Proof Must Have Statement of what is to be proven. "Proof:" to indicate where the proof starts Clear indication of flow Clear indication of reason for each step Careful notation, completeness and order Clear indication of the conclusion

Number Theory - Ch 3 Definitions Z --- integers Q - rational numbers (quotients of integers) rQ  a,bZ, (r = a/b) ^ (b  0) Irrational = not rational R --- real numbers superscript of + --- positive portion only superscript of - --- negative portion only other superscripts: Zeven, Zodd , Q>5 "closure" of these sets for an operation Z closed under what operations?

Integer Definitions even integer odd integer prime integer (Z>1) n Zeven  k  Z, n = 2k odd integer n  Zodd  k  Z, n = 2k+1 prime integer (Z>1) n Zprime  r,sZ+, (n=r*s) (r=1)v(s=1) composite integer (Z>1) n  Zcomposite  r,sZ+, n=r*s ^(r1)^(s1)

Constructive Proof of Existence If we want to prove: nZeven, p,q, r,sZprime n = p+q ^ n = r+s ^pr^ ps^ qr^ qs let n=10 n Zeven by definition of even Let p = 5 and the q = 5 p,q Zprime by definition of prime 10 = 5+5 Let r = 3 and s = 7 r,s Zprime by definition of prime 10 = 3+7 and all of the inequalities hold Make sure to mention the appendix that has rules of algebra.

Methods of Proving Universally Quantified Statements Method of Exhaustion prove for each and every member of the domain rZ+ where 23<r<29   p,q Z+ (r = p*q)^(p<=q) Generalizing from the "generic particular" suppose x is a particular but arbitrarily chosen element of the domain show that x satisfies the property i.e. rZ, r Zeven  r2 Zeven Exhaustion: Let r be an arbitrary member of Z 24 = 4*6 Assume r Zeven 25 = 5*5 exists k in evens such that r = 2k (def of even) 26 = 2*13 when squaring both sides of r = 2k, we get r2 = 4k2 27 = 3*9 by factoring out a 2 we get r2 = 2(2k2) 28=2*14 2k2 is even because 2k2 is an integer by closure therefore r2 is even r in Zeven -> r2 in Zeven

Examples of Generalizing from the "Generic Particular" The product of any two odd integers is also odd. m,n Z, [(m  Zodd  n Zodd ) m*n Zodd ] The product of any two rationals is also rational. m,nQ, m*n Q

Disproof by Counter Example  rZ, r2Z+  rZ+ Counter Example: r2= 9 ^ r = -3 r2Z+ since 9 Z+ so the antecedent is true but rZ+ since -3Z+ so the consequent is false this means the implication is false for r = -3 so this is a valid counter example When a counter example is given you must always justify that it is a valid counter example by showing the algebra (or other interpretation needed) to support your claim

Division definitions d | n  kZ, n = d*k n is divisible by d n is a multiple of d d is a divisor of n d divides n standard factored form n = p1e1 * p2e2 * p3e3 * …* pkek

Proof using the Contrapositive For all positive integers, if n does not divide a number to which d is a factor, then n can not divide d.

Proof using the Contrapositive For all positive integers, if n does not divide a number to which d is a factor, then n can not divide d. n,d,cZ+, ndc  nd

Proof using the Contrapositive For all positive integers, if n does not divide a number to which d is a factor, then n can not divide d. n,d,cZ+, ndc  nd n,d,cZ+, n|d  n|dc proof: let n, d and c be arbitrary in Z+ assume n|d there exists a k in Z such that d=nk by def of divides multiplying both sides by c gives dc = nkc kc in Z by closure of Z in multiplication n|dc by definition of divides n|d -> n|dc by closing conditional world w/out contra forall n,d,c in Z+, n|d -> n|dc by generalizing from GP

more integer definitions div and mod operators n div d --- integer quotient for n mod d --- integer remainder for (n div d = q) ^ (n mod d = r)  n = d*q+r where n Z0, d Z+, r Z, q Z, 0 r<d relating “mod” to “divides” d|n  0 = n mod d  0 d n definition of equivalence in a mod x d y  d|(x-y) [note: their remainders are equal] sometimes written as x  y mod d meaning (x  y) mod d examples of equivalence in a modulous 122,12 mod 10 & 10,13 mod 3

Quotient Remainder Theorem nZ dZ+ q,rZ (n=dq+r) ^ (0  r < d) Proving definition of equiv in a mod by using the quotient remainder theorem This means prove that if [m d n], then [d|(n-m)] where m,nZ and dZ+ Using what you know about mod operator from programming that the remainder is the same exists k,l,r in Z such that n = dk+r and m = dl+r n=dk+r n-dk = r m = dl + r m-dl = r n-dk = m-dl n-m = dk-dl n-m = d(k-l) k-l in Z by closure so d| (n-m) by definition of divides

Proofs using this definition mZ+ a,bZ a m b   k Z a=b+km mZ+ a,b,c,dZ a m b ^ c m d  a+c m b+d #1 #2 m in Z+ and a,b in Z are GP m in Z+, a,b,c,d in Z are GP Assume a m-equiv b assume a m-equiv b and c m-equiv d m|(a-b) by def of equiv in a mod m | (a-b) and m|(c-d) by def of equiv in a mod exists k in Z such that (a-b) = mk exists k in Z such that a-b = km a = b + km by algebra exists x in Z such that c-d = xm by def of divides Assume exists k in Z a=b+mk a-b = mk (a-b)+(c-d) = km+xm by adding equals to equals Since k is an integer, (a+c)-(b+d) = (k+x)m by algebra m|(a-b) by def of divides Since k+x in Z by closure of Z in addition a m-equiv b by def of equiv in a mod m| (a+c)-(b+d) by definition of divides (a+c) m-equiv (b+d) by def of equiv in a mod

Proof by Division into Cases nZ 3n  n2 3 1 #1 #2 forall n in Z [n/2] is Let n be arbitrary in Z n/2 if n is even We know by the quotient remainder theorem that or (n-1/2) if n is odd there exists a q and r such that proof: n = 3q+0 or n = 3q+1 or n=3q+2 Case 1:Assume n is even exists k in Z n = 2k Taking these three cases: by def of even Case 1: (assume: n=3q+0) exists x in Z where x = floor n/2 n=3q and 3|n which means the antecedent is false x <= n/2 < x+1 and the implication is true x <= 2k/2 < x+1 x <= k < x+1 x = k Case 2: (assume n = 3q+1) Case 2: Assume n is odd since n=3q+1, n^2 (3q+1)^2=9q^2+6q+1=3(3q^2+2q)+1 exists w in Z n = 2w+1 n^2-1 = 3(3q^2+2q) by def of odd Since 2q^2+2q is in Z by closure of Z in * and + exists x in Z when x = floor n/2 3| n^2 –1 by the definition of divides x <= n/2<x+1 and n^2 3-equiv to 1 by def of equiv in a mod x <=(2w+1)/2<x+1 x<= w+1/2 < x+1 Case 3: (assume n = 3q+2) x <= w < x+1 since n=3q+2,n^2=(3q+2)^2=9q^2+18q+4=3(3q^2+6q+1)+1 x = w n^2 – 1 = 3(3q^2+6q+1) n = 2w+1 3q^2+6q+1 is in Z by closure of Z in * and + n-1 = 2w 3|n^2-1 by definition of divides (n-1)/2 = w n^2 equiv to 1 by def of equiv in a mod (n-1)/2 = x implication true in all three cases

Floor and Ceiling Definitions n is the floor of x where xR ^ n Z x = n  n  x < n+1 n is the ceiling of x where xR ^ n Z x = n  n-1 < x  n

Floor/Ceiling Proofs x,yR x+y = x + y xR yZ x+y = x +y #1 disproof by counter example #2 proof by GP let x = 2.5 and let y = 2.5 let x be arbitrary in R and let y be arbitrary in Z then floor(x) = 2 and floor(y) =2 by the definition of floor of x and floor(x)+floor(y) = 4 exists n in Z such that n <= x<n+1 adding y to all sides of the inequality we get x+y = 5.0 n+y <= x+y < n+y+1 then floor(x+y) = 5 by closure n+y and n+1+y are in Z so this is the definition of the floor of x+y which are not equal

Proof by Division into Cases (again) The floor of (n/2) is either a) n/2 when n is even or b) (n-1)/2 when n is odd #1 #2 forall n in Z [n/2] is Let n be arbitrary in Z n/2 if n is even We know by the quotient remainder theorem that or (n-1/2) if n is odd there exists a q and r such that proof: n = 3q+0 or n = 3q+1 or n=3q+2 Case 1:Assume n is even exists k in Z n = 2k Taking these three cases: by def of even Case 1: (assume: n=3q+0) exists x in Z where x = floor n/2 n=3q and 3|n which means the antecedent is false x <= n/2 < x+1 and the implication is true x <= 2k/2 < x+1 x <= k < x+1 x = k Case 2: (assume n = 3q+1) Case 2: Assume n is odd since n=3q+1, n^2 (3q+1)^2=9q^2+6q+1=3(3q^2+2q)+1 exists w in Z n = 2w+1 n^2-1 = 3(3q^2+2q) by def of odd Since 2q^2+2q is in Z by closure of Z in * and + exists x in Z when x = floor n/2 3| n^2 –1 by the definition of divides x <= n/2<x+1 and n^2 3-equiv to 1 by def of equiv in a mod x <=(2w+1)/2<x+1 x<= w+1/2 < x+1 Case 3: (assume n = 3q+2) x <= w < x+1 since n=3q+2,n^2=(3q+2)^2=9q^2+18q+4=3(3q^2+6q+1)+1 x = w n^2 – 1 = 3(3q^2+6q+1) n = 2w+1 3q^2+6q+1 is in Z by closure of Z in * and + n-1 = 2w 3|n^2-1 by definition of divides (n-1)/2 = w n^2 equiv to 1 by def of equiv in a mod (n-1)/2 = x implication true in all three cases

Prime Factored Form kZ, p1,p2,…pkZprime, e1,e2,…ekZ+, n = p1e1 * p2e2 * p3e3 * …* pkek Unique Factorization Theorem (Theorem 3.3.3) given any integer n>1 kZ, p1,p2,…pkZprime, e1,e2,…ekZ+, where the p’s are distinct and any other expression of n is identical to this except maybe in the order of the factors Standard Factored Form pi < pi+1 mZ,8*7*6*5*4*3*2*m=17*16*15*14*13*12*11*10 Does 17|m ??

Steps Toward Proving the Unique Factorization Theorem Every integer greater than or equal to 2 has at least one prime that divides it For all integers greater than 1, if a|b, then a (b+1) There are an infinite number of primes #1 Proof by div into cases #2 let a and be be arbitrary in Z where a > 1 case 1: n is prime assume a|b it has a prime that divides it exists k in Z such that b = ak (itself) assume a | b+1 case 2: n is not prime exists m in Z such that b+1=am we need chapter 4 to prove this ak+1 = am by substitution 1 = am-ak=a(m-k) 1/a = m-k 1/a is an integer only when a = 1 but m-k is an integer by closure Contradiction a not| b+1 #3 assume there are a finite number of primes lets say the number of primes is m We can take that list of primes and find their product p1*p2*p3*…*pm = k k is an integer by closure of integers in multiplication Lets look at k+1 since p1|k, p1 not| k+1 since p2|k, p2 not|k+1 … but by the first proof on this page there must be an integer that divides k+1 that isn’t in this list of m integers contradiction so there must be a finite number of primes

Using the Unique Factorization Theorem Prove that the Prove: aZ+qZprime q|a2  q |a #1 Assume the cuberoot(3) is rational there exists a,b in Z such that cuberoot(3) = a/b and b neq 0 cubroot(3) = a/b b(cuberoot(3)) = a 3*b^2 = a^2 by the prime factorization theorem a can be prime factorized to p1^e1*p2^e2*…px^ex b can be prime factorized to q1^d1*q2^d2*…qy^dy a^2 = p1^2e1*p2^2e2*…px^2ex b^2 = q1^2d1*q2^2d2*…qy^2dy By the unique prime factorization theorem one of the p’s must be 3 factor all of the three’s out a^2 = 3^2ei(a1^2) where a1 is the left over product of primes and it is important to know that 3 not| a1 3*b^2=3^2ei(a1^2) cancel a 3 from both sides and we get b^2=3^(2ei-1)(a1^2) Since ei >=1, there is still a 3 left on the right so one of the q’s must be a 3 factor all of the three’s out to get b^2 = 3^2dj(b1^2) where b1 is the left over product of primes and it is important to know that 3 not|b1 by substitution we get 3^2dj(b1^2) = 3^(2ei-1)(a1^2) by the unique prime factorization theorem 2dj = 2ei-1 contradiction since one is odd the other is even --------------------------------------------------------- Prove: a,b,cZ+pZprime [(a2 = b3 = c) ^ (p|c)] (p6 | c) (a^2=b^3=c) and (p|c) since p|c – exists k in Z such that c=pk Since a^2 = c a can be prime factorized and a^2 has 2 p’s since b^3 = c and b can be prime factorized b^3 has 3 p’s since c must have a multiple of 2 p’s and a multiple of 3 p’s it must have a multiple of 6 p’s

Summary of Proof Methods Constructive Proof of Existence Proof by Exhaustion Proof by Generalizing from the Generic Particular Proof by Contraposition Proof by Contradiction Proof by Division into Cases

Errors in Proofs Arguing from example for universal proof. Misuse of Variables Jumping to the Conclusion (missing steps) Begging the Question Using "if" about something that is known