the definition of an even number) Therefore x 2 =( 2  y) 2 =(2  y)  (2  y) =2  (2  y 2 ) So, if z= 2  y 2 then x 2 =2z Therefore x 2 is even.

9. EE1J2 - Slide 9 Proof of  x:(P(x 2 )  P(x)) In this case we’ll use ‘Equivalence of Contrapositive’ I.e. a  b   b   a So, proving P(x 2 )  P(x) is equivalent to proving  P(x)   P(x 2 )

10. EE1J2 - Slide 10 Proof of  P(x)   P(x 2 ) Assume  P(x). In other words x is odd Then there exists y  ℤ such that x=2  y+1 So, x 2 =( 2  y+1) 2 =4  y 2 + 4  y + 1 Let z=2  y 2 + 2  y Then x 2 =2  z + 1, so x 2 is odd. I.e.  P(x 2 ) is true So  P(x)   P(x 2 ), hence P(x 2 )  P(x)

11. EE1J2 - Slide 11 Mathematical Induction Let P(n) be the predicate This is true for particular values of n, e.g:

12. EE1J2 - Slide 12 Mathematical Induction Want to show that P(n) is true for all integers n I.e.:  n:P(n) The easiest way to prove such a statement is to use the method of proof by induction

13. EE1J2 - Slide 13 Mathematical Induction If P(1) is true …and for all k, P(k)  P(k+1) Then P(n) is true for all values of n This is the principle of mathematical induction

14. EE1J2 - Slide 14 Proof by Induction Intuitively… 1. Show that P(k)  P(k+1) for any k 2. Show that P(1) is true 3. Then from 1 and 2, P(2) is true 4. …and from 1 and 3, P(3) is true 5. …and from 1 and 4, P(4) is true 6. etc

15. EE1J2 - Slide 15 Example 1 Case n=1:, so P(1) is true Now assume P(k) is true Need to show P(k+1) is true I.e. need to show that

16. EE1J2 - Slide 16 Example 1 (continued) Step 1: Write S(k+1) in terms of S(k) Step 2: Use the fact that P(k) is true Step 3: Manipulate to get the right formula

17. EE1J2 - Slide 17 Example 1 (concluded) So, Therefore P(k+1) is true Therefore, by the principle of mathematical induction,  n:P(n)

18. EE1J2 - Slide 18 Example 2 Let P(n) be the predicate defined by: P(n): ‘n 3 -n is divisible by 3’ Show that  n:P(n) Case n=1: n 3 -n=1-1=0, which is divisible by 3. Hence P(1) is true Now assume that P(k) holds Need to prove that P(k+1) holds

19. EE1J2 - Slide 19 Example 2 (continued) Case n=k+1: (k+1) 3 - (k+1) = (k 3 +3k 2 +3k+1) - (k+1) = (k 3 -k)+(3k 2 +3k) = (k 3 -k) + 3(k 2 +k) Divisible by 3 since P(k) is true Clearly divisible by 3

20. EE1J2 - Slide 20 Example 2 (concluded) Hence if k 3 -k is divisible by 3, then (k+1) 3 -(k+1) is also divisible by 3 In other words,  k:(P(k)  P(k+1)) Also, P(1) is true Therefore, by mathematical induction,  n:P(n)

21. EE1J2 - Slide 21 Example 3 Let P(n) be the predicate ‘if S is a finite set such that |S| = n, then |P(S)|=2 n ’ Claim:  n:P(n) This is another way of saying that for any finite set S, |P(S)| = 2 |S| (See lecture 7, slide18)

22. EE1J2 - Slide 22 Example 3 (continued) Case n=1: If |S|=1, then S has just one element, x say. Then S = {x} In this case P(S)={ ,S}, so |P(S)|=2=2 1 =2 |S| Hence P(1) is true Now assume P(k) is true for some k In other words, if S is a set such that |S| = k, then |P(S)|=2 k

23. EE1J2 - Slide 23 Example 3 (continued) Let S k+1 be a set such that |S k+1 |=k+1 Write S k+1 = {x 1, x 2,…,x k+1 } = S k  {x k+1 }, where S k = {x 1, x 2,…,x k } We can write P(S k+1 ) = P k  P k+1, where: P k is the set of subsets of S which don’t include x k+1 P k+1 is the set of subsets of S which do include x k+1 But P k =P(S k ), so | P k |=|P(S k )|=2 k by assumption

24. EE1J2 - Slide 24 Example 3 (continued) Now consider P k+1, the set of subsets which include x k+1 Every subset in P k+1 must arise by taking a subset of S k and adding x k+1 to it …so | P k+1 | = |P(S k )| = 2 k Hence | P(S k+1 )| = |P k |  |P k+1 | = 2 k + 2 k = 2  2 k = 2 k+1

25. EE1J2 - Slide 25 Example 3 (concluded) Hence P(1) is true, and if P(k) is true then P(k+1) is also true Hence, by mathematical induction,  n:P(n)

26. EE1J2 - Slide 26 Notes Mathematical induction can only be used to prove arguments for which the domain of discourse is the positive, whole numbers. For example, if P(b) is the predicate: the roots of the equation x 2 +bx+1 are given by Then  b:P(b) is true, but cannot be proved by induction

27. EE1J2 - Slide 27 Summary Properties of the integers Proof by induction